Re: [algogeeks] Re: probability
@Sukun: these probabilities should have been given in the questn.. since it's not given and there are 3 possible answers, i considered probability of each one of them to b correct as 1/3 (since there are 3 possible answers and only one answer is correct) On Sun, Sep 9, 2012 at 12:48 AM, Sukun Tarachandani suku...@gmail.comwrote: How do you get P(0.25 being correct) = P(0.5 being correct) = P(0.6 being correct) = 1/3? On Saturday, September 8, 2012 4:42:51 PM UTC+5:30, Shruti wrote: shouldn't it b done like this : P(correct answer when choosing randomly )= P(choosing 0.25)*P(0.25 being correct ans)+ P(choosing 0.60)*P(0.60 being correct ans) + P(choosing 0.50)*P(0.50 being correct ans) = 2/4*1/3 + 1/4*1/3 + 1/4*1/3 [since 3 possible answers, hence prob of an ans being correct=1/3] taking 1/3 common, =1/3(2/4+1/4+1/4) =1/3 pls correct me if i'm wrong On Fri, Sep 7, 2012 at 7:54 PM, isandeep isand...@gmail.com wrote: Ans : 0.5 there is two cases : i) if correct ans is 0.25 probability will be 2/4 = 0.5 ii) if correct ans is 0.6 or 0.50 probability will be 1/3 = 0.33 since 0.33 is not in option so correct answer will be 0.5. On Friday, September 7, 2012 6:05:08 PM UTC+5:30, noname wrote: [image: -]14Answers http://www.careercup.com/question?id=14553727 What is the probability of being the answer correct for this question, when the answer is chosen randomly: a. 0.25 b. 0.60 c. 0.25 d. 0.50 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/** msg/algogeeks/-/SE8PpC7PLwgJhttps://groups.google.com/d/msg/algogeeks/-/SE8PpC7PLwgJ . To post to this group, send email to algo...@googlegroups.com. To unsubscribe from this group, send email to algogeeks+...@** googlegroups.com. For more options, visit this group at http://groups.google.com/** group/algogeeks?hl=en http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/z2HuFFlPSWYJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: probability
shouldn't it b done like this : P(correct answer when choosing randomly )= P(choosing 0.25)*P(0.25 being correct ans)+ P(choosing 0.60)*P(0.60 being correct ans) + P(choosing 0.50)*P(0.50 being correct ans) = 2/4*1/3 + 1/4*1/3 + 1/4*1/3 [since 3 possible answers, hence prob of an ans being correct=1/3] taking 1/3 common, =1/3(2/4+1/4+1/4) =1/3 pls correct me if i'm wrong On Fri, Sep 7, 2012 at 7:54 PM, isandeep isandee...@gmail.com wrote: Ans : 0.5 there is two cases : i) if correct ans is 0.25 probability will be 2/4 = 0.5 ii) if correct ans is 0.6 or 0.50 probability will be 1/3 = 0.33 since 0.33 is not in option so correct answer will be 0.5. On Friday, September 7, 2012 6:05:08 PM UTC+5:30, noname wrote: [image: -]14Answers http://www.careercup.com/question?id=14553727 What is the probability of being the answer correct for this question, when the answer is chosen randomly: a. 0.25 b. 0.60 c. 0.25 d. 0.50 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/SE8PpC7PLwgJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: probability
It cannot be either 0.25 or 0.50 http://math.stackexchange.com/questions/76491/multiple-choice-question-about-the-probability-of-a-random-answer-to-itself-bein On Friday, September 7, 2012 6:05:08 PM UTC+5:30, noname wrote: [image: -]14Answers http://www.careercup.com/question?id=14553727 What is the probability of being the answer correct for this question, when the answer is chosen randomly: a. 0.25 b. 0.60 c. 0.25 d. 0.50 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/Yp8D_rik3DcJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: probability
How do you get P(0.25 being correct) = P(0.5 being correct) = P(0.6 being correct) = 1/3? On Saturday, September 8, 2012 4:42:51 PM UTC+5:30, Shruti wrote: shouldn't it b done like this : P(correct answer when choosing randomly )= P(choosing 0.25)*P(0.25 being correct ans)+ P(choosing 0.60)*P(0.60 being correct ans) + P(choosing 0.50)*P(0.50 being correct ans) = 2/4*1/3 + 1/4*1/3 + 1/4*1/3 [since 3 possible answers, hence prob of an ans being correct=1/3] taking 1/3 common, =1/3(2/4+1/4+1/4) =1/3 pls correct me if i'm wrong On Fri, Sep 7, 2012 at 7:54 PM, isandeep isand...@gmail.com javascript: wrote: Ans : 0.5 there is two cases : i) if correct ans is 0.25 probability will be 2/4 = 0.5 ii) if correct ans is 0.6 or 0.50 probability will be 1/3 = 0.33 since 0.33 is not in option so correct answer will be 0.5. On Friday, September 7, 2012 6:05:08 PM UTC+5:30, noname wrote: [image: -]14Answers http://www.careercup.com/question?id=14553727 What is the probability of being the answer correct for this question, when the answer is chosen randomly: a. 0.25 b. 0.60 c. 0.25 d. 0.50 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/SE8PpC7PLwgJ. To post to this group, send email to algo...@googlegroups.comjavascript: . To unsubscribe from this group, send email to algogeeks+...@googlegroups.com javascript:. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/z2HuFFlPSWYJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: probability
Ans : 0.5 there is two cases : i) if correct ans is 0.25 probability will be 2/4 = 0.5 ii) if correct ans is 0.6 or 0.50 probability will be 1/3 = 0.33 since 0.33 is not in option so correct answer will be 0.5. On Friday, September 7, 2012 6:05:08 PM UTC+5:30, noname wrote: [image: -]14Answers http://www.careercup.com/question?id=14553727 What is the probability of being the answer correct for this question, when the answer is chosen randomly: a. 0.25 b. 0.60 c. 0.25 d. 0.50 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/SE8PpC7PLwgJ. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: probability of winning with two cards
@Don and Sundi..
As Don pointed out, all we are looking for is:
sum of a1 sum of a2
sum of a1 sum of a3
Assumption:
1) The 2 cards picked for a particular player are unique.
2) Cards are numbered : 1,..., 12, 13.
Hence, the following code should give the answer for the a1's
probability to win:
for( int i =1; i 23; ++i)
{
sampleCount+= a[i]*b[i-1]*b[i-1];
}
probability= sampleCount/ 474552;
sampleCount will be: 145650
probability = 0.306921
On Jan 23, 2:36 pm, Lucifer sourabhd2...@gmail.com wrote:
@Don..
Yup, it seems I misread it ... :) .. Thanks
On Jan 23, 9:17 am, Don dondod...@gmail.com wrote:
I think that you are misreading the problem. A1 wins if his sum is
larger than A2's sum and larger than A3's sum. A1's sum doesn't have
to be larger than A2+A3.
Don
On Jan 22, 5:18 pm, Lucifer sourabhd2...@gmail.com wrote:
@sundi..
Lets put is this way..
Probability of (a1 wins + a1 draws + a1 losses) = 1,
Now, sample count a1 wins = 46298 ( using the above given code)
Hence, the probability (win) = 46298/474552 = .097561
[ @ Don - as i mentioned in my previous post that i had initially
missed a factor 2, hence the above calculated value shall justify
that]
Based on explanation given in the previous post, you can use the same
approach and find out the sample count for a1's draw and loss..
Add the following code snippet to calculate the same:
// Draws ( a1 = a2 + a3 )
int sampleCount2 = 0;
for(int i =0; i 23; ++i)
{
for(int j = 0; j i; ++j)
{
if(i - j == j)
sampleCount2 += a[j] * a[j] * a[i];
else if (i - j j)
sampleCount2 += 2 * a[j] * a[i-j] * a[i];
}
}
// Losses ( a1 a2 + a3 )
int sampleCount3 = 0;
for(int i =0; i 23; ++i)
{
for(int j = 0; j 23; ++j)
{
if (i - j + 1 ==j)
{
sampleCount3 += a[j] * a[j] * a[i];
sampleCount3 += 2 * a[j] * (b[22] - b[i-j+1]) * a[i];
}
else if(i - j + 1 j)
{
// this is a special case as both i and j are smaller than
// (i - j + 1)
if ( i==0 j ==0 )
sampleCount3 = a[j] * a[j] * a[i];
sampleCount3 += 2 * a[j] * (b[22] - b[i-j]) * a[i];
}
else
{
sampleCount3 += a[j] * a[j] * a[i];
sampleCount3 += 2 * a[j] * (b[22] - b[j]) * a[i];
}
}
}
On executing the given snippet you will get:
sampleCount2 = 10184 (draw)
samepleCount3 = 418070 ( loss)
Now, for the probability to be 1:
sampleCount + sampleCount2 + sampleCount3 should be 78^3 (474552)..
Now,
46298 + 10184 + 418070 = 474552 which is equal to (78^3)..
On Jan 23, 2:34 am, Sundi sundi...@gmail.com wrote:
Hi Lucifer,
Have you checked the sum of probability of (a winning + b
winning + c winning + draw)==1 ?
On Jan 22, 2:38 pm, Lucifer sourabhd2...@gmail.com wrote:
@above
editing mistake.. (btw the working code covers it)
/*
int j =*1*;
for(int i = 0; i 12 ; i+=2)
{
A[i] = A[i+1] = A[22-i] = A[21-i] = j;
++j;}
*/
On Jan 22, 6:53 pm, Lucifer sourabhd2...@gmail.com wrote:
@Don..
Well i will explain the approach that i took to arrive at the
probability..
Well yes u are correct in saying that it doesn't make a lot of sense
but then the no. of wins by a1 keeping in mind that a1 a2 + a3 is
much less than a1 = a2 + a3..
Or may be I have gone wrong in calculating the same..
Please let me know if u find some issue in the below given
explanation..
now, the given nos. are 1,2,3,4,.13..
Hence, the possible pair sums are 3,4,5,,25...
The total no. of pairs that can be formed are 13 C 2 = 78.
Now, for each pair within 3-25 (including both extremes) lets find
the
no. of ways we get to the particular sum.
i.e for 3 its 1 ( 1 + 2)
for 7 its 3 (1 + 6, 2 + 5, 3 + 4)
Lets take an array A[23], to store the count of occurrences for pair
sums (3 - 25)
A[i] - will store the no. of ways of getting 'i+3' pair-sum
A[i] values will be:
i -
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
18 19 20 21 22
A[i] -
1 1 2 2 3 3 4 4 5 5 6 6 6 5 5 4 4
3 3 2 2 1 1
Now, the above can be generated by using the following code:
Lets say the input is stored in X[R] = {1,2, ..., 13}
Here R is 13..
Lets say, the array A is initialized with 0.
for (int i = 0; i R-1; ++i)
for ( int j = 1; i R; ++j)
++A[ X[i] + X[j] ];
Well, as the nos. are continuous , hence we can minimize the
initialization
[algogeeks] Re: probability of winning with two cards
@Don,
I think i misunderstood the question again.. :)..
One of major things that i went wrong with was that for me the deck
consisted of 13*3=39 cards..( basically an assumption made based on
the way i understood the question)
Thanks for the explanation..
On Jan 23, 9:17 pm, Don dondod...@gmail.com wrote:
Close. You actually have to be sure that all 6 cards dealt to the
players are unique.
For instance, if I get 3 points, if you don't require that all the
cards dealth in the game are unique, you would conclude that there is
a very small, but positive probability that I will win. In reality, 3
points means that my two cards are a 1 and a 2. Thus there are not 4
1's left, and I am sure not to win.
One way to do that is to select 2 cards for P1. Then select 2 cards
for P2, making sure that neither one was dealt to P1. If the sum of P1
is greater than the sum of P2, select 2 cards from the remaining deck
for P3. If the sum of P1 is greater than the sum of P3, add one to
your counter. I'm sure that there is a faster way to code it, but on
my computer this ran in about 2 seconds.
There are 52!/(46!*8) ways that the cards can be dealt. 565,271,160
of those result in P1 winning.
Thus P(p1 wins) = 0.30850919745967...
Don
On Jan 23, 7:34 am, Lucifer sourabhd2...@gmail.com wrote:
@Don and Sundi..
As Don pointed out, all we are looking for is:
sum of a1 sum of a2
sum of a1 sum of a3
Assumption:
1) The 2 cards picked for a particular player are unique.
2) Cards are numbered : 1,..., 12, 13.
Hence, the following code should give the answer for the a1's
probability to win:
for( int i =1; i 23; ++i)
{
sampleCount+= a[i]*b[i-1]*b[i-1];
}
probability= sampleCount/ 474552;
sampleCount will be: 145650
probability = 0.306921
On Jan 23, 2:36 pm, Lucifer sourabhd2...@gmail.com wrote:
@Don..
Yup, it seems I misread it ... :) .. Thanks
On Jan 23, 9:17 am, Don dondod...@gmail.com wrote:
I think that you are misreading the problem. A1 wins if his sum is
larger than A2's sum and larger than A3's sum. A1's sum doesn't have
to be larger than A2+A3.
Don
On Jan 22, 5:18 pm, Lucifer sourabhd2...@gmail.com wrote:
@sundi..
Lets put is this way..
Probability of (a1 wins + a1 draws + a1 losses) = 1,
Now, sample count a1 wins = 46298 ( using the above given code)
Hence, the probability (win) = 46298/474552 = .097561
[ @ Don - as i mentioned in my previous post that i had initially
missed a factor 2, hence the above calculated value shall justify
that]
Based on explanation given in the previous post, you can use the same
approach and find out the sample count for a1's draw and loss..
