0

If it is assumed that all 5) poker hands are equally likely, what is the probability of being dealt :

two pairs? (This occurs when the cards have denominations a, a, b, b, c, where a, b, and c are all distinct.)

my answer was (13c1)(4c2)(12c1)(4c2)(11c1)(4c1) / (52c5)

however this was wrong and supposedly the answer was (13c1)(4c2)(4c2)(11c1)(4c1) / (52c5)

I cannot intuitively understand why my solution was incorrect because I feel like they are both explaining the same thing however yield different solutions.

big_d
  • 1
  • 1
    your answer appears to be correct. the "supposed" answer doesn't try to pick the rank of the second pair. – Dan Uznanski Oct 13 '24 at 18:40
  • 1
    Neither of those is correct. Yours is off by a factor of $2$ since you count $AABBC$ and $BBAAC$ as if they were distinct. The second is wrong since the first term ought to be $\binom {13}{2}$ not $\binom {13}1$ as you are choosing two ranks, not one. (Note: I suspect that you simply mistyped the second one, or else your source had a typo). – lulu Oct 13 '24 at 18:43
  • Note that the duplicate coincides with your question, except that it has the correct factor of $\binom {13}2$ in the official method. – lulu Oct 13 '24 at 18:48

0 Answers0