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).
 A: Think about what your sample space and event space are in this situation.
The sample space can be thought of as the number of ways to choose $26$ cards. And your event space is the number of ways to choose $13$ black and $13$ red cards.
Then our resulting probability is the ratio of the event to the sample space:
$$\frac{\binom{26}{13}\cdot\binom{26}{13}}{\binom{52}{26}}.$$
A: We have a total of $52$ cards and as we can choose any $26$ of them, the number of ways are equal to $\binom {52}{26} $. Now there are $26$ black cards and $26$ red cards so, the probability of choosing 13 black and red cards are both equal to $\binom {26}{13} $ . Hence the probability is equivalent to $\frac {\binom{26}{13}^{2}}{\binom{52}{26}} $. Hope it helps.
A: 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 $\frac {{26 \choose 13}^2}{52 \choose 26}\approx 0.218$
A: I have simulated this simple problem and my results to me appear logical but do not reflect the answers given here. This is reflected in another question if anyone is interested.
You can go to: Probability of equal no. of red/black cards from selection - simulation vs. answers discrepancy.
I would love to know how to resolve this discrepancy!
A: 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
