@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 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 -
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