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If the player is immediately to the left of the dealer, I think the answer is:

52 x (52-p) x (52-2p) x ... x (52-(n-1)p),

and if the player is the kth player, I think the answer is:

(52-k+1) x (52-p-k+1) x (52-2p-k+1) x ... x (52-(n-1)p-k+1).

Am I right?

I'm assuming that the pack has 52 cards, and that dealing starts immediately to the left of the dealer and goes round clockwise.

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    There are various issues here, such as whether the order in which the cards arrive matter. If not, you might want to divide your results by $1 \times 2 \times 3 \times \cdots \times p$2011-03-27
  • 0
    It's the same for all players (assuming that $n$ is a multiple of $k$).2011-03-27

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