Add the following code snippet to calculate the same:
// Draws ( a1 = a2 + a3 )
int sampleCount2 = 0;
for(int i =0; i 23; ++i)
{
for(int j = 0; j i; ++j)
{
if(i - j == j)
sampleCount2 += a[j] * a[j] * a[i];
else if (i - j j)
sampleCount2 += 2 * a[j] * a[i-j] * a[i];
}
}
// Losses ( a1 a2 + a3 )
int sampleCount3 = 0;
for(int i =0; i 23; ++i)
{
for(int j = 0; j 23; ++j)
{
if (i - j + 1 ==j)
{
sampleCount3 += a[j] * a[j] * a[i];
sampleCount3 += 2 * a[j] * (b[22] - b[i-j+1]) * a[i];
}
else if(i - j + 1 j)
{
// this is a special case as both i and j are smaller than
// (i - j + 1)
if ( i==0 j ==0 )
sampleCount3 = a[j] * a[j] * a[i];
sampleCount3 += 2 * a[j] * (b[22] - b[i-j]) * a[i];
}
else
{
sampleCount3 += a[j] * a[j] * a[i];
sampleCount3 += 2 * a[j] * (b[22] - b[j]) * a[i];
}
}
}
On executing the given snippet you will get:
sampleCount2 = 10184 (draw)
samepleCount3 = 418070 ( loss)
Now, for the probability to be 1:
sampleCount + sampleCount2 + sampleCount3 should be 78^3 (474552)..
Now,
46298 + 10184 + 418070 = 474552 which is equal to (78^3)..
On Jan 23, 2:34 am, Sundi sundi...@gmail.com wrote:
Hi Lucifer,
Have you checked the sum of probability of (a winning + b
winning + c winning + draw)==1 ?
On Jan 22, 2:38 pm, Lucifer sourabhd2...@gmail.com wrote:
@above
editing mistake.. (btw the working code covers it)
/*
int j =*1*;
for(int i = 0; i 12 ; i+=2)
{
A[i] = A[i+1] = A[22-i] = A[21-i] = j;
++j;}
*/
On Jan 22, 6:53 pm, Lucifer sourabhd2...@gmail.com wrote:
@Don..
Well i will explain the approach that i took to arrive at the
probability..
Well yes u are correct in saying that it
[algogeeks] Re: probability of winning with two cards
@Don..
Well i will explain the approach that i took to arrive at the
probability..
Well yes u are correct in saying that it doesn't make a lot of sense
but then the no. of wins by a1 keeping in mind that a1 a2 + a3 is
much less than a1 = a2 + a3..
Or may be I have gone wrong in calculating the same..
Please let me know if u find some issue in the below given
explanation..
now, the given nos. are 1,2,3,4,.13..
Hence, the possible pair sums are 3,4,5,,25...
The total no. of pairs that can be formed are 13 C 2 = 78.
Now, for each pair within 3-25 (including both extremes) lets find the
no. of ways we get to the particular sum.
i.e for 3 its 1 ( 1 + 2)
for 7 its 3 (1 + 6, 2 + 5, 3 + 4)
Lets take an array A[23], to store the count of occurrences for pair
sums (3 - 25)
A[i] - will store the no. of ways of getting 'i+3' pair-sum
A[i] values will be:
i -
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
18 19 20 21 22
A[i] -
1 1 2 2 3 3 4 4 5 56 66 5544
332 211
Now, the above can be generated by using the following code:
Lets say the input is stored in X[R] = {1,2, ..., 13}
Here R is 13..
Lets say, the array A is initialized with 0.
for (int i = 0; i R-1; ++i)
for ( int j = 1; i R; ++j)
++A[ X[i] + X[j] ];
Well, as the nos. are continuous , hence we can minimize the
initialization operation as follows
(based on fact that there is a pattern and holds for any set of
continuous nos. from 1 to K)
// initialze pair counts.. ( 3 ... 25)
int j =0;
for(int i = 0; i 12 ; i+=2)
{
A[i] = A[i+1] = A[22-i] = A[21-i] = j;
++j;
}
a[12] = 6;
[ the above code is specifically written for 1 to 13 (K), but you can
generalize it based on ur need.
All you need to do is take care of the last initialization statement
a[12] = 6; based on value K (13).]
Now, A[i] basically represent the no. of ways we can get 'i+3' - lets
say this is a1's current pick.
Now, for sum of a2 and a3's pick to be smaller than a1's we can do the
following:
If A[i] is picked by a1, then let a2 pick A[p] where p i, then the
possible picks by a3 would be
from anywhere b/w A[p] to A[i - p - 1].
Here, there is a catch .. we need to insure that i - p -1 = p
otherwise the range for a3's pick would be invalid.
Also, the above explanation is based on the assumption that a2 =a3.
Hence, to complete figure out all the possibilities of a1, a2 and a3,
we need to do the following..
For a given pick by a1 say A[i], then the no. of possiblites such that
a1 a2 + a3 would be:
1) if a2=a3, A[p] * A[p] * A[i]
2) if a2!=a3 , 2 * A[p] *( cumulative sum of A[p+1] to A[i - p -1])
*A[i]
[ a factor of 2 is multiplied to remove the
assumption a2 a3 ]
Now, once we get the total no. of possibities by the above given
equation, the probability would be:
(No. of possiblites) / (78^3) ..
[ 78 - 13 C 2]
Code:
int a[23];// to store the count
int b[23];// to store the cumulative count
int k = 1;
// initialze pair counts.. ( 3 ... 25)
for(int i = 0; i 12; i+=2)
{
a[i] = a[i+1] = a[22-i] = a[21-i] = k;
++k;
}
a[12] = 6;
b[0]=a[0];
//cumulative sum
for(int i = 1; i 23; i+=1)
{
b[i] = b[i-1] + a[i];
}
// calculate possibilities..
// i =0 (3 :minimum sum pair)... i=22 (25 : max sum pair)
int sampleCount = 0;
for(int i =0; i 23; ++i)
{
for(int j = 0; j i; ++j)
{
if(i - j - 1 = j)
{
sampleCount += a[j] * a[j] * a[i];
if (i - j - 1 j)
sampleCount += 2 * a[j] * (b[i-j-1] - b[j]) * a[i];
}
}
}
int R = 78*78*78;
printf(probability = %f , (float)sampleCount / R);
Don, as I mentioned in the start that there is possibility i might
have gone wrong in calculation. The fact being that i missed the
factor 2 when i wrote the code.
But, then the main point here is that whether the approach is correct
or not.
-
On Jan 20, 3:41 am, Don dondod...@gmail.com wrote:
You are saying that a1 wins roughly 1 in 20 times? How does that make
any sence?
Don
On Jan 19, 2:35 pm, Lucifer sourabhd2...@gmail.com wrote:
@correction:
Probalilty (a1 wins) = 24575/474552 = .051786
On Jan 20, 1:30 am, Lucifer sourabhd2...@gmail.com wrote:
hoping that the cards are numbered 1,2,3,,13..
Probalilty (a1 wins) = 21723/474552 = .045776
On Jan 20, 12:47 am, Don dondod...@gmail.com wrote:
P= 8800/28561 ~= 0.308112461...
On Jan 18, 7:40 pm, Sundi sundi...@gmail.com wrote:
there are 52 cards.. there are 3 players a1,a2,a3 each player is given
2 cards each one of A=2...J=11,Q=12,K=13..a user wins if his sum of
cards is greater then the other two players sum.
find the probability of a1 being the winner?- Hide quoted text -
- Show quoted text -
--
You received this message
[algogeeks] Re: probability of winning with two cards
@above
editing mistake.. (btw the working code covers it)
/*
int j =*1*;
for(int i = 0; i 12 ; i+=2)
{
A[i] = A[i+1] = A[22-i] = A[21-i] = j;
++j;
}
*/
On Jan 22, 6:53 pm, Lucifer sourabhd2...@gmail.com wrote:
@Don..
Well i will explain the approach that i took to arrive at the
probability..
Well yes u are correct in saying that it doesn't make a lot of sense
but then the no. of wins by a1 keeping in mind that a1 a2 + a3 is
much less than a1 = a2 + a3..
Or may be I have gone wrong in calculating the same..
Please let me know if u find some issue in the below given
explanation..
now, the given nos. are 1,2,3,4,.13..
Hence, the possible pair sums are 3,4,5,,25...
The total no. of pairs that can be formed are 13 C 2 = 78.
Now, for each pair within 3-25 (including both extremes) lets find the
no. of ways we get to the particular sum.
i.e for 3 its 1 ( 1 + 2)
for 7 its 3 (1 + 6, 2 + 5, 3 + 4)
Lets take an array A[23], to store the count of occurrences for pair
sums (3 - 25)
A[i] - will store the no. of ways of getting 'i+3' pair-sum
A[i] values will be:
i -
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
18 19 20 21 22
A[i] -
1 1 2 2 3 3 4 4 5 5 6 6 6 5 5 4 4
3 3 2 2 1 1
Now, the above can be generated by using the following code:
Lets say the input is stored in X[R] = {1,2, ..., 13}
Here R is 13..
Lets say, the array A is initialized with 0.
for (int i = 0; i R-1; ++i)
for ( int j = 1; i R; ++j)
++A[ X[i] + X[j] ];
Well, as the nos. are continuous , hence we can minimize the
initialization operation as follows
(based on fact that there is a pattern and holds for any set of
continuous nos. from 1 to K)
// initialze pair counts.. ( 3 ... 25)
int j =0;
for(int i = 0; i 12 ; i+=2)
{
A[i] = A[i+1] = A[22-i] = A[21-i] = j;
++j;}
a[12] = 6;
[ the above code is specifically written for 1 to 13 (K), but you can
generalize it based on ur need.
All you need to do is take care of the last initialization statement
a[12] = 6; based on value K (13).]
Now, A[i] basically represent the no. of ways we can get 'i+3' - lets
say this is a1's current pick.
Now, for sum of a2 and a3's pick to be smaller than a1's we can do the
following:
If A[i] is picked by a1, then let a2 pick A[p] where p i, then the
possible picks by a3 would be
from anywhere b/w A[p] to A[i - p - 1].
Here, there is a catch .. we need to insure that i - p -1 = p
otherwise the range for a3's pick would be invalid.
Also, the above explanation is based on the assumption that a2 =a3.
Hence, to complete figure out all the possibilities of a1, a2 and a3,
we need to do the following..
For a given pick by a1 say A[i], then the no. of possiblites such that
a1 a2 + a3 would be:
1) if a2=a3, A[p] * A[p] * A[i]
2) if a2!=a3 , 2 * A[p] *( cumulative sum of A[p+1] to A[i - p -1])
*A[i]
[ a factor of 2 is multiplied to remove the
assumption a2 a3 ]
Now, once we get the total no. of possibities by the above given
equation, the probability would be:
(No. of possiblites) / (78^3) ..
[ 78 - 13 C 2]
Code:
int a[23];// to store the count
int b[23];// to store the cumulative count
int k = 1;
// initialze pair counts.. ( 3 ... 25)
for(int i = 0; i 12; i+=2)
{
a[i] = a[i+1] = a[22-i] = a[21-i] = k;
++k;}
a[12] = 6;
b[0]=a[0];
//cumulative sum
for(int i = 1; i 23; i+=1)
{
b[i] = b[i-1] + a[i];
}
// calculate possibilities..
// i =0 (3 :minimum sum pair)... i=22 (25 : max sum pair)
int sampleCount = 0;
for(int i =0; i 23; ++i)
{
for(int j = 0; j i; ++j)
{
if(i - j - 1 = j)
{
sampleCount += a[j] * a[j] * a[i];
if (i - j - 1 j)
sampleCount += 2 * a[j] * (b[i-j-1] - b[j]) * a[i];
}
}
}
int R = 78*78*78;
printf(probability = %f , (float)sampleCount / R);
Don, as I mentioned in the start that there is possibility i might
have gone wrong in calculation. The fact being that i missed the
factor 2 when i wrote the code.
But, then the main point here is that whether the approach is correct
or not.
-
On Jan 20, 3:41 am, Don dondod...@gmail.com wrote:
You are saying that a1 wins roughly 1 in 20 times? How does that make
any sence?
Don
On Jan 19, 2:35 pm, Lucifer sourabhd2...@gmail.com wrote:
@correction:
Probalilty (a1 wins) = 24575/474552 = .051786
On Jan 20, 1:30 am, Lucifer sourabhd2...@gmail.com wrote:
hoping that the cards are numbered 1,2,3,,13..
Probalilty (a1 wins) = 21723/474552 = .045776
On Jan 20, 12:47 am, Don dondod...@gmail.com wrote:
P= 8800/28561 ~= 0.308112461...
On Jan 18, 7:40 pm, Sundi
[algogeeks] Re: probability of winning with two cards
Hi Lucifer,
Have you checked the sum of probability of (a winning + b
winning + c winning + draw)==1 ?
On Jan 22, 2:38 pm, Lucifer sourabhd2...@gmail.com wrote:
@above
editing mistake.. (btw the working code covers it)
/*
int j =*1*;
for(int i = 0; i 12 ; i+=2)
{
A[i] = A[i+1] = A[22-i] = A[21-i] = j;
++j;}
*/
On Jan 22, 6:53 pm, Lucifer sourabhd2...@gmail.com wrote:
@Don..
Well i will explain the approach that i took to arrive at the
probability..
Well yes u are correct in saying that it doesn't make a lot of sense
but then the no. of wins by a1 keeping in mind that a1 a2 + a3 is
much less than a1 = a2 + a3..
Or may be I have gone wrong in calculating the same..
Please let me know if u find some issue in the below given
explanation..
now, the given nos. are 1,2,3,4,.13..
Hence, the possible pair sums are 3,4,5,,25...
The total no. of pairs that can be formed are 13 C 2 = 78.
