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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).

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    $\begingroup$ $\frac{\binom{26}{13}\cdot\binom{26}{13}}{\binom{26+26}{13+13}}$ $\endgroup$ – barak manos Nov 21 '16 at 13:36
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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 $\frac {{26 \choose 13}^2}{52 \choose 26}\approx 0.218$

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    $\begingroup$ This is not correct. The chance of getting two reds and two blacks from an eight card deck is $\frac {{4 \choose 2}^2}{8 \choose 4}=\frac {36}{70}$ $\endgroup$ – Ross Millikan May 8 '17 at 20:57
  • $\begingroup$ @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....) $\endgroup$ – Barry Cipra May 9 '17 at 1:32
  • $\begingroup$ 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. $\endgroup$ – Ross Millikan May 9 '17 at 2:18
  • $\begingroup$ @RossMillikan, thanks! (But I'm still embarrassed, especially since it looks like the OP is no longer around.) $\endgroup$ – Barry Cipra May 9 '17 at 2:42
  • $\begingroup$ This came up because it was referenced in another question $\endgroup$ – Ross Millikan May 9 '17 at 2:44
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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:

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

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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.

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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!

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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

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