I'm trying to understand why my method doesn't work to answer the question: What is the probability of having 3 cards of the same suit and $2$ cards of the same suit (but a different suit than the first three) in a $5$ card hand from a standard $52$ card deck?

The method that seems to work uses combinations. You do: $$ {4 \choose 1}\cdot{13 \choose 3}\cdot{3 \choose 1}\cdot{13 \choose 2} = 267,696 $$ To find the number of successful possibilities and then $$ {52 \choose 5} = 2,598,960 $$ To find the number of total possibilities. So the answer is: $$ \frac{267,696}{2,598,960} $$

I'm wondering why figuring out the individual card chances and then multiplying them doesn't work. My method goes like this: $$ \frac{12}{51}\cdot\frac{11}{50}\cdot\frac{39}{49}\cdot\frac{12}{48} $$

After you've picked your first card you have a $\frac{12}{51}$ chance of getting a matching second card, then you've a $\frac{11}{50}$ chance of getting your third card to match.

For the other two cards you've a $\frac{39}{49}$ chance, because you still have three suits left. You need the next one to match, so now you have a $\frac{12}{48}$ chance.

Shouldn't multiplying all of those individual probabilities yield the same answer as the first method? I realize that there are different ways to order the individual probabilities, but when they're all multiplying, shouldn't it all work out to be the same?

I've noticed that multiplying the answer to my method by $10$ makes it equal to the first method's answer. Is there a reason that works?


1 Answer 1


Your calculation is the chance that you will draw three matching cards first, followed by a pair. But when you are looking for a full house, the second card doesn't have to match the first, so you have missed some possibilities. The factor $10$ is because there are ${5 \choose 3}=10$ different ways to choose the positions of the five draws that will be the three of a kind.

  • $\begingroup$ Thanks for the answer! I think I'm starting to understand why it would work like that. But why is it 5C3 instead of the 5P5? I'd always thought that you multiplied by the permutations of all the individual ways to choose. $\endgroup$
    – user74928
    Apr 29, 2013 at 4:10
  • 1
    $\begingroup$ @user74928: because the three of a kind are interchangeable, so you are choosing which three positions in the draw order they occupy. Your calculation already allows for changing the order of the three matching cards and of the two cards in the pair. $\endgroup$ Apr 29, 2013 at 13:34

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