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I would like to calculate the number of combinations for a specific setup to happen in a card game where the whole deck is dealt. So you have a deck of d cards, you deal the deck in h hands (i.e. to h players). The order of cards in each hand does not matter ; the order of the hands does not matter either (we are interested in the game setup, not who gets what).

In my specific case, I'm looking at 5 hands for a deck of 37 cards (2 8-cards hands and 3 7-cards hands).

Zonko
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  • Basically you have the correct answer, the only question I still think is open for discussion is the $h!$ term in the denominator, which assumes an equivalence between all hands, I am not sure if this should apply to your description in which the hands differ in the number of cards, but that depends on the exact definition of the dealing process (if it's always players 1 and 2 that get 8 cards then 5! should be replaced by 2!3!). – Shahar Even-Dar Mandel Sep 07 '15 at 14:10
  • Ok, I think I'll rephrase this question as a self-answered question. – Zonko Sep 07 '15 at 15:26

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the number of combinations for each hand is the binomial coefficient $$\left( \begin{array}{c} \text{nb cards remaining in deck} \\ \text{nb cards in hand} \end{array} \right)$$

As this accounts for an ordering of the hands, to disregard the ordering, We'll later divide the product of all hands combinations by the number of permutations of h hands, which is $h!$.

If we note $H_{i}$ the number of cards in hand $i$, we can see that the product of all hands combinations simplifies itself : $$\dbinom{d}{H_{1}} \times \dbinom{d}{H_{2}} \times ... \times \dbinom{d}{H_{i}}$$ $$=\frac{d!}{H_{1}!\times(d-H_{1})!}\times\frac{(d-H_{1})!}{H_{2}!\times(d-H_{1}-H_{2})!}...$$ $$=\frac{d!}{\prod{H_i !}}$$

And we divide this by $h!$ as explained before : $$\frac{d!}{h! \times \prod{H_i !}}$$

In your case, for 5 hands and a deck of 37 cards, that'd be $$\frac{37!}{5!\times8!^2 \times 7!^3}$$ which is $551,090,558,931,404,263,200$ combinations $(\approx 5\times 10^{20})$.

Zonko
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