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).

The OP should be commended for approaching the problem by first thinking about a smaller analog that's easy to solve explicitly -- and then for rejecting an idea because it gives the wrong answer for the smaller analog.  That is exactly the right thing to do when faced with a problem that seems complicated by its very size.
You have to choose two red cards and two black cards.  There are $4 \choose 2$ ways to choose the red cards and $4 \choose 2$ ways to choose the black ones.  There are $8 \choose 4$ ways to choose the four cards if you don't care about the colors, so the chance you get two reds and two blacks is $\frac {{4 \choose 2}^2}{8 \choose 4}=\frac {36}{70}.$  For $13$ reds out of $26$ cards drawn from a standard deck it is $\frac {{26 \choose 13}^2}{52 \choose 26}\approx 0.218$
