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If I draw five cards from a deck of 51 cards, how many possible arrangements possible if i want 2 cards with the same value ie 2 jacks?

My method in solving this was to do the following: (52)(3)(50C3)

Because the first card doesn't matter and the next one matters and the rest of the three don't matter at all.

I have a very strong feeling that I was overcounting and my method was flawed

i was wondering if I was making any mistakes?

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    $\begingroup$ Is it exactly two cards with the same value? At least two cards with the same value? Is the value fixed beforehand (say, you're only interested in hands which contain exactly/at least tow jacks), or is any repeated value okay? $\endgroup$ – Fimpellizieri May 24 '17 at 4:47
  • $\begingroup$ I agree with @Fimpellizieri you have to specify more clearly. $\endgroup$ – Iti Shree May 24 '17 at 4:49
  • $\begingroup$ Did you hide an ace from that deck? :-P $\endgroup$ – lesath82 May 24 '17 at 6:28
  • $\begingroup$ Did you mean from a deck of $\color{red}{52}$ cards? $\endgroup$ – N. F. Taussig May 24 '17 at 8:41
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Do you want exactly one pair, or just any hand with two cards equal (so you allow two pairs or a three-of-a-kind etc.)

Of the $\binom{52}5$ poker hands, there are $$13\times\binom{12}3\times 6\times 4^3$$ which are one pair hands. This is $13$ possible denominations for the pair, $\binom{12}3$ for the singletons, $6=\binom42$ choices for the suits of the pair, and $4^3$ choices for the suits of the singletons.

The number of hands with at least one repeated value is $$\binom{52}5-\binom{13}5 4^5.$$

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