Now, for each pair within 3-25 (including both extremes) lets find the
no. of ways we get to the particular sum.
i.e for 3 its 1 ( 1 + 2)
for 7 its 3 (1 + 6, 2 + 5, 3 + 4)
Lets take an array A[23], to store the count of occurrences for pair
sums (3 - 25)
A[i] - will store the no. of ways of getting 'i+3' pair-sum
A[i] values will be:
i -
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
18 19 20 21 22
A[i] -
1 1 2 2 3 3 4 4 5 5 6 6 6 5 5 4 4
3 3 2 2 1 1
Now, the above can be generated by using the following code:
Lets say the input is stored in X[R] = {1,2, ..., 13}
Here R is 13..
Lets say, the array A is initialized with 0.
for (int i = 0; i R-1; ++i)
for ( int j = 1; i R; ++j)
++A[ X[i] + X[j] ];
Well, as the nos. are continuous , hence we can minimize the
initialization operation as follows
(based on fact that there is a pattern and holds for any set of
continuous nos. from 1 to K)
// initialze pair counts.. ( 3 ... 25)
int j =0;
for(int i = 0; i 12 ; i+=2)
{
A[i] = A[i+1] = A[22-i] = A[21-i] = j;
++j;}
a[12] = 6;
[ the above code is specifically written for 1 to 13 (K), but you can
generalize it based on ur need.
All you need to do is take care of the last initialization statement
a[12] = 6; based on value K (13).]
Now, A[i] basically represent the no. of ways we can get 'i+3' - lets
say this is a1's current pick.
Now, for sum of a2 and a3's pick to be smaller than a1's we can do the
following:
If A[i] is picked by a1, then let a2 pick A[p] where p i, then the
possible picks by a3 would be
from anywhere b/w A[p] to A[i - p - 1].
Here, there is a catch .. we need to insure that i - p -1 = p
otherwise the range for a3's pick would be invalid.
Also, the above explanation is based on the assumption that a2 =a3.
Hence, to complete figure out all the possibilities of a1, a2 and a3,
we need to do the following..
For a given pick by a1 say A[i], then the no. of possiblites such that
a1 a2 + a3 would be:
1) if a2=a3, A[p] * A[p] * A[i]
2) if a2!=a3 , 2 * A[p] *( cumulative sum of A[p+1] to A[i - p -1])
*A[i]
[ a factor of 2 is multiplied to remove the
assumption a2 a3 ]
Now, once we get the total no. of possibities by the above given
equation, the probability would be:
(No. of possiblites) / (78^3) ..
[ 78 - 13 C 2]
Code:
int a[23];// to store the count
int b[23];// to store the cumulative count
int k = 1;
// initialze pair counts.. ( 3 ... 25)
for(int i = 0; i 12; i+=2)
{
a[i] = a[i+1] = a[22-i] = a[21-i] = k;
++k;}
a[12] = 6;
b[0]=a[0];
//cumulative sum
for(int i = 1; i 23; i+=1)
{
b[i] = b[i-1] + a[i];
}
// calculate possibilities..
// i =0 (3 :minimum sum pair)... i=22 (25 : max sum pair)
int sampleCount = 0;
for(int i =0; i 23; ++i)
{
for(int j = 0; j i; ++j)
{
if(i - j - 1 = j)
{
sampleCount += a[j] * a[j] * a[i];
if (i - j - 1 j)
sampleCount += 2 * a[j] * (b[i-j-1] - b[j]) * a[i];
}
}
}
int R = 78*78*78;
printf(probability = %f , (float)sampleCount / R);
Don, as I mentioned in the start that there is possibility i might
have gone wrong in calculation. The fact being that i missed the
factor 2 when i wrote the code.
But, then the main point here is that whether the approach is correct
or not.
-
On Jan 20, 3:41 am, Don dondod...@gmail.com wrote:
You are saying that a1 wins roughly 1 in 20 times? How does that make
any sence?
Don
On Jan 19, 2:35 pm, Lucifer sourabhd2...@gmail.com wrote:
@correction:
Probalilty (a1 wins) = 24575/474552 = .051786
[algogeeks] Re: probability of winning with two cards
@sundi..
Lets put is this way..
Probability of (a1 wins + a1 draws + a1 losses) = 1,
Now, sample count a1 wins = 46298 ( using the above given code)
Hence, the probability (win) = 46298/474552 = .097561
[ @ Don - as i mentioned in my previous post that i had initially
missed a factor 2, hence the above calculated value shall justify
that]
Based on explanation given in the previous post, you can use the same
approach and find out the sample count for a1's draw and loss..
Add the following code snippet to calculate the same:
// Draws ( a1 = a2 + a3 )
int sampleCount2 = 0;
for(int i =0; i 23; ++i)
{
for(int j = 0; j i; ++j)
{
if(i - j == j)
sampleCount2 += a[j] * a[j] * a[i];
else if (i - j j)
sampleCount2 += 2 * a[j] * a[i-j] * a[i];
}
}
// Losses ( a1 a2 + a3 )
int sampleCount3 = 0;
for(int i =0; i 23; ++i)
{
for(int j = 0; j 23; ++j)
{
if (i - j + 1 ==j)
{
sampleCount3 += a[j] * a[j] * a[i];
sampleCount3 += 2 * a[j] * (b[22] - b[i-j+1]) * a[i];
}
else if(i - j + 1 j)
{
// this is a special case as both i and j are smaller than
// (i - j + 1)
if ( i==0 j ==0 )
sampleCount3 = a[j] * a[j] * a[i];
sampleCount3 += 2 * a[j] * (b[22] - b[i-j]) * a[i];
}
else
{
sampleCount3 += a[j] * a[j] * a[i];
sampleCount3 += 2 * a[j] * (b[22] - b[j]) * a[i];
}
}
}
On executing the given snippet you will get:
sampleCount2 = 10184 (draw)
samepleCount3 = 418070 ( loss)
Now, for the probability to be 1:
sampleCount + sampleCount2 + sampleCount3 should be 78^3 (474552)..
Now,
46298 + 10184 + 418070 = 474552 which is equal to (78^3)..
On Jan 23, 2:34 am, Sundi sundi...@gmail.com wrote:
Hi Lucifer,
Have you checked the sum of probability of (a winning + b
winning + c winning + draw)==1 ?
On Jan 22, 2:38 pm, Lucifer sourabhd2...@gmail.com wrote:
@above
editing mistake.. (btw the working code covers it)
/*
int j =*1*;
for(int i = 0; i 12 ; i+=2)
{
A[i] = A[i+1] = A[22-i] = A[21-i] = j;
++j;}
*/
On Jan 22, 6:53 pm, Lucifer sourabhd2...@gmail.com wrote:
@Don..
Well i will explain the approach that i took to arrive at the
probability..
Well yes u are correct in saying that it doesn't make a lot of sense
but then the no. of wins by a1 keeping in mind that a1 a2 + a3 is
much less than a1 = a2 + a3..
Or may be I have gone wrong in calculating the same..
Please let me know if u find some issue in the below given
explanation..
now, the given nos. are 1,2,3,4,.13..
Hence, the possible pair sums are 3,4,5,,25...
The total no. of pairs that can be formed are 13 C 2 = 78.
Now, for each pair within 3-25 (including both extremes) lets find the
no. of ways we get to the particular sum.
i.e for 3 its 1 ( 1 + 2)
for 7 its 3 (1 + 6, 2 + 5, 3 + 4)
Lets take an array A[23], to store the count of occurrences for pair
sums (3 - 25)
A[i] - will store the no. of ways of getting 'i+3' pair-sum
A[i] values will be:
i -
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
18 19 20 21 22
A[i] -
1 1 2 2 3 3 4 4 5 5 6 6 6 5 5 4 4
3 3 2 2 1 1
Now, the above can be generated by using the following code:
Lets say the input is stored in X[R] = {1,2, ..., 13}
Here R is 13..
Lets say, the array A is initialized with 0.
for (int i = 0; i R-1; ++i)
for ( int j = 1; i R; ++j)
++A[ X[i] + X[j] ];
Well, as the nos. are continuous , hence we can minimize the
initialization operation as follows
(based on fact that there is a pattern and holds for any set of
continuous nos. from 1 to K)
// initialze pair counts.. ( 3 ... 25)
int j =0;
for(int i = 0; i 12 ; i+=2)
{
A[i] = A[i+1] = A[22-i] = A[21-i] = j;
++j;}
a[12] = 6;
[ the above code is specifically written for 1 to 13 (K), but you can
generalize it based on ur need.
All you need to do is take care of the last initialization statement
a[12] = 6; based on value K (13).]
Now, A[i] basically represent the no. of ways we can get 'i+3' - lets
say this is a1's current pick.
Now, for sum of a2 and a3's pick to be smaller than a1's we can do the
following:
If A[i] is picked by a1, then let a2 pick A[p] where p i, then the
possible picks by a3 would be
from anywhere b/w A[p] to A[i - p - 1].
Here, there is a catch .. we need to insure that i - p -1 = p
otherwise the range for a3's pick would be invalid.
Also, the above explanation is based on the assumption that a2 =a3.
Hence, to complete figure out all the possibilities of a1, a2 and a3,
we need to do the following..
For
[algogeeks] Re: probability of winning with two cards
I think that you are misreading the problem. A1 wins if his sum is
larger than A2's sum and larger than A3's sum. A1's sum doesn't have
to be larger than A2+A3.
Don
On Jan 22, 5:18 pm, Lucifer sourabhd2...@gmail.com wrote:
@sundi..
Lets put is this way..
Probability of (a1 wins + a1 draws + a1 losses) = 1,
Now, sample count a1 wins = 46298 ( using the above given code)
Hence, the probability (win) = 46298/474552 = .097561
[ @ Don - as i mentioned in my previous post that i had initially
missed a factor 2, hence the above calculated value shall justify
that]
Based on explanation given in the previous post, you can use the same
approach and find out the sample count for a1's draw and loss..
Add the following code snippet to calculate the same:
// Draws ( a1 = a2 + a3 )
int sampleCount2 = 0;
for(int i =0; i 23; ++i)
{
for(int j = 0; j i; ++j)
{
if(i - j == j)
sampleCount2 += a[j] * a[j] * a[i];
else if (i - j j)
sampleCount2 += 2 * a[j] * a[i-j] * a[i];
}
}
// Losses ( a1 a2 + a3 )
int sampleCount3 = 0;
for(int i =0; i 23; ++i)
{
for(int j = 0; j 23; ++j)
{
if (i - j + 1 ==j)
{
sampleCount3 += a[j] * a[j] * a[i];
sampleCount3 += 2 * a[j] * (b[22] - b[i-j+1]) * a[i];
}
else if(i - j + 1 j)
{
// this is a special case as both i and j are smaller than
// (i - j + 1)
if ( i==0 j ==0 )
sampleCount3 = a[j] * a[j] * a[i];
sampleCount3 += 2 * a[j] * (b[22] - b[i-j]) * a[i];
}
else
{
sampleCount3 += a[j] * a[j] * a[i];
sampleCount3 += 2 * a[j] * (b[22] - b[j]) * a[i];
}
}
}
On executing the given snippet you will get:
sampleCount2 = 10184 (draw)
samepleCount3 = 418070 ( loss)
Now, for the probability to be 1:
sampleCount + sampleCount2 + sampleCount3 should be 78^3 (474552)..
Now,
46298 + 10184 + 418070 = 474552 which is equal to (78^3)..
On Jan 23, 2:34 am, Sundi sundi...@gmail.com wrote:
Hi Lucifer,
Have you checked the sum of probability of (a winning + b
winning + c winning + draw)==1 ?
On Jan 22, 2:38 pm, Lucifer sourabhd2...@gmail.com wrote:
@above
editing mistake.. (btw the working code covers it)
/*
int j =*1*;
for(int i = 0; i 12 ; i+=2)
{
A[i] = A[i+1] = A[22-i] = A[21-i] = j;
++j;}
*/
On Jan 22, 6:53 pm, Lucifer sourabhd2...@gmail.com wrote:
@Don..
Well i will explain the approach that i took to arrive at the
probability..
Well yes u are correct in saying that it doesn't make a lot of sense
but then the no. of wins by a1 keeping in mind that a1 a2 + a3 is
much less than a1 = a2 + a3..
Or may be I have gone wrong in calculating the same..
Please let me know if u find some issue in the below given
explanation..
now, the given nos. are 1,2,3,4,.13..
Hence, the possible pair sums are 3,4,5,,25...
The total no. of pairs that can be formed are 13 C 2 = 78.
Now, for each pair within 3-25 (including both extremes) lets find the
no. of ways we get to the particular sum.
i.e for 3 its 1 ( 1 + 2)
for 7 its 3 (1 + 6, 2 + 5, 3 + 4)
Lets take an array A[23], to store the count of occurrences for pair
sums (3 - 25)
A[i] - will store the no. of ways of getting 'i+3' pair-sum
A[i] values will be:
i -
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
18 19 20 21 22
A[i] -
1 1 2 2 3 3 4 4 5 5 6 6 6 5 5 4 4
3 3 2 2 1 1
Now, the above can be generated by using the following code:
Lets say the input is stored in X[R] = {1,2, ..., 13}
Here R is 13..
Lets say, the array A is initialized with 0.
for (int i = 0; i R-1; ++i)
for ( int j = 1; i R; ++j)
++A[ X[i] + X[j] ];
Well, as the nos. are continuous , hence we can minimize the
initialization operation as follows
(based on fact that there is a pattern and holds for any set of
continuous nos. from 1 to K)
// initialze pair counts.. ( 3 ... 25)
int j =0;
for(int i = 0; i 12 ; i+=2)
{
A[i] = A[i+1] = A[22-i] = A[21-i] = j;
++j;}
a[12] = 6;
[ the above code is specifically written for 1 to 13 (K), but you can
generalize it based on ur need.
All you need to do is take care of the last initialization statement
a[12] = 6; based on value K (13).]
Now, A[i] basically represent the no. of ways we can get 'i+3' - lets
say this is a1's current pick.
