[seqfan] Re: Distinct poker hands
Max Alekseyev
maxale at gmail.com
Fri Oct 15 06:12:34 CEST 2010
General formula for n ranks, m suites, and k-card hands is:
the coefficient of x^k in B_m(0!*(1+x)^n, 1!*(1+x^2)^n, ...,
(m-1)!*(1+x^m)^n) / m!,
where B_m() is the m-th complete Bell polynomial.
In particular,
S3(n) = [x^5] ( (1+x)^(3*n) + 3*(1+x^2)^n*(1+x)^n + 2*(1+x^3)^n ) / 6
= C(3*n,5)/6 + (C(n,0)*C(n,5) + C(n,1)*C(n,3) + C(n,2)*C(n,1))/2
= n * (n-1) * (41*n^3 - 89*n^2 + 86*n - 24) / 120
Similar explicit formulae can be obtained for the other Si(n) as well.
Regards,
Max
On Mon, Oct 4, 2010 at 10:46 PM, David Scambler <dscambler at bmm.com> wrote:
> In poker (usually) the suite does not matter except insofar as there are cards with matching suites. For a normal deck of 13 ranks and 4 suites there are C(52, 5) 5-card hands, but when suite symmetries are removed there are only 134459 distinct hands.
>
> Keeping 4 suites and 5-card hands, varying the number n of ranks per suite generates the following number of distinct hands:
>
> S4(n) = 0, 0, 6, 57, 272,901,2376,5362, 10808,19998,34602,56727,88968,134459,196924, 280728,390928,533324,714510,941925,1223904,...
>
> Varying the number of suites and 5-card hands,
>
> 1 suite: S1(n) = A000389 = C(n,5)
> 2 suites: S2(n) = A053132(n+2), n>=3 otherwise zero = C(2n-4,5)
> 3 suites: S3(n) = 0,0,2,27,152,551,1536,3598,7448,14058,24702,40997,64944,98969,145964, 209328,293008,401540,540090,714495,931304,...
> 5 suites: S5(n) = 0, 1,12,78,328,1027,2628,5824,11600,21285,36604,59730, 93336,140647,205492,292356,406432,...
>
> Adding a joker to the normal deck
>
> Sj1(n) = 0,1,15,99,405,1231,3072,6671,13070,23661,40237,65043, 100827,150891,219142,310143,...
>
>
> Adding two indistinguishable jokers to the normal deck:
>
> Sj2(n) = 0,2,21,119,453,1326,3238,6937,13470,24234,41027,66099, 102203,152646,221340,312853,...
>
>
> Cheers
> dave
>
>
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