# 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).

• $\frac{\binom{26}{13}\cdot\binom{26}{13}}{\binom{26+26}{13+13}}$ – barak manos Nov 21 '16 at 13:36

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 $\frac {{26 \choose 13}^2}{52 \choose 26}\approx 0.218$

• This is not correct. The chance of getting two reds and two blacks from an eight card deck is $\frac {{4 \choose 2}^2}{8 \choose 4}=\frac {36}{70}$ – Ross Millikan May 8 '17 at 20:57
• @RossMillikan, oh good grief, how embarrassing! Would you like to fix it for me? (I've had too much wine tonight to trust any edit I do right now....) – Barry Cipra May 9 '17 at 1:32
• See if you like what I did. I believe the $4 \choose 2$ is not the positions in the order but the selection of the two red cards out of four available, which corresponds to my denominator of $8 \choose 4$ as the ways to choose four cards. – Ross Millikan May 9 '17 at 2:18
• @RossMillikan, thanks! (But I'm still embarrassed, especially since it looks like the OP is no longer around.) – Barry Cipra May 9 '17 at 2:42
• This came up because it was referenced in another question – Ross Millikan May 9 '17 at 2:44

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:

$$\frac{\binom{26}{13}\cdot\binom{26}{13}}{\binom{52}{26}}.$$

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.

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.

I would love to know how to resolve this discrepancy!

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