Now, for sum of a2 and a3's pick to be smaller than a1's we can do the
following:
If A[i] is picked by a1, then let a2 pick A[p] where p i, then the
possible picks by a3 would
[algogeeks] Re: probability of winning with two cards
:)... Lets say you have two players a and b one card is distrbuted to each player if the card with a is higher then a wins else a loses. probability of a winning: total num of cards with combos where a wins a is the first player: 2-1,3-1...13-1 3-2...13-2 .. 13-12 this sum equals = (13+12+...1)*4, 4 suites are possible therefore probability=sum/52C2. I dont think it is half in this case. similarly extending for 3 players with 2 cards each i dont think its going to be 1/3 becoz of the condition given. I might be wrong. Please correct me if so. On Jan 19, 1:08 am, Prateek Jain prateek10011...@gmail.com wrote: ya obviously...it is 1/3.no need of any data is required On 1/18/12, sunny agrawal sunny816.i...@gmail.com wrote: isn't the answer will be 1/3, without any calculations :) On Thu, Jan 19, 2012 at 7:10 AM, Sundi sundi...@gmail.com wrote: there are 52 cards.. there are 3 players a1,a2,a3 each player is given 2 cards each one of A=2...J=11,Q=12,K=13..a user wins if his sum of cards is greater then the other two players sum. find the probability of a1 being the winner? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Sunny Aggrawal B.Tech. V year,CSI Indian Institute Of Technology,Roorkee -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: probability of winning with two cards
@Sunny: The probability of a1 being the winner is not 1/3 because of ties. I.e., if a1 = a2 a3, then a1 and a2 are tied and there is no winner. What we can say with no calculations is that P(a1 winning) = (1 - P(no winner)) / 3. Dave On Jan 18, 10:52 pm, sunny agrawal sunny816.i...@gmail.com wrote: isn't the answer will be 1/3, without any calculations :) On Thu, Jan 19, 2012 at 7:10 AM, Sundi sundi...@gmail.com wrote: there are 52 cards.. there are 3 players a1,a2,a3 each player is given 2 cards each one of A=2...J=11,Q=12,K=13..a user wins if his sum of cards is greater then the other two players sum. find the probability of a1 being the winner? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Sunny Aggrawal B.Tech. V year,CSI Indian Institute Of Technology,Roorkee -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: probability of winning with two cards
P= 8800/28561 ~= 0.308112461... On Jan 18, 7:40 pm, Sundi sundi...@gmail.com wrote: there are 52 cards.. there are 3 players a1,a2,a3 each player is given 2 cards each one of A=2...J=11,Q=12,K=13..a user wins if his sum of cards is greater then the other two players sum. find the probability of a1 being the winner? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: probability of winning with two cards
hoping that the cards are numbered 1,2,3,,13.. Probalilty (a1 wins) = 21723/474552 = .045776 On Jan 20, 12:47 am, Don dondod...@gmail.com wrote: P= 8800/28561 ~= 0.308112461... On Jan 18, 7:40 pm, Sundi sundi...@gmail.com wrote: there are 52 cards.. there are 3 players a1,a2,a3 each player is given 2 cards each one of A=2...J=11,Q=12,K=13..a user wins if his sum of cards is greater then the other two players sum. find the probability of a1 being the winner? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: probability of winning with two cards
@correction: Probalilty (a1 wins) = 24575/474552 = .051786 On Jan 20, 1:30 am, Lucifer sourabhd2...@gmail.com wrote: hoping that the cards are numbered 1,2,3,,13.. Probalilty (a1 wins) = 21723/474552 = .045776 On Jan 20, 12:47 am, Don dondod...@gmail.com wrote: P= 8800/28561 ~= 0.308112461... On Jan 18, 7:40 pm, Sundi sundi...@gmail.com wrote: there are 52 cards.. there are 3 players a1,a2,a3 each player is given 2 cards each one of A=2...J=11,Q=12,K=13..a user wins if his sum of cards is greater then the other two players sum. find the probability of a1 being the winner? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: probability of winning with two cards
You are saying that a1 wins roughly 1 in 20 times? How does that make any sence? Don On Jan 19, 2:35 pm, Lucifer sourabhd2...@gmail.com wrote: @correction: Probalilty (a1 wins) = 24575/474552 = .051786 On Jan 20, 1:30 am, Lucifer sourabhd2...@gmail.com wrote: hoping that the cards are numbered 1,2,3,,13.. Probalilty (a1 wins) = 21723/474552 = .045776 On Jan 20, 12:47 am, Don dondod...@gmail.com wrote: P= 8800/28561 ~= 0.308112461... On Jan 18, 7:40 pm, Sundi sundi...@gmail.com wrote: there are 52 cards.. there are 3 players a1,a2,a3 each player is given 2 cards each one of A=2...J=11,Q=12,K=13..a user wins if his sum of cards is greater then the other two players sum. find the probability of a1 being the winner?- Hide quoted text - - Show quoted text - -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: probability question
Take it simple silly ... for each 10 min interval, if man comes in first 2 min, he'll catch the 1st train, if he comes in next 8 min, he'll catch the 2nd train. hence for harbor line - (2/10) 0.2 and for main line 0.8. On Wed, Aug 31, 2011 at 10:37 PM, ravi rravi...@gmail.com wrote: A man goes to the station every day to catch the first train that comes ?? = man catches the first train that comes after him On Wed, Aug 31, 2011 at 10:30 PM, Dave dave_and_da...@juno.com wrote: @Ankul: According to the problem statement, the first train is the one that arrives at 5:00 a.m. Dave On Aug 31, 11:26 pm, Ankuj Gupta ankuj2...@gmail.com wrote: I could not get it. What does first train mean here? On Sep 1, 1:08 am, Don dondod...@gmail.com wrote: Assuming that the man arrives at a random time during the 24-hour day, there are 228 minutes in the day when the next train will be the harbour line (2 minutes of every 10 for 19 hours). For the other 1212 minutes the main line will be the next train. Therefore, the probability of catching the main line train is 0.841666... Don On Aug 31, 8:37 am, swetha rahul swetharahu...@gmail.com wrote: In a railway station, there are two trains going. One in the harbour line and one in the main line, each having a frequency of 10 minutes. The main line service starts at 5 o'clock and the harbour line starts at 5.02A.M. A man goes to the station every day to catch the first train that comes.What is the probability of the man catching the first train?- Hide quoted text - - Show quoted text - -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- -- Ravi Ramadasu | Computer Science and Engineering, IIT Bombay | rravi...@gmail.com | +919757074652 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Piyush Agarwal Final Year Undergraduate Department of Computer Engineering Malaviya National Institute of Technology Jaipur -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: probability ques
@Aditya: If you are comfortable with working with fractions of area,
with area being a real number, then why not length, also being a real
number? The probability of a randomly selected point in the interval
[1, 100] satisfying a + 100/a 50 is exactly sqrt(2100)/99.
Dave
On Sep 1, 3:43 am, Aditya Virmani virmanisadi...@gmail.com wrote:
the space in tht case is restricted... my point is...number of points in a 4
m2 area wud be exactly 4 times of the number of points in 1 m2 area... so we
r actually talking over area in ur qn...now if i were to say...tht find the
probability tht it will hit the target at coordinate (0,0) ... tht wud be
close to 0...
On Thu, Sep 1, 2011 at 2:57 AM, Dave dave_and_da...@juno.com wrote:
@Aditya. There are an infinite number of points in that target, too.
But we don't have any trouble saying that 1/4 of them are in the
bullseye.
Dave
On Aug 31, 4:07 pm, Aditya Virmani virmanisadi...@gmail.com wrote:
@DAVE again u r considering a finite space... in the above case...but how
wud u take the space in real number thing...with no particular info
thr
r infinite real number frm 1-100...if i cud change the qn to find the
probability the chosen number is in the range a to a+1 ...thn it cud be
aptly answered
On Tue, Aug 30, 2011 at 10:10 AM, Dave dave_and_da...@juno.com wrote:
@AnikKumar: Most people normally wouldn't have difficulty with
probabilities on the real numbers. E.g., there is a target with two
regions, the bullseye with radius 1 and a concentric region with
radius 2. What is the probability of a randomly-thrown dart hitting
the bullseye, given that it hits the target? Most people would say
that since the area of the bullseye is 1/4 the area of the target, the
probability is 1/4. Wouldn't you say that, too?
Dave
On Aug 29, 11:15 pm, AnilKumar B akumarb2...@gmail.com wrote:
Agree with Don.
But what if we want to find probability of on real line?
How we can consider R as sample space?
Is that Sample space should be COUNTABLE and FINITE?
*By the quadratic formula, a is 2.08712 or 47.9128.
The range is 45.8256.
A falls in the range of 1..100 or 99. So the probability is
47.9128/99*
*
*
*Here you are considering Sample space as length of the interval,
right?
but
i think it should be cardinal({x/x belongs to Q and x belongs to
[1,100]}).*
On Fri, Aug 26, 2011 at 2:04 AM, Aditya Virmani
virmanisadi...@gmail.comwrote:
+1 Don... nthin is specified fr the nature of numbers if thy can be
rational or thy hav to be only natural/integral numbers...
On Wed, Aug 24, 2011 at 9:33 PM, Don dondod...@gmail.com wrote:
First find the endpoints of the region where the condition is met:
a + 100/a = 50
a^2 - 50a + 100 = 0
By the quadratic formula, a is 2.08712 or 47.9128.
The range is 45.8256.
A falls in the range of 1..100 or 99. So the probability is
47.9128/99
= 0.48397
Don
On Aug 23, 11:56 am, ramya reddy rmy.re...@gmail.com wrote:
Let 'a' be a number between 1 and 100. what is the probability
of
choosing
'a' such that a+ (100/a) 50
--
Regards
Ramya
*
*
*Try to learn something about everything and everything about
something*
--
You received this message because you are subscribed to the Google
Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
--
You received this message because you are subscribed to the Google
Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.-Hidequoted text -
- Show quoted text -
--
You received this message because you are subscribed to the Google
Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.-Hide quoted text -
- Show quoted text -
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.-
[algogeeks] Re: probability question
0.8 On Aug 31, 6:37 pm, swetha rahul swetharahu...@gmail.com wrote: In a railway station, there are two trains going. One in the harbour line and one in the main line, each having a frequency of 10 minutes. The main line service starts at 5 o'clock and the harbour line starts at 5.02A.M. A man goes to the station every day to catch the first train that comes.What is the probability of the man catching the first train? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: probability question
In my experience, the probability that a train stays on schedule to within 5 minutes is about 0.01, so I'm going to say that the probability is about 0.51. Don On Aug 31, 8:37 am, swetha rahul swetharahu...@gmail.com wrote: In a railway station, there are two trains going. One in the harbour line and one in the main line, each having a frequency of 10 minutes. The main line service starts at 5 o'clock and the harbour line starts at 5.02A.M. A man goes to the station every day to catch the first train that comes.What is the probability of the man catching the first train? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: probability question
@Swetha: My probability of reaching any train station by 5:00 a.m is zero. So I would never catch the first train. Now if the trains started at 8 or 9 a.m., then I might have some probability of catching the first one. :-) (In case it isn't obvious, I'm focusing on the wording of the question at the end of the puzzle. It would be better if it asked for the probability of catching a train on the main line instead of asking about the probability of catching the first train, which leaves at 5:00 a.m.) Dave On Aug 31, 8:37 am, swetha rahul swetharahu...@gmail.com wrote: In a railway station, there are two trains going. One in the harbour line and one in the main line, each having a frequency of 10 minutes. The main line service starts at 5 o'clock and the harbour line starts at 5.02A.M. A man goes to the station every day to catch the first train that comes.What is the probability of the man catching the first train? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: probability question
i think ur question is not clear dear acc to ur question the man should reacch the station exactly at 5 or before 5 to get the first train . then prob will be 1 otherwise prob will be 0. correct me if i am wrong. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: probability question
Assuming that the man arrives at a random time during the 24-hour day, there are 228 minutes in the day when the next train will be the harbour line (2 minutes of every 10 for 19 hours). For the other 1212 minutes the main line will be the next train. Therefore, the probability of catching the main line train is 0.841666... Don On Aug 31, 8:37 am, swetha rahul swetharahu...@gmail.com wrote: In a railway station, there are two trains going. One in the harbour line and one in the main line, each having a frequency of 10 minutes. The main line service starts at 5 o'clock and the harbour line starts at 5.02A.M. A man goes to the station every day to catch the first train that comes.What is the probability of the man catching the first train? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: probability ques
@DAVE again u r considering a finite space... in the above case...but how
wud u take the space in real number thing...with no particular info thr
r infinite real number frm 1-100...if i cud change the qn to find the
probability the chosen number is in the range a to a+1 ...thn it cud be
aptly answered
On Tue, Aug 30, 2011 at 10:10 AM, Dave dave_and_da...@juno.com wrote:
@AnikKumar: Most people normally wouldn't have difficulty with
probabilities on the real numbers. E.g., there is a target with two
regions, the bullseye with radius 1 and a concentric region with
radius 2. What is the probability of a randomly-thrown dart hitting
the bullseye, given that it hits the target? Most people would say
that since the area of the bullseye is 1/4 the area of the target, the
probability is 1/4. Wouldn't you say that, too?
Dave
On Aug 29, 11:15 pm, AnilKumar B akumarb2...@gmail.com wrote:
Agree with Don.
But what if we want to find probability of on real line?
How we can consider R as sample space?
Is that Sample space should be COUNTABLE and FINITE?
*By the quadratic formula, a is 2.08712 or 47.9128.
The range is 45.8256.
A falls in the range of 1..100 or 99. So the probability is 47.9128/99*
*
*
*Here you are considering Sample space as length of the interval, right?
but
i think it should be cardinal({x/x belongs to Q and x belongs to
[1,100]}).*
On Fri, Aug 26, 2011 at 2:04 AM, Aditya Virmani
virmanisadi...@gmail.comwrote:
+1 Don... nthin is specified fr the nature of numbers if thy can be
rational or thy hav to be only natural/integral numbers...
