I saw this:
Probability of getting a full house
and the top answer makes sense to me.
However, why can't I also do this?
Pick a suite. $\binom{4}{1}$.
Take 3 cards from that suite. $\binom{13}{3}$.
Pick a different suite. $\binom{3}{1}$.
Take 2 cards from that suite. $\binom{13}{2}$.
So desired hands are $\binom{4}{1}\binom{13}{3}\binom{3}{1}\binom{13}{2}$.
This is much larger than their answer.

Their answer is $\binom{13}{1}\binom{4}3\binom{12}1\binom{4}2$.

  • 1
    $\begingroup$ But that will not satisfy the condition for a full house $\endgroup$ – AnotherJohnDoe Oct 12 '17 at 12:15
  • 2
    $\begingroup$ I think you're confused as to what a full house is; simply google "full house cards" and see, Three matching cards of one rank and two matching cards of another rank. You're making that Three matching cards of one suite and two matching cards of another suite, which is not the same thing. $\endgroup$ – vrugtehagel Oct 12 '17 at 12:16

It's not cards from same suite like heart or spade, it's same type, like $2, 7,$ J, K, etc. so $13$ types.


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