Probability of drawing exactly 13 black & 13 red cards from deck of 52 We have a normal deck of $52$ cards and we draw $26$. What's the probability of drawing exactly $13$ black and $13$ red cards?
Here's what I have so far. Consider a simplified deck of $8$ (with $4$ $B$'s and $4$ $R$'s), we have 6 permutations of $BBRR,RRBB,RBRB,RBBR,BRBR,BRRB$, each with probability $p=\frac{4^23^2}{(8*7*6*5)}$, therefore the overall probability is $6p = 0.5143$. I could extend this method to 52 if I knew how to find the number of multi-set permutations, but I'm not sure how to get that. I thought it's $\frac{nPr}{n_B!n_R!}$ but this gives $8!/(8-4)!/4!^2 = 2.9166$ for the 8 card example, which is incorrect (so I made a mistake).

So then taking the original question to the general case, the probability of selecting equal reds and blacks in a sample of $Y$ cards taken from a deck of $X$ cards (assuming both $X$ and $Y$ are even and $X > Y$) is:
$\frac {\binom{X/2}{Y/2}^{2}}{\binom{X}{Y}}$
At least that's my summation! Cheers - Dave
