# Conditional probability within a poker hand

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.
• 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? – jeliot Oct 31 '12 at 2:18
• @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. – joriki Oct 31 '12 at 2:21
• Sorry, I'm not sure why I'm having so much trouble understanding! Are you saying it should be: $\binom{4}{1}*(\binom{12}{1}^4+3*\binom{12}{2}\binom{12}{1}^2)$ – jeliot Oct 31 '12 at 2:33
• @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.) – joriki Oct 31 '12 at 2:36