On Wed, Aug 24, 2011 at 9:33 PM, Don dondod...@gmail.com wrote:
First find the endpoints of the region where the condition is met:
a + 100/a = 50
a^2 - 50a + 100 = 0
By the quadratic formula, a is 2.08712 or 47.9128.
The range is 45.8256.
A falls in the range of 1..100 or 99. So the probability is 47.9128/99
= 0.48397
Don
On Aug 23, 11:56 am, ramya reddy rmy.re...@gmail.com wrote:
Let 'a' be a number between 1 and 100. what is the probability of
choosing
'a' such that a+ (100/a) 50
--
Regards
Ramya
*
*
*Try to learn something about everything and everything about
something*
--
You received this message because you are subscribed to the Google
Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
--
You received this message because you are subscribed to the Google
Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.- Hide quoted text -
- Show quoted text -
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: probability ques
@Aditya. There are an infinite number of points in that target, too.
But we don't have any trouble saying that 1/4 of them are in the
bullseye.
Dave
On Aug 31, 4:07 pm, Aditya Virmani virmanisadi...@gmail.com wrote:
@DAVE again u r considering a finite space... in the above case...but how
wud u take the space in real number thing...with no particular info thr
r infinite real number frm 1-100...if i cud change the qn to find the
probability the chosen number is in the range a to a+1 ...thn it cud be
aptly answered
On Tue, Aug 30, 2011 at 10:10 AM, Dave dave_and_da...@juno.com wrote:
@AnikKumar: Most people normally wouldn't have difficulty with
probabilities on the real numbers. E.g., there is a target with two
regions, the bullseye with radius 1 and a concentric region with
radius 2. What is the probability of a randomly-thrown dart hitting
the bullseye, given that it hits the target? Most people would say
that since the area of the bullseye is 1/4 the area of the target, the
probability is 1/4. Wouldn't you say that, too?
Dave
On Aug 29, 11:15 pm, AnilKumar B akumarb2...@gmail.com wrote:
Agree with Don.
But what if we want to find probability of on real line?
How we can consider R as sample space?
Is that Sample space should be COUNTABLE and FINITE?
*By the quadratic formula, a is 2.08712 or 47.9128.
The range is 45.8256.
A falls in the range of 1..100 or 99. So the probability is 47.9128/99*
*
*
*Here you are considering Sample space as length of the interval, right?
but
i think it should be cardinal({x/x belongs to Q and x belongs to
[1,100]}).*
On Fri, Aug 26, 2011 at 2:04 AM, Aditya Virmani
virmanisadi...@gmail.comwrote:
+1 Don... nthin is specified fr the nature of numbers if thy can be
rational or thy hav to be only natural/integral numbers...
On Wed, Aug 24, 2011 at 9:33 PM, Don dondod...@gmail.com wrote:
First find the endpoints of the region where the condition is met:
a + 100/a = 50
a^2 - 50a + 100 = 0
By the quadratic formula, a is 2.08712 or 47.9128.
The range is 45.8256.
A falls in the range of 1..100 or 99. So the probability is 47.9128/99
= 0.48397
Don
On Aug 23, 11:56 am, ramya reddy rmy.re...@gmail.com wrote:
Let 'a' be a number between 1 and 100. what is the probability of
choosing
'a' such that a+ (100/a) 50
--
Regards
Ramya
*
*
*Try to learn something about everything and everything about
something*
--
You received this message because you are subscribed to the Google
Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
--
You received this message because you are subscribed to the Google
Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.-Hide quoted text -
- Show quoted text -
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.- Hide quoted text -
- Show quoted text -
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: probability question
I could not get it. What does first train mean here? On Sep 1, 1:08 am, Don dondod...@gmail.com wrote: Assuming that the man arrives at a random time during the 24-hour day, there are 228 minutes in the day when the next train will be the harbour line (2 minutes of every 10 for 19 hours). For the other 1212 minutes the main line will be the next train. Therefore, the probability of catching the main line train is 0.841666... Don On Aug 31, 8:37 am, swetha rahul swetharahu...@gmail.com wrote: In a railway station, there are two trains going. One in the harbour line and one in the main line, each having a frequency of 10 minutes. The main line service starts at 5 o'clock and the harbour line starts at 5.02A.M. A man goes to the station every day to catch the first train that comes.What is the probability of the man catching the first train? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: probability ques
Agree with Don.
But what if we want to find probability of on real line?
How we can consider R as sample space?
Is that Sample space should be COUNTABLE and FINITE?
*By the quadratic formula, a is 2.08712 or 47.9128.
The range is 45.8256.
A falls in the range of 1..100 or 99. So the probability is 47.9128/99*
*
*
*Here you are considering Sample space as length of the interval, right? but
i think it should be cardinal({x/x belongs to Q and x belongs to [1,100]}).*
On Fri, Aug 26, 2011 at 2:04 AM, Aditya Virmani virmanisadi...@gmail.comwrote:
+1 Don... nthin is specified fr the nature of numbers if thy can be
rational or thy hav to be only natural/integral numbers...
On Wed, Aug 24, 2011 at 9:33 PM, Don dondod...@gmail.com wrote:
First find the endpoints of the region where the condition is met:
a + 100/a = 50
a^2 - 50a + 100 = 0
By the quadratic formula, a is 2.08712 or 47.9128.
The range is 45.8256.
A falls in the range of 1..100 or 99. So the probability is 47.9128/99
= 0.48397
Don
On Aug 23, 11:56 am, ramya reddy rmy.re...@gmail.com wrote:
Let 'a' be a number between 1 and 100. what is the probability of
choosing
'a' such that a+ (100/a) 50
--
Regards
Ramya
*
*
*Try to learn something about everything and everything about something*
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: probability ques
@AnikKumar: Most people normally wouldn't have difficulty with
probabilities on the real numbers. E.g., there is a target with two
regions, the bullseye with radius 1 and a concentric region with
radius 2. What is the probability of a randomly-thrown dart hitting
the bullseye, given that it hits the target? Most people would say
that since the area of the bullseye is 1/4 the area of the target, the
probability is 1/4. Wouldn't you say that, too?
Dave
On Aug 29, 11:15 pm, AnilKumar B akumarb2...@gmail.com wrote:
Agree with Don.
But what if we want to find probability of on real line?
How we can consider R as sample space?
Is that Sample space should be COUNTABLE and FINITE?
*By the quadratic formula, a is 2.08712 or 47.9128.
The range is 45.8256.
A falls in the range of 1..100 or 99. So the probability is 47.9128/99*
*
*
*Here you are considering Sample space as length of the interval, right? but
i think it should be cardinal({x/x belongs to Q and x belongs to [1,100]}).*
On Fri, Aug 26, 2011 at 2:04 AM, Aditya Virmani
virmanisadi...@gmail.comwrote:
+1 Don... nthin is specified fr the nature of numbers if thy can be
rational or thy hav to be only natural/integral numbers...
On Wed, Aug 24, 2011 at 9:33 PM, Don dondod...@gmail.com wrote:
First find the endpoints of the region where the condition is met:
a + 100/a = 50
a^2 - 50a + 100 = 0
By the quadratic formula, a is 2.08712 or 47.9128.
The range is 45.8256.
A falls in the range of 1..100 or 99. So the probability is 47.9128/99
= 0.48397
Don
On Aug 23, 11:56 am, ramya reddy rmy.re...@gmail.com wrote:
Let 'a' be a number between 1 and 100. what is the probability of
choosing
'a' such that a+ (100/a) 50
--
Regards
Ramya
*
*
*Try to learn something about everything and everything about something*
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.- Hide quoted text -
- Show quoted text -
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: probability ques
Hi Guys, Comment required from NIT Warangal ppls. Who is this guy?? Who claims this ... http://timesofindia.indiatimes.com/tech/careers/job-trends/Facebook-hires-NIT-Warangal-student-for-Rs-45-lakh/articleshow/9793300.cms -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: probability ques
+1 Don... nthin is specified fr the nature of numbers if thy can be rational or thy hav to be only natural/integral numbers... On Wed, Aug 24, 2011 at 9:33 PM, Don dondod...@gmail.com wrote: First find the endpoints of the region where the condition is met: a + 100/a = 50 a^2 - 50a + 100 = 0 By the quadratic formula, a is 2.08712 or 47.9128. The range is 45.8256. A falls in the range of 1..100 or 99. So the probability is 47.9128/99 = 0.48397 Don On Aug 23, 11:56 am, ramya reddy rmy.re...@gmail.com wrote: Let 'a' be a number between 1 and 100. what is the probability of choosing 'a' such that a+ (100/a) 50 -- Regards Ramya * * *Try to learn something about everything and everything about something* -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: probability ques
First find the endpoints of the region where the condition is met: a + 100/a = 50 a^2 - 50a + 100 = 0 By the quadratic formula, a is 2.08712 or 47.9128. The range is 45.8256. A falls in the range of 1..100 or 99. So the probability is 47.9128/99 = 0.48397 Don On Aug 23, 11:56 am, ramya reddy rmy.re...@gmail.com wrote: Let 'a' be a number between 1 and 100. what is the probability of choosing 'a' such that a+ (100/a) 50 -- Regards Ramya * * *Try to learn something about everything and everything about something* -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: probability ques
sorry i thought it was a+ 100/a50 because same question was asked by national instrument 2 days ago in this case it'll be [3-47] 45 no so prob will be 45/100 On Aug 24, 8:21 am, Ankur Goel goel.anku...@gmail.com wrote: Dude how it can be 50 if u do 50 + 100/50 which is 52 50 On Wed, Aug 24, 2011 at 8:42 AM, Manoj Bagari manojbag...@gmail.com wrote: possible value for a can be 1,2,48,49,50 and all (50-100] no. so prob will be 55/100 On Tue, Aug 23, 2011 at 10:41 PM, priya ramesh love.for.programm...@gmail.com wrote: try substituting values for a. for a=1, 2 it's 50 for a=3, 4, 5, 6, 7... it's less than 50 when a approaches 48, 49... it's 50 the max value of a that satisfies the condition is 47. p= total numbers between 3 and 47 (inclusive of 3 and 47) / total numbers between 1 and 100 = 45/98 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.- Hide quoted text - - Show quoted text - -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: probability ques
@Dheeraj: Because there are 98 numbers between 1 and 100. They are 2, 3, 4, ..., 99. It is a matter of semantics. 1 and 100 are not between 1 and 100. Dave On Aug 23, 11:21 pm, Dheeraj Sharma dheerajsharma1...@gmail.com wrote: y its 45/98 and not 45/100 ?? On Wed, Aug 24, 2011 at 9:41 AM, Manoj Bagari manojbag...@gmail.com wrote: oh sorry i thought it was a+100/a50 because this same question asked by national instrument 2 day ago. in this case passibe cases wil be 44 [3-47} no so it'll be 44/100 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- *Dheeraj Sharma* Comp Engg. NIT Kurukshetra +91 8950264227 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: probability ques
ok thanks On Wed, Aug 24, 2011 at 10:02 AM, Dave dave_and_da...@juno.com wrote: @Dheeraj: Because there are 98 numbers between 1 and 100. They are 2, 3, 4, ..., 99. It is a matter of semantics. 1 and 100 are not between 1 and 100. Dave On Aug 23, 11:21 pm, Dheeraj Sharma dheerajsharma1...@gmail.com wrote: y its 45/98 and not 45/100 ?? On Wed, Aug 24, 2011 at 9:41 AM, Manoj Bagari manojbag...@gmail.com wrote: oh sorry i thought it was a+100/a50 because this same question asked by national instrument 2 day ago. in this case passibe cases wil be 44 [3-47} no so it'll be 44/100 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- *Dheeraj Sharma* Comp Engg. NIT Kurukshetra +91 8950264227 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- *Dheeraj Sharma* Comp Engg. NIT Kurukshetra +91 8950264227 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: probability! tough one to crack!
a wonderful explaination divye!! thank you so much :) :) -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: probability! tough one to crack!
@Priya: Consider one of the squares in the grid. It has an area of 4 square inches. If the coin lands so that its center is within 1/2 inch of any grid line, the coin will touch the line. The area of the region within 1/2 inch of the boundary is 3 square inches. Therefore, 3/4 of the time the coin will touch a grid line, leaving 1/4 as the probability of not touching any of the grid lines. Dave On Aug 19, 11:50 am, priya ramesh love.for.programm...@gmail.com wrote: A 1 inch diameter coin is thrown on a table covered with a grid of lines two inches apart. What is the probability the coin lands in a square without touching any of the lines of the grid? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: probability! tough one to crack!
@Dave : me too didnt get the meaning you want to convey, plz throw some light. Sanjay Kumar B.Tech Final Year Department of Computer Engineering National Institute of Technology Kurukshetra Kurukshetra - 136119 Haryana, India On Fri, Aug 19, 2011 at 10:09 AM, priya ramesh love.for.programm...@gmail.com wrote: @dave: You are great!! The ans is indeed 1/4. I dint understand this sentence... The area of the region within 1/2 inch of the boundary is 3 square inches. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: probability! tough one to crack!
The requirements will be satisfied if the centre of the coin is such that the coin just touches the square. This is possible only when the centre of coin is in a smaller square of 1 inch side. Hence the result. On Fri, Aug 19, 2011 at 10:41 PM, Sanjay Rajpal srn...@gmail.com wrote: @Dave : me too didnt get the meaning you want to convey, plz throw some light. Sanjay Kumar B.Tech Final Year Department of Computer Engineering National Institute of Technology Kurukshetra Kurukshetra - 136119 Haryana, India On Fri, Aug 19, 2011 at 10:09 AM, priya ramesh love.for.programm...@gmail.com wrote: @dave: You are great!! The ans is indeed 1/4. I dint understand this sentence... The area of the region within 1/2 inch of the boundary is 3 square inches. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Romil -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: probability! tough one to crack!
why this sentence?? 3/4 of the time the coin will touch a grid line?? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: probability! tough one to crack!
got it. Sanju :) On Fri, Aug 19, 2011 at 10:25 AM, Romil ... vamosro...@gmail.comwrote: Because this is the answer. Rest of the times it will not touch any of the grid lines. On Fri, Aug 19, 2011 at 10:50 PM, priya ramesh love.for.programm...@gmail.com wrote: why this sentence?? 3/4 of the time the coin will touch a grid line?? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Romil -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: probability! tough one to crack!
