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I'm working on a card game, which uses a non-standard deck of cards. Since I'm still tweaking the layout of the deck, I've been using variables as follows:

Hand size: $H$
Number of suits: $S$
Number of ranks: $R$
Number of copies of each card: $C$

Thus, the total number of unique cards in the deck is $S*R$, and the absolute total number of cards in the deck is $S*R*C$.

Since there are duplicates for each card, I'm trying to find the number of unique hands of size $H$ that are possible for given $S$, $R$, $C$, and $H$. If I'm remembering correctly, ${S*R*C\choose H}$ would over count the number of unique hands in this case.

How would I handle this calculation with duplicate cards?

EDIT: As an aside, how would calculations of unique number of winning hands be different with duplicate cards? I'm thinking of having general types of winning hands, such as $N$-flushes, $N$-straights, $N$-of-a-kinds, etc (restricted to $1 \le N \le H$, naturally).

Mark LeMoine
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  • Using generating functions, you are looking for the coefficient of $x^H$ in the expansion of $\left(\dfrac{1-x^{C+1}}{1-x}\right)^{SR}$ – Henry Jan 10 '20 at 11:12

1 Answers1

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Number the unique cards $1$ through $RS$. Then you’re asking for the number of solutions of $$x_1+x_2+\ldots+x_{RS}=H\tag{1}$$ in non-negative integers, subject to the restriction that each $x_k\le C$. (In other words, for a given solution to $(1)$, $x_k$ is the number of copies of card $k$ the hand.) Without the restriction this would be a straightforward stars and bars problem, and the answer would be $$\binom{H+RS-1}{RS-1}=\binom{H+RS-1}H\;.$$ Unfortunately, the restriction eliminates many of these naïve solutions, and an inclusion-exclusion calculation is needed to take it into account. The result is that the number of distinct hands is

$$\sum_i(-1)^i\binom{RS}i\binom{H+RS-1-i(C+1)}{RS-1}\;.\tag{2}$$

To the best of my knowledge there is no nice closed form for this.

(See also this essentially identical question. The answer by true blue anil gives an indication of how $(2)$ is arrived at.)

For the question about winning hands, you probably want to take into account the number of ways in which a given winning hand may be made. For an $H$-card flush, for instance, you can have any $H$ cards from any of the $S$ suits, and each suit effectively has $CR$ cards, so there are $S\binom{CR}H$ $H$-card flushes, all equally likely. The fact that some of the cards may be identical doesn’t affect this. Similarly, in counting straights you effectively have $CS$ cards of each rank.

Community
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Brian M. Scott
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  • Thanks Brian, this looks very promising. I'm assuming $i$ in the summation above is from $0$ to $RS$? – Mark LeMoine May 17 '12 at 23:27
  • @Mark: It may stop short of $RS$: you don’t want $H+RS-1-i(C+1)$ to be negative. – Brian M. Scott May 17 '12 at 23:42
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    I have trouble applying the formula. In my example, H=7, RS=4, C=6. This should give me $\binom{7+4-1}{4} - 4 = 206$ combinations, because I have the "standard" 210 combinations and 4 of them are invalid. Substituting in your formula gives me for $i=1$ the term $-1 \binom{4}{1} \binom{7+4-1-1(6+1)}{4-1} = -4$. For $i=2$, I get a negative term in the binomial coefficient and stop there, as suggested in your comment. This gives me a total of -4 combinations. Where did I go wrong? – rumtscho Apr 27 '16 at 22:10