Given a 5 card poker hand from a standard deck, I'm looking to calculate the probability of getting: all 1 suit, 2 different suits, 3 different suits or 4 different suits, with the additional information that the hand contains exactly one queen. I'm familiar with how I would like to set up the question: $\frac{Pr(A\cap B)}{Pr(A)}$, with $A$ representing "getting exactly one Queen" and $B$ representing "containing exactly x suits (with x from 1-4)". The problem I'm running into is finding this $Pr(A\cap B)$! I believe I might want to set it up something like this, (but I may be headed in the wrong direction!). From what I can tell, this is how I might like to set these up:

$Pr(A\cap B)$ given $x$=1: $\binom{4}{1}\binom{12}{4}$: choose 4 cards of the same suit as the Queen, choose the suit.

$Pr(A\cap B)$ given $x$=2: $12*[\binom{12}{3}\binom{12}{1}+\binom{12}{4}+\binom{12}{2}^2+\binom{12}{3}\binom{12}{1}]$: Here are all the combinations I can see happening: choose the Queen, pick 3 more cards of the same suit, then pick a final card of a different suit; pick the Queen, then 4 cards of a different suit; pick the Queen, pick 2 cards of one suit and 2 of the other; pick the Queen, pick another card of this suit, then pick 3 cards of the other suit. Multiply by 12 to account for the different combinations of suits.

$Pr(A\cap B)$ given $x$=3: This one seems it would be most complicated, so I'd like to simply do $1-Pr(Everything Else)$

$Pr(A\cap B)$ given $x$=4: $\binom{4}{1}*(\binom{12}{1}^4+\binom{12}{2}\binom{12}{1}^2)$: Here we pick an Queen, then either four cards (one of each suit) or 4 cards of representing 3 suits, letting the Queen constitute the fourth.

Summing these up (and dividing by $Pr(A)$)

1 suit: .0025

2 suit: .156

3 suit: .686

4 suit: .1554

Looking at these numbers I'm a little disturbed - the 3 suit probability seems awfully high! Am I under-counting at some point (when I calculate probability without taking into account the Queen, the '4 suit' possibility is much greater; perhaps there is an error there)? Thanks!


You missed a factor $3$ for the three possible suits to pick two cards from in the second term for $x=4$; if you take that into account the numbers are in better agreement with intuition.

| cite | improve this answer | |
  • $\begingroup$ just to verify, are you saying it should be $\binom{4}{1}*(\binom{12}{1}^4+\binom{12}{3}\binom{12}{1}^2)$ ? This gives a .269 probability, which seems pretty reasonable; does my logic seem reasonable to you in general? $\endgroup$ – jeliot Oct 31 '12 at 2:18
  • $\begingroup$ @jeliot: No, I literally meant a factor $3$, just like I wrote it. Your binomial coefficients were correct, $2$ cards from $12$ once and $1$ card from $12$ twice; you were just missing the fact that this can be done for each of the three choices of a suit to pick the $2$ cards from. Yes, the rest of it all seems right to me. $\endgroup$ – joriki Oct 31 '12 at 2:21
  • $\begingroup$ Sorry, I'm not sure why I'm having so much trouble understanding! Are you saying it should be: $\binom{4}{1}*(\binom{12}{1}^4+3*\binom{12}{2}\binom{12}{1}^2)$ $\endgroup$ – jeliot Oct 31 '12 at 2:33
  • $\begingroup$ @jeliot: Yes. Sorry if I wasn't clear enough -- you seemed to be so good at this in general that I thought it would be enough just to point out the mistake. (By the way, asterisks are sort of old-school and computer-keyboardy; $\TeX$ has \cdot.) $\endgroup$ – joriki Oct 31 '12 at 2:36
  • $\begingroup$ Yeah, I think its getting a bit late over here :) Thanks again for the help! $\endgroup$ – jeliot Oct 31 '12 at 3:01

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.