For those of you who want an explanation of Dave's answer, please refer to the diagram below. | 0.5 in |-| 0.5 in | xxx - xxx 0.5 in xxx - xx...xx | xx...xx | xx...xx | xx...xx | xxx - xxx 0.5 in xxx - Since the radius of the coin is 0.5 in, if the center of the coin falls in the x area, it will cross a grid line. So, probability of the coin not crossing the grid lines is area of dots / area of square = (1 inch square / 4 inch square) = 1/4 -- DK http://www.divye.in http://twitter.com/divyekapoor http://gplus.to/divyekapoor -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/osMGCscKC70J. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: probability! tough one to crack!
GOOD one dave and thanks divye for a wonderful explanation On Fri, Aug 19, 2011 at 11:44 PM, DK divyekap...@gmail.com wrote: For those of you who want an explanation of Dave's answer, please refer to the diagram below. | 0.5 in |-| 0.5 in | xxx - xxx 0.5 in xxx - xx...xx | xx...xx | xx...xx | xx...xx | xxx - xxx 0.5 in xxx - Since the radius of the coin is 0.5 in, if the center of the coin falls in the x area, it will cross a grid line. So, probability of the coin not crossing the grid lines is area of dots / area of square = (1 inch square / 4 inch square) = 1/4 -- DK http://www.divye.in http://twitter.com/divyekapoor http://gplus.to/divyekapoor -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To view this discussion on the web visit https://groups.google.com/d/msg/algogeeks/-/osMGCscKC70J. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- **Regards SAGAR PAREEK COMPUTER SCIENCE AND ENGINEERING NIT ALLAHABAD -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: probability tough one!
Is that exactly 2 or atleast 2? P(atleast 2)=1-P(no 2 people )=1-(364*363*362*.*317/365^49) On Aug 17, 5:24 pm, priya ramesh love.for.programm...@gmail.com wrote: nothing is specified. I guess it's 365 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: probability tough one!
can yo explain it pl? On Wed, Aug 17, 2011 at 6:11 PM, Aditya Jain aditya2...@gmail.com wrote: Is that exactly 2 or atleast 2? P(atleast 2)=1-P(no 2 people )=1-(364*363*362*.*317/365^49) On Aug 17, 5:24 pm, priya ramesh love.for.programm...@gmail.com wrote: nothing is specified. I guess it's 365 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: probability tough one!
Take it as: P(atleast 2) = 1-P(no 2 have same b'day) = 1- ((365C50)/50!) where C represents the combinations On Wed, Aug 17, 2011 at 6:14 PM, sukran dhawan sukrandha...@gmail.comwrote: can yo explain it pl? On Wed, Aug 17, 2011 at 6:11 PM, Aditya Jain aditya2...@gmail.com wrote: Is that exactly 2 or atleast 2? P(atleast 2)=1-P(no 2 people )=1-(364*363*362*.*317/365^49) On Aug 17, 5:24 pm, priya ramesh love.for.programm...@gmail.com wrote: nothing is specified. I guess it's 365 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Romil -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: probability tough one!
there is something anamoly about this birthday probability caculation. Search in google..you may find it. Regards, Adi Srikanth. Mob No 9887233349 Personal Pages: adisrikanth.co.nr On Wed, Aug 17, 2011 at 6:25 PM, Romil ... vamosro...@gmail.com wrote: Take it as: P(atleast 2) = 1-P(no 2 have same b'day) = 1- ((365C50)/50!) where C represents the combinations On Wed, Aug 17, 2011 at 6:14 PM, sukran dhawan sukrandha...@gmail.comwrote: can yo explain it pl? On Wed, Aug 17, 2011 at 6:11 PM, Aditya Jain aditya2...@gmail.comwrote: Is that exactly 2 or atleast 2? P(atleast 2)=1-P(no 2 people )=1-(364*363*362*.*317/365^49) On Aug 17, 5:24 pm, priya ramesh love.for.programm...@gmail.com wrote: nothing is specified. I guess it's 365 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Romil -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: probability tough one!
The question is directly taken from coreman...read it,the best is explained there On Wed, Aug 17, 2011 at 7:10 PM, Adi Srikanth adisriika...@gmail.com wrote: there is something anamoly about this birthday probability caculation. Search in google..you may find it. Regards, Adi Srikanth. Mob No 9887233349 Personal Pages: adisrikanth.co.nr On Wed, Aug 17, 2011 at 6:25 PM, Romil ... vamosro...@gmail.com wrote: Take it as: P(atleast 2) = 1-P(no 2 have same b'day) = 1- ((365C50)/50!) where C represents the combinations On Wed, Aug 17, 2011 at 6:14 PM, sukran dhawan sukrandha...@gmail.com wrote: can yo explain it pl? On Wed, Aug 17, 2011 at 6:11 PM, Aditya Jain aditya2...@gmail.com wrote: Is that exactly 2 or atleast 2? P(atleast 2)=1-P(no 2 people )=1-(364*363*362*.*317/365^49) On Aug 17, 5:24 pm, priya ramesh love.for.programm...@gmail.com wrote: nothing is specified. I guess it's 365 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Romil -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Saurabh Singh B.Tech (Computer Science) MNNIT ALLAHABAD -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: probability tough one!
I don't have coreman. If you have an e book can you plz upload it?? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: probability tough one!
Sorry the problems are not same.I should have read the problem more carefully.Anyways I would recommend its high time you get a hard copy of coreman.. Trying for a solution now. On Wed, Aug 17, 2011 at 7:19 PM, priya ramesh love.for.programm...@gmail.com wrote: I don't have coreman. If you have an e book can you plz upload it?? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Saurabh Singh B.Tech (Computer Science) MNNIT ALLAHABAD -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
I think there is some ambiguity in the question. (All this time you don't know you were tossing a fair coin or not). 1) Does the above statement mean that the thower don't know whether he or she threw a fair coin even after throwing? Or is the thrower not informed beforehand that one of them is not a fair coin? 2) Does the coin count reduce after every throw or should it be put back? 3) Depending on 1) and 2), there will be different answers. On Aug 9, 12:13 am, Maddy madhu.mitha...@gmail.com wrote: I think the answer is 17/80, because as you say the 5 trials are independent.. but the fact that a head turns up in all the 5 trials, give some information about our original probability of choosing the coins. in case we had obtained a tail in the first trial, we can be sure its the fair coin, and so the consecutive trials would become independent.. but since that is not the case, every head is going to increase the chance of choosing the biased coin(initially), and hence affect the probability of the next head.. before the first trial probability of landing a head is 3/5, but once u see the first head, the probability of landing a head on the second trial changes to 4/5*1/4+1/5, and so on..that is, there is a higher probability that we chose a biased coin, rather than the fair coin. hope its clear.. On Aug 7, 11:36 pm, sumit gaur sumitgau...@gmail.com wrote: (3/5) On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not).- Hide quoted text - - Show quoted text - -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
I'm little late but I too got 17/18. On Tue, Aug 16, 2011 at 10:47 PM, Jacob Ridley jridley2...@gmail.comwrote: I think there is some ambiguity in the question. (All this time you don't know you were tossing a fair coin or not). 1) Does the above statement mean that the thower don't know whether he or she threw a fair coin even after throwing? Or is the thrower not informed beforehand that one of them is not a fair coin? 2) Does the coin count reduce after every throw or should it be put back? 3) Depending on 1) and 2), there will be different answers. On Aug 9, 12:13 am, Maddy madhu.mitha...@gmail.com wrote: I think the answer is 17/80, because as you say the 5 trials are independent.. but the fact that a head turns up in all the 5 trials, give some information about our original probability of choosing the coins. in case we had obtained a tail in the first trial, we can be sure its the fair coin, and so the consecutive trials would become independent.. but since that is not the case, every head is going to increase the chance of choosing the biased coin(initially), and hence affect the probability of the next head.. before the first trial probability of landing a head is 3/5, but once u see the first head, the probability of landing a head on the second trial changes to 4/5*1/4+1/5, and so on..that is, there is a higher probability that we chose a biased coin, rather than the fair coin. hope its clear.. On Aug 7, 11:36 pm, sumit gaur sumitgau...@gmail.com wrote: (3/5) On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not).- Hide quoted text - - Show quoted text - -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- regards, chinna. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
A=p(biased coin/5 heads)=8/9 probability that the coin is biased given 5 heads (bayes theorem) B=p(unbiased coin/5 heads)=1/9 P(6th head)=A*1+B*1/2=17/18 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability
Let f(x)= no. of ways of getting a sum x when pair of dice is thrown. S()=sum . tot=S( f(i) ) 2=i=12 So the solution is : S( f(i)*(tot-f(i)) )/(6^4). Is daT fine.. ? -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability
total possible outcomes= 6*6*6*6
possibility of sum gives 2 in both pair -- 1*1*1*1 =1
possibility of sum gives 3 in both pair -- 2*1*2*1 =4
{because the possibilities are (1,2)(1,2), (1,2)(2,1), (2,1)(1,2),
(2,1)(2,1)}
possibility of 4 -- 9 { 2,2 1,3 and 3,1}
possibility of 5 -- 16 {1,4 2,3 3,2 4,1}
possibility of 6 -- 25{1,5 2,4 3,3 4,2 5,1}
possibility of 7 -- 36{1,6 2,5 3,4 4,3 5,2 6,1}
possibility of 8 -- 25{2,6 3,5 4,4 5,3 6,2}
9-- 16
10 -- 9
11 -- 4
12 -- 1
so probability required = (1+4+9+16+25+36+25+16+9+4+1)/(6*6*6*6)
=146/1296
=73/648
On Fri, Aug 12, 2011 at 1:19 AM, Hurricane ashman...@gmail.com wrote:
Let f(x)= no. of ways of getting a sum x when pair of dice is thrown.
S()=sum .
tot=S( f(i) ) 2=i=12
So the solution is : S( f(i)*(tot-f(i)) )/(6^4).
Is daT fine.. ?
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability
sorry that was for equal case : for unequal case 1-(73/648)= 575/648
On Fri, Aug 12, 2011 at 2:28 AM, Prakash D cegprak...@gmail.com wrote:
total possible outcomes= 6*6*6*6
possibility of sum gives 2 in both pair -- 1*1*1*1 =1
possibility of sum gives 3 in both pair -- 2*1*2*1 =4
{because the possibilities are (1,2)(1,2), (1,2)(2,1), (2,1)(1,2),
(2,1)(2,1)}
possibility of 4 -- 9 { 2,2 1,3 and 3,1}
possibility of 5 -- 16 {1,4 2,3 3,2 4,1}
possibility of 6 -- 25{1,5 2,4 3,3 4,2 5,1}
possibility of 7 -- 36{1,6 2,5 3,4 4,3 5,2 6,1}
possibility of 8 -- 25{2,6 3,5 4,4 5,3 6,2}
9-- 16
10 -- 9
11 -- 4
12 -- 1
so probability required = (1+4+9+16+25+36+25+16+9+4+1)/(6*6*6*6)
=146/1296
=73/648
--
You received this message because you are subscribed to the Google Groups
Algorithm Geeks group.
To post to this group, send email to algogeeks@googlegroups.com.
To unsubscribe from this group, send email to
algogeeks+unsubscr...@googlegroups.com.
