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I have a standard deck of 52 cards. How do I find the number of hands of 13 cards that contain 4 cards of the same rank? (A,2-10, J,Q,K)

My initial thought was to choose 10 cards from 49, because 9 are normal cards and the 10th can be any random starting card that will determine what the next 3 cards are. The choose 3 cards from 3 as they must all be the same rank. This must be done for each rank, so it is brought to the 13th power.

$${49 \choose 10}{3 \choose 3}^{13} = {49 \choose 10}^{13}$$

I'm very confused with this question, is this on the right track?

Thanks

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  • $\begingroup$ I think it is easier to first count how any ways to get exactly 4 Aces: that is 4 aces, and 1 non-ace. Then, multiple by the total number of ranks. $\endgroup$ – JavaMan Mar 24 at 2:41
  • $\begingroup$ So it would be ((48 choose 9))^13? $\endgroup$ – jacob Mar 24 at 3:38
  • $\begingroup$ I just realized you want 13-card hands, so strike my comment above. $\endgroup$ – JavaMan Mar 24 at 4:15
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Your approach fails because you are assuming that the last three cards match the tenth. There are many more ways for the hand to have four of a kind.

The number of hands that include $4$ aces is ${48 \choose 9}$ because you have that many ways to choose the rest of the cards. It is tempting to multiply this by $13$ for the number of hands that include four of a kind, but you count hands with two four of a kinds twice, so you need the inclusion-exclusion principle.

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  • $\begingroup$ In that case, can I do $$({48 \choose 9}^{13}) / 2$$ to eliminate the double? $\endgroup$ – jacob Mar 24 at 4:15
  • $\begingroup$ No. you need to compute the number of hands that have two four of a kinds. How many ways to choose the two four of a kinds? How many ways to choose the rest of the cards? Then you have the problem of three four of a kinds. That is a small correction $\endgroup$ – Ross Millikan Mar 24 at 4:17
  • $\begingroup$ Can you please give me more information on how to do what you just suggested? Im really confused on this part $\endgroup$ – jacob Mar 24 at 4:41
  • $\begingroup$ You would then think the number of ways to get two four of a kinds would be ${13 \choose 2}{44\choose 5}$ because you choose the ranks for the four of a kind and then choose the other cards. Now you have the problem of the few hands that have three four of a kinds. If you have understood this you can do that part. $\endgroup$ – Ross Millikan Mar 24 at 4:44
  • $\begingroup$ So it would be $$13 * {48\choose 9} - {13\choose 2}{44\choose 5}$$? Or is it supposed to be to the 13th power? $\endgroup$ – jacob Mar 24 at 10:43

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