0
$\begingroup$

Probability of 5 Card Poker Hand with Only Two Suits:

Pick a card from any suit: 52

Pick a card from a different suit: 39

Pick three more cards from either of the first two suits: $24*23*22$

So, $52*39*24*23*22$ should be the number of ordered 5 card poker hands consisting of only two suits.

To get unordered hands, divide by the number of ways to organize 5 things, that is, 5!.

Why is $\frac{52*39*24*23*22}{5!}$ not the right answer?

I understand one can get the answer by doing the following:

Pick two suits: 6

Pick 5 cards from the two suits: $26 ~C~ 5$

Subtract the sets of the above that have only one of each suit: $2(13 ~ C ~ 5)$

Thus, a correct answer would be 6*(26 C 5 - 2(13 C 5))

Which could also be written $6*((26*25*24*23*22/5!) - 2*(13*12*11*10*9/5!))$

I get why that works, I just need help understanding why my first method does not, or how I could modify it so it would work? Any thoughts?

$\endgroup$
  • $\begingroup$ What you're trying to calculate would more unambiguously be referred to as the probability of a hand having exactly two suits. I would tend to interpret "a hand having only two suits" as "a hand having at most two suits". $\endgroup$ – joriki Feb 4 '13 at 19:24
  • $\begingroup$ You are right, I should have written this as "exactly two suits." $\endgroup$ – jsjwooowooo Feb 5 '13 at 0:10
1
$\begingroup$

You are forcing the first two cards to be different suits and missing SSHHS for example.

$\endgroup$
  • $\begingroup$ I am trying to first find the number of ordered hands, that is, just all the possibilities. And then since we want unordered, divide by the number of ways to order the selected cards. $\endgroup$ – jsjwooowooo Feb 5 '13 at 0:12
  • $\begingroup$ @jsjwooowooo: I see that. But the fact that you only count $39$ for the second card requires that it be of a different suit than the first. You would have to add in $52*12*39*23*22$ for hands where you draw two of the first suit first, then another suit, plus some more terms for the other orders. $\endgroup$ – Ross Millikan Feb 5 '13 at 0:30
  • $\begingroup$ I see what you are saying now! Any advice on how to adapt this method for the problem? $\endgroup$ – jsjwooowooo Feb 5 '13 at 4:38
  • $\begingroup$ @jsjwooowooo: add up the numbers for the possibilities when you get the first card of the second suit. My last comment gives the second card and third card. For the fourth, it is $52*12*11*39*22$ and the fifth $52*12*11*10*39$. Then divide by $5!$. You should be there, but I didn't check. $\endgroup$ – Ross Millikan Feb 5 '13 at 4:46

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.