For more options, visit this group at
http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
@dave: yes it seems so that 17/18 is correct...I deduced it from the cond prob formula.. I have a minor doubt in general why prob( 2nd toss is a head given that a head occurred in the first toss ) doesnt seem same as p( head in first toss and head in second toss with fair coin) +p(head in first toss and head in second toss with unfair coin)? is it due to the fact that we are not looking at the same sample space in both cases?i am not able to visualise the difference in general..this is also the reason why most of the people said earlier 17/80 as the answer moreover, if the question was exactly the same except in that it was NOT mentioned that heads occurred previously , what would the prob of getting a head in the second toss? would it be P( of getting tail in first toss and head in second toss given that fair coin is chosen) +P( of getting head in first toss and head in second toss given that fair coin is chosen) +P( getting heads in first toss and heads in second toss given that unfair coin is chosen) ? this for any toss turns out to be 3/5 can u explain the logic abt why it always gives 3/5? On Tue, Aug 9, 2011 at 7:37 AM, raj kumar megamonste...@gmail.com wrote: plz reply am i right or wrong -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- People often say that motivation doesn't last. Well, neither does bathing - that's why we recommend it daily. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
it is (1/5)/( (4/5 *(1/2)^6) + (1/5 * 1)) = 80/85 = 16/17 On Aug 7, 10:54 pm, Nitish Garg nitishgarg1...@gmail.com wrote: Should be (4/5 *(1/2)^6) + (1/5 * 1) = 17/80 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
go through the posts before posting anything :) On Tue, Aug 9, 2011 at 6:29 PM, arpit.gupta arpitg1...@gmail.com wrote: it is (1/5)/( (4/5 *(1/2)^6) + (1/5 * 1)) = 80/85 = 16/17 On Aug 7, 10:54 pm, Nitish Garg nitishgarg1...@gmail.com wrote: Should be (4/5 *(1/2)^6) + (1/5 * 1) = 17/80 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
ans is 16/17 + 1/2*1/17 = 33/34 On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
@Arpit: No. The probability of getting 6 consecutive heads is 1/5 * 1 + 4/5 * (1/2)^6 ) = 17/80, while the probability of getting 5 consecutive heads is 1/5 * 1 + 4/5 * (1/2)^6 ) = 9/40. Thus, the probability of getting a head on the sixth roll given that you have gotten heads on all five previous rolls is (17/80) / (9/40), which is 17/18. Dave On Aug 9, 7:59 am, arpit.gupta arpitg1...@gmail.com wrote: it is (1/5)/( (4/5 *(1/2)^6) + (1/5 * 1)) = 80/85 = 16/17 On Aug 7, 10:54 pm, Nitish Garg nitishgarg1...@gmail.com wrote: Should be (4/5 *(1/2)^6) + (1/5 * 1) = 17/80 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
@Arun: The probability of getting a head on the first toss is 1/5 * 1 + 4/5 * (1/2) ) = 3/5, while the probability of getting 2 consecutive heads is 1/5 * 1 + 4/5 * (1/2)^2 ) = 2/5. Thus, the probability of getting a head on the second roll given that you have gotten a head on the first roll is (2/5) / (3/5), which is 2/3. If you didn't know the outcome of the first roll, the probability of heads on the second roll would still be 3/5. Dave On Aug 9, 2:57 am, Arun Vishwanathan aaron.nar...@gmail.com wrote: @dave: yes it seems so that 17/18 is correct...I deduced it from the cond prob formula.. I have a minor doubt in general why prob( 2nd toss is a head given that a head occurred in the first toss ) doesnt seem same as p( head in first toss and head in second toss with fair coin) +p(head in first toss and head in second toss with unfair coin)? is it due to the fact that we are not looking at the same sample space in both cases?i am not able to visualise the difference in general..this is also the reason why most of the people said earlier 17/80 as the answer moreover, if the question was exactly the same except in that it was NOT mentioned that heads occurred previously , what would the prob of getting a head in the second toss? would it be P( of getting tail in first toss and head in second toss given that fair coin is chosen) +P( of getting head in first toss and head in second toss given that fair coin is chosen) +P( getting heads in first toss and heads in second toss given that unfair coin is chosen) ? this for any toss turns out to be 3/5 can u explain the logic abt why it always gives 3/5? On Tue, Aug 9, 2011 at 7:37 AM, raj kumar megamonste...@gmail.com wrote: plz reply am i right or wrong -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- People often say that motivation doesn't last. Well, neither does bathing - that's why we recommend it daily. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
@dave- calculation mistake on my part - method is right. getting 17/18 only thanks anyways. On Aug 9, 6:26 pm, Dave dave_and_da...@juno.com wrote: @Arun: The probability of getting a head on the first toss is 1/5 * 1 + 4/5 * (1/2) ) = 3/5, while the probability of getting 2 consecutive heads is 1/5 * 1 + 4/5 * (1/2)^2 ) = 2/5. Thus, the probability of getting a head on the second roll given that you have gotten a head on the first roll is (2/5) / (3/5), which is 2/3. If you didn't know the outcome of the first roll, the probability of heads on the second roll would still be 3/5. Dave On Aug 9, 2:57 am, Arun Vishwanathan aaron.nar...@gmail.com wrote: @dave: yes it seems so that 17/18 is correct...I deduced it from the cond prob formula.. I have a minor doubt in general why prob( 2nd toss is a head given that a head occurred in the first toss ) doesnt seem same as p( head in first toss and head in second toss with fair coin) +p(head in first toss and head in second toss with unfair coin)? is it due to the fact that we are not looking at the same sample space in both cases?i am not able to visualise the difference in general..this is also the reason why most of the people said earlier 17/80 as the answer moreover, if the question was exactly the same except in that it was NOT mentioned that heads occurred previously , what would the prob of getting a head in the second toss? would it be P( of getting tail in first toss and head in second toss given that fair coin is chosen) +P( of getting head in first toss and head in second toss given that fair coin is chosen) +P( getting heads in first toss and heads in second toss given that unfair coin is chosen) ? this for any toss turns out to be 3/5 can u explain the logic abt why it always gives 3/5? On Tue, Aug 9, 2011 at 7:37 AM, raj kumar megamonste...@gmail.com wrote: plz reply am i right or wrong -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- People often say that motivation doesn't last. Well, neither does bathing - that's why we recommend it daily. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
The probability of getting n consecutive heads is P(n heads) = 1/5 * 1 + 4/5 * (1/2)^n, Thus, the probability of getting a head on the n+1st roll given that you have gotten heads on all n previous rolls is P(n+1 heads | n heads) = P(n+1) / P(n) = ( 1/5 * 1 + 4/5 * (1/2)^(n+1) ) / ( 1/5 * 1 + 4/5 * (1/2)^n ). Multiplying numerator and denominator by 5* 2^(n-1) and recognizing 4 as 2^2 gives P(n+1 heads | n heads) = (2^(n-1) + 1) / (2^(n-1) + 2). Dave On Aug 9, 12:30 am, programming love love.for.programm...@gmail.com wrote: @Dave: I guess 17/18 is correct. Since we have to *calculate the probability of getting a head in the 6th flip given that first 5 flips are a head*. Can you please explain how you got the values of consequent flips when you said this? *In fact, the probability is 3/5 for the first flip. After a head is flipped, the probability of a head is 2/3. After two heads have been flipped, it is 3/4. After 3 heads, it is 5/6. After 4 heads, the probability is 9/10, and after 5 heads, the probability is 17/18.* -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
@dave - method is right, calculation mistake on my part, getting 17/18 only. thanks anyways. On Aug 9, 6:26 pm, Dave dave_and_da...@juno.com wrote: @Arun: The probability of getting a head on the first toss is 1/5 * 1 + 4/5 * (1/2) ) = 3/5, while the probability of getting 2 consecutive heads is 1/5 * 1 + 4/5 * (1/2)^2 ) = 2/5. Thus, the probability of getting a head on the second roll given that you have gotten a head on the first roll is (2/5) / (3/5), which is 2/3. If you didn't know the outcome of the first roll, the probability of heads on the second roll would still be 3/5. Dave On Aug 9, 2:57 am, Arun Vishwanathan aaron.nar...@gmail.com wrote: @dave: yes it seems so that 17/18 is correct...I deduced it from the cond prob formula.. I have a minor doubt in general why prob( 2nd toss is a head given that a head occurred in the first toss ) doesnt seem same as p( head in first toss and head in second toss with fair coin) +p(head in first toss and head in second toss with unfair coin)? is it due to the fact that we are not looking at the same sample space in both cases?i am not able to visualise the difference in general..this is also the reason why most of the people said earlier 17/80 as the answer moreover, if the question was exactly the same except in that it was NOT mentioned that heads occurred previously , what would the prob of getting a head in the second toss? would it be P( of getting tail in first toss and head in second toss given that fair coin is chosen) +P( of getting head in first toss and head in second toss given that fair coin is chosen) +P( getting heads in first toss and heads in second toss given that unfair coin is chosen) ? this for any toss turns out to be 3/5 can u explain the logic abt why it always gives 3/5? On Tue, Aug 9, 2011 at 7:37 AM, raj kumar megamonste...@gmail.com wrote: plz reply am i right or wrong -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- People often say that motivation doesn't last. Well, neither does bathing - that's why we recommend it daily. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
The statement You randomly pulled one coin from the bag and tossed tells that all the events of tossing the coin are independent hence ans is 3/5 On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
@Dave: Thanks for the explanation :) -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
@Ritu: We are flipping one coin five times. Are you saying that you don't learn anything about the coin by flipping it? Would you learn something if any one of the five flips turned up tails? After a tails, would you say that the probability of a subsequent head is still 3/5? Dave On Aug 9, 11:19 am, ritu ritugarg.c...@gmail.com wrote: The statement You randomly pulled one coin from the bag and tossed tells that all the events of tossing the coin are independent hence ans is 3/5 On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
@dave: thank you.. nice explanation :) On Wed, Aug 10, 2011 at 3:24 AM, Dave dave_and_da...@juno.com wrote: @Ritu: We are flipping one coin five times. Are you saying that you don't learn anything about the coin by flipping it? Would you learn something if any one of the five flips turned up tails? After a tails, would you say that the probability of a subsequent head is still 3/5? Dave On Aug 9, 11:19 am, ritu ritugarg.c...@gmail.com wrote: The statement You randomly pulled one coin from the bag and tossed tells that all the events of tossing the coin are independent hence ans is 3/5 On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
@dave : nice explanationsthank you for pointing out :) On Wed, Aug 10, 2011 at 3:39 AM, Prakash D cegprak...@gmail.com wrote: @dave: thank you.. nice explanation :) On Wed, Aug 10, 2011 at 3:24 AM, Dave dave_and_da...@juno.com wrote: @Ritu: We are flipping one coin five times. Are you saying that you don't learn anything about the coin by flipping it? Would you learn something if any one of the five flips turned up tails? After a tails, would you say that the probability of a subsequent head is still 3/5? Dave On Aug 9, 11:19 am, ritu ritugarg.c...@gmail.com wrote: The statement You randomly pulled one coin from the bag and tossed tells that all the events of tossing the coin are independent hence ans is 3/5 On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards, Shachindra A C -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
(3/5) On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
@brijesh *first five times* is mentioned intentionally to mislead i think. I vote for 3/5. Moreover, 17/80 doesn't make sense also. Plz correct me if I am wrong. On Mon, Aug 8, 2011 at 12:06 PM, sumit gaur sumitgau...@gmail.com wrote: (3/5) On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards, Shachindra A C -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
I think 17/80 is wrong because if you say that while calculating the answer 3/5, you havent included the first 5 cases, then even after including it will only increase the probability of getting the biased coin in hand and thus increasing the overall probability of getting the heads and 17/80 is a way lesser than 3/5 although i am not sure about 3/5 even coz of the reasoning i just gave also, when you are calculating 4/5*1/2^6 you are not getting any benefit out of the first five tosses, like, they must have gave you some positive response towards that yes, you will get the head even next time, but doing this you are actually decreasing the probability as compared to the one you could have get without those 5 cases On Mon, Aug 8, 2011 at 1:06 PM, Shachindra A C sachindr...@gmail.comwrote: @brijesh *first five times* is mentioned intentionally to mislead i think. I vote for 3/5. Moreover, 17/80 doesn't make sense also. Plz correct me if I am wrong. On Mon, Aug 8, 2011 at 12:06 PM, sumit gaur sumitgau...@gmail.com wrote: (3/5) On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards, Shachindra A C -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- --- Puneet Goyal Student of B. Tech. III Year (Software Engineering) Delhi Technological University, Delhi --- -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
I think the answer is 17/80, because as you say the 5 trials are independent.. but the fact that a head turns up in all the 5 trials, give some information about our original probability of choosing the coins. in case we had obtained a tail in the first trial, we can be sure its the fair coin, and so the consecutive trials would become independent.. but since that is not the case, every head is going to increase the chance of choosing the biased coin(initially), and hence affect the probability of the next head.. before the first trial probability of landing a head is 3/5, but once u see the first head, the probability of landing a head on the second trial changes to 4/5*1/4+1/5, and so on..that is, there is a higher probability that we chose a biased coin, rather than the fair coin. hope its clear.. On Aug 7, 11:36 pm, sumit gaur sumitgau...@gmail.com wrote: (3/5) On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
The answer is 17 in 18, because flipping 5 heads in a row is evidence that the probability is high that we have the coin with two heads. Don On Aug 7, 12:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
@don: i too get yr answer 17/18 using conditional probability...does that make sense??i guess this is first new answer lol On Mon, Aug 8, 2011 at 9:29 PM, Don dondod...@gmail.com wrote: The answer is 17 in 18, because flipping 5 heads in a row is evidence that the probability is high that we have the coin with two heads. Don On Aug 7, 12:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- People often say that motivation doesn't last. Well, neither does bathing - that's why we recommend it daily. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
Consider the 5 * 64 possible outcomes for the selection of coin and six flips, each one happening with equal probability. Of those 320 possible outcomes, 4*62 are excluded by knowing that the first 5 flips are heads. That leaves 64 outcomes with the rigged coin and 2 outcomes with each of the fair coins, for a total of 72 outcomes. 68 of those are heads, so the answer to the puzzle is 68 of 72, or 17 of 18. Don On Aug 8, 2:36 am, Shachindra A C sachindr...@gmail.com wrote: @brijesh *first five times* is mentioned intentionally to mislead i think. I vote for 3/5. Moreover, 17/80 doesn't make sense also. Plz correct me if I am wrong. On Mon, Aug 8, 2011 at 12:06 PM, sumit gaur sumitgau...@gmail.com wrote: (3/5) On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards, Shachindra A C -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
