@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 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
>
> ...
>
> read more »
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