answer is 3/5. 17/80 is the answer for 6 consecutive heads. On Tue, Aug 9, 2011 at 2:07 AM, Don dondod...@gmail.com wrote: Consider the 5 * 64 possible outcomes for the selection of coin and six flips, each one happening with equal probability. Of those 320 possible outcomes, 4*62 are excluded by knowing that the first 5 flips are heads. That leaves 64 outcomes with the rigged coin and 2 outcomes with each of the fair coins, for a total of 72 outcomes. 68 of those are heads, so the answer to the puzzle is 68 of 72, or 17 of 18. Don On Aug 8, 2:36 am, Shachindra A C sachindr...@gmail.com wrote: @brijesh *first five times* is mentioned intentionally to mislead i think. I vote for 3/5. Moreover, 17/80 doesn't make sense also. Plz correct me if I am wrong. On Mon, Aug 8, 2011 at 12:06 PM, sumit gaur sumitgau...@gmail.com wrote: (3/5) On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards, Shachindra A C -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
@shady: 3/5 can be the answer to such a question: what is prob of getting head on nth toss if we have 4 coins fair and one biased...then at nth toss u choose 4/5 1/5 prob and then u get 3/5 @shady , don: i did this: P( 6th head | 5 heads occured)= P( 6 heads )/ P( 5 heads) answr u get is 17/18..i cud be wrong please correct if so On Mon, Aug 8, 2011 at 10:45 PM, shady sinv...@gmail.com wrote: answer is 3/5. 17/80 is the answer for 6 consecutive heads. On Tue, Aug 9, 2011 at 2:07 AM, Don dondod...@gmail.com wrote: Consider the 5 * 64 possible outcomes for the selection of coin and six flips, each one happening with equal probability. Of those 320 possible outcomes, 4*62 are excluded by knowing that the first 5 flips are heads. That leaves 64 outcomes with the rigged coin and 2 outcomes with each of the fair coins, for a total of 72 outcomes. 68 of those are heads, so the answer to the puzzle is 68 of 72, or 17 of 18. Don On Aug 8, 2:36 am, Shachindra A C sachindr...@gmail.com wrote: @brijesh *first five times* is mentioned intentionally to mislead i think. I vote for 3/5. Moreover, 17/80 doesn't make sense also. Plz correct me if I am wrong. On Mon, Aug 8, 2011 at 12:06 PM, sumit gaur sumitgau...@gmail.com wrote: (3/5) On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards, Shachindra A C -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- People often say that motivation doesn't last. Well, neither does bathing - that's why we recommend it daily. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
Man, I feel so stupid. Yes, it is a case of conditional probability. We have to calculate the probability of six heads, given that 5 heads have occured. So answer is 17/18. On Tue, Aug 9, 2011 at 1:47 AM, Arun Vishwanathan aaron.nar...@gmail.comwrote: @shady: 3/5 can be the answer to such a question: what is prob of getting head on nth toss if we have 4 coins fair and one biased...then at nth toss u choose 4/5 1/5 prob and then u get 3/5 @shady , don: i did this: P( 6th head | 5 heads occured)= P( 6 heads )/ P( 5 heads) answr u get is 17/18..i cud be wrong please correct if so On Mon, Aug 8, 2011 at 10:45 PM, shady sinv...@gmail.com wrote: answer is 3/5. 17/80 is the answer for 6 consecutive heads. On Tue, Aug 9, 2011 at 2:07 AM, Don dondod...@gmail.com wrote: Consider the 5 * 64 possible outcomes for the selection of coin and six flips, each one happening with equal probability. Of those 320 possible outcomes, 4*62 are excluded by knowing that the first 5 flips are heads. That leaves 64 outcomes with the rigged coin and 2 outcomes with each of the fair coins, for a total of 72 outcomes. 68 of those are heads, so the answer to the puzzle is 68 of 72, or 17 of 18. Don On Aug 8, 2:36 am, Shachindra A C sachindr...@gmail.com wrote: @brijesh *first five times* is mentioned intentionally to mislead i think. I vote for 3/5. Moreover, 17/80 doesn't make sense also. Plz correct me if I am wrong. On Mon, Aug 8, 2011 at 12:06 PM, sumit gaur sumitgau...@gmail.com wrote: (3/5) On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Regards, Shachindra A C -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- People often say that motivation doesn't last. Well, neither does bathing - that's why we recommend it daily. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- Shuaib http://www.bytehood.com http://twitter.com/ShuaibKhan -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
answer should be 3/5 think like that tossing 5 times will not help you predict the outcome of sixth toss. Therefore that information is meaningless. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
@Vinay: What if you tossed 100 consecutive heads? Would that be enough to convince you that you had the double-headed coin? If so, then doesn't tossing 5 consecutive heads give you at least an inkling that you might have it? Wouldn't you then think that there would be a higher probability of getting a head on the sixth toss than there was on the first toss (3/5)? Don's conditional probability answer 17/18 is the right answer. Dave On Aug 8, 5:04 pm, vinay aggarwal vinayiiit2...@gmail.com wrote: answer should be 3/5 think like that tossing 5 times will not help you predict the outcome of sixth toss. Therefore that information is meaningless. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
3/5. As the question doesn't ask anything about the sequence. Had the question been Find the probability that all 6 are H then it would have been 17/80. On 9 August 2011 04:07, Dave dave_and_da...@juno.com wrote: @Vinay: What if you tossed 100 consecutive heads? Would that be enough to convince you that you had the double-headed coin? If so, then doesn't tossing 5 consecutive heads give you at least an inkling that you might have it? Wouldn't you then think that there would be a higher probability of getting a head on the sixth toss than there was on the first toss (3/5)? Don's conditional probability answer 17/18 is the right answer. Dave On Aug 8, 5:04 pm, vinay aggarwal vinayiiit2...@gmail.com wrote: answer should be 3/5 think like that tossing 5 times will not help you predict the outcome of sixth toss. Therefore that information is meaningless. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- ___ Please do not print this e-mail until urgent requirement. Go Green!! Save Papers = Save Trees -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
@Dipankar: You are correct about the answer to your alternative question being 17/80, but your answer 3/5 says that you don't think you have learned anything by the five heads flips. Don has given a good explanation as to why the answer is 17/18, but you apparently refuse to accept it. There is none so blind as one who will not see. Dave On Aug 8, 9:26 pm, Dipankar Patro dip10c...@gmail.com wrote: 3/5. As the question doesn't ask anything about the sequence. Had the question been Find the probability that all 6 are H then it would have been 17/80. On 9 August 2011 04:07, Dave dave_and_da...@juno.com wrote: @Vinay: What if you tossed 100 consecutive heads? Would that be enough to convince you that you had the double-headed coin? If so, then doesn't tossing 5 consecutive heads give you at least an inkling that you might have it? Wouldn't you then think that there would be a higher probability of getting a head on the sixth toss than there was on the first toss (3/5)? Don's conditional probability answer 17/18 is the right answer. Dave On Aug 8, 5:04 pm, vinay aggarwal vinayiiit2...@gmail.com wrote: answer should be 3/5 think like that tossing 5 times will not help you predict the outcome of sixth toss. Therefore that information is meaningless. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- ___ Please do not print this e-mail until urgent requirement. Go Green!! Save Papers = Save Trees -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
it's 3/5 On Tue, Aug 9, 2011 at 8:29 AM, Dave dave_and_da...@juno.com wrote: @Dipankar: You are correct about the answer to your alternative question being 17/80, but your answer 3/5 says that you don't think you have learned anything by the five heads flips. Don has given a good explanation as to why the answer is 17/18, but you apparently refuse to accept it. There is none so blind as one who will not see. Dave On Aug 8, 9:26 pm, Dipankar Patro dip10c...@gmail.com wrote: 3/5. As the question doesn't ask anything about the sequence. Had the question been Find the probability that all 6 are H then it would have been 17/80. On 9 August 2011 04:07, Dave dave_and_da...@juno.com wrote: @Vinay: What if you tossed 100 consecutive heads? Would that be enough to convince you that you had the double-headed coin? If so, then doesn't tossing 5 consecutive heads give you at least an inkling that you might have it? Wouldn't you then think that there would be a higher probability of getting a head on the sixth toss than there was on the first toss (3/5)? Don's conditional probability answer 17/18 is the right answer. Dave On Aug 8, 5:04 pm, vinay aggarwal vinayiiit2...@gmail.com wrote: answer should be 3/5 think like that tossing 5 times will not help you predict the outcome of sixth toss. Therefore that information is meaningless. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- ___ Please do not print this e-mail until urgent requirement. Go Green!! Save Papers = Save Trees -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
@all those who gave Should be (4/5 *(1/2)^6) + (1/5 * 1) = 17/80 when it's already given that 5 heads have turned up already then why abut are you adding that probability you all are considering it as finding the probability of finding 6 consecutive heads. since all tosses are independent the answer should be 3/5. the point that 5 heads have turned up already may points that the coin selected is biased in that case pr(6)=1; now the answer depends on the interviewer according to me it should be 3/5 thanks -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
Pls check the ques 8th This may remove misunderstanding... http://www.folj.com/puzzles/difficult-logic-problems.htm On Tue, Aug 9, 2011 at 10:21 AM, raj kumar megamonste...@gmail.com wrote: @all those who gave Should be (4/5 *(1/2)^6) + (1/5 * 1) = 17/80 when it's already given that 5 heads have turned up already then why abut are you adding that probability you all are considering it as finding the probability of finding 6 consecutive heads. since all tosses are independent the answer should be 3/5. the point that 5 heads have turned up already may points that the coin selected is biased in that case pr(6)=1; now the answer depends on the interviewer according to me it should be 3/5 thanks -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- **Regards SAGAR PAREEK COMPUTER SCIENCE AND ENGINEERING NIT ALLAHABAD -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
@Coder: You (and others) are saying that the probability of a head is 3/5 on the first flip, and that it doesn't change after any number of heads are flipped. Notice, however, that if the first flip were tails, you wouldn't say that the probability of getting heads on the next flip is 3/5. You would have learned that one of the four fair coins was chosen. So even though the probability of a head was 3/5 on the first flip, it changes to 1/2 on all flips subsequent to a tail. Since the probabililty changes if a tail is flipped, what makes you think it doesn't change if a head is flipped. In fact, the probability is 3/5 for the first flip. After a head is flipped, the probability of a head is 2/3. After two heads have been flipped, it is 3/4. After 3 heads, it is 5/6. After 4 heads, the probability is 9/10, and after 5 heads, the probability is 17/18. Dave On Aug 8, 11:23 pm, coder dumca coder.du...@gmail.com wrote: it's 3/5 On Tue, Aug 9, 2011 at 8:29 AM, Dave dave_and_da...@juno.com wrote: @Dipankar: You are correct about the answer to your alternative question being 17/80, but your answer 3/5 says that you don't think you have learned anything by the five heads flips. Don has given a good explanation as to why the answer is 17/18, but you apparently refuse to accept it. There is none so blind as one who will not see. Dave On Aug 8, 9:26 pm, Dipankar Patro dip10c...@gmail.com wrote: 3/5. As the question doesn't ask anything about the sequence. Had the question been Find the probability that all 6 are H then it would have been 17/80. On 9 August 2011 04:07, Dave dave_and_da...@juno.com wrote: @Vinay: What if you tossed 100 consecutive heads? Would that be enough to convince you that you had the double-headed coin? If so, then doesn't tossing 5 consecutive heads give you at least an inkling that you might have it? Wouldn't you then think that there would be a higher probability of getting a head on the sixth toss than there was on the first toss (3/5)? Don's conditional probability answer 17/18 is the right answer. Dave On Aug 8, 5:04 pm, vinay aggarwal vinayiiit2...@gmail.com wrote: answer should be 3/5 think like that tossing 5 times will not help you predict the outcome of sixth toss. Therefore that information is meaningless. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- ___ Please do not print this e-mail until urgent requirement. Go Green!! Save Papers = Save Trees -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en. -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
@Raj. Granted that the first flip has a 3/5 probability of getting a head. But if it produces a tail, would you say that the second flip also has a 3/5 probability of getting a head? Or have you learned something from the tail? If you learn something from a tail, why don't you learn something from a head? Dave On Aug 8, 11:51 pm, raj kumar megamonste...@gmail.com wrote: @all those who gave Should be (4/5 *(1/2)^6) + (1/5 * 1) = 17/80 when it's already given that 5 heads have turned up already then why abut are you adding that probability you all are considering it as finding the probability of finding 6 consecutive heads. since all tosses are independent the answer should be 3/5. the point that 5 heads have turned up already may points that the coin selected is biased in that case pr(6)=1; now the answer depends on the interviewer according to me it should be 3/5 thanks -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
Just to resolve the issue what will be the probability of getting 6 consecutive heads -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
no then it will be 1/2 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
@Raj: After getting 5 consecutive heads, the probability of getting a 6th head is 17/18. Dave On Aug 9, 12:17 am, raj kumar megamonste...@gmail.com wrote: Just to resolve the issue what will be the probability of getting 6 consecutive heads -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
@Raj. Good. So now answer my last question? Dave On Aug 9, 12:21 am, raj kumar megamonste...@gmail.com wrote: no then it will be 1/2 -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
@Dave: I guess 17/18 is correct. Since we have to *calculate the probability of getting a head in the 6th flip given that first 5 flips are a head*. Can you please explain how you got the values of consequent flips when you said this? *In fact, the probability is 3/5 for the first flip. After a head is flipped, the probability of a head is 2/3. After two heads have been flipped, it is 3/4. After 3 heads, it is 5/6. After 4 heads, the probability is 9/10, and after 5 heads, the probability is 17/18.* -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
http://math.arizona.edu/~jwatkins/f-condition.pdf see this link now ithink the answer should be 65/66 bcoz the probability of selectting double headed coin after n heads =2^n/2^n+1 and fair coin is =1/2^n+1 so for 6th head it should be :2^n/2^n+1*1+((1/2^n+1)*1/2) -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
Re: [algogeeks] Re: Probability Puzzle
plz reply am i right or wrong -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
0.6? On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
sry...its wrong On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
[algogeeks] Re: Probability Puzzle
Can anyone explain the approach how to solve this . I think all tosses are independent so it should be 3/5. why is this in- correct On Aug 7, 10:55 pm, saurabh chhabra saurabh131...@gmail.com wrote: sry...its wrong On Aug 7, 10:34 pm, Algo Lover algolear...@gmail.com wrote: A bag contains 5 coins. Four of them are fair and one has heads on both sides. You randomly pulled one coin from the bag and tossed it 5 times, heads turned up all five times. What is the probability that you toss next time, heads turns up. (All this time you don't know you were tossing a fair coin or not). -- You received this message because you are subscribed to the Google Groups Algorithm Geeks group. To post to this group, send email to algogeeks@googlegroups.com. To unsubscribe from this group, send email to algogeeks+unsubscr...@googlegroups.com. For more options, visit this group at http://groups.google.com/group/algogeeks?hl=en.
