We have actually a typical deck of \$52\$ cards and we attract \$26\$. What"s the probability of drawing exactly \$13\$ black and \$13\$ red cards?

Here"s what I have actually so far. Consider a simplified deck of \$8\$ (with \$4\$ \$B\$"s and \$4\$ \$R\$"s), we have 6 permutations the \$BBRR,RRBB,RBRB,RBBR,BRBR,BRRB\$, each with probability \$p=frac4^23^2(8*7*6*5)\$, thus the overall probability is \$6p = 0.5143\$. I could extend this technique to 52 if ns knew how to find the variety of multi-set permutations, yet I"m no sure how to gain that. I believed it"s \$fracnPrn_B!n_R!\$ yet this provides \$8!/(8-4)!/4!^2 = 2.9166\$ for the 8 map example, i m sorry is incorrect (so i made a mistake).

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We have a complete of \$52\$ cards and as we deserve to choose any kind of \$26\$ of them, the variety of ways room equal to \$inom 5226 \$. Currently there space \$26\$ black color cards and \$26\$ red cards so, the probability of choosing 13 black and also red cards are both equal to \$inom 2613 \$ . Hence the probability is indistinguishable to \$frac inom2613^2inom5226 \$. Hope it helps.

Think around what your sample room and event space are in this situation.

The sample an are can be believed of as the variety of ways to select \$26\$ cards. And your event an are is the variety of ways to choose \$13\$ black and also \$13\$ red cards.

Then our resulting probability is the ratio of the event to the sample space:

\$\$fracinom2613cdotinom2613inom5226.\$\$

The OP need to be commended because that approaching the problem by very first thinking around a smaller sized analog that"s easy to solve explicitly -- and also then because that rejecting an idea because it provides the not correct answer for the smaller sized analog. That is exactly the appropriate thing to carry out when faced with a difficulty that seems facility by its an extremely size.

You have actually to pick two red cards and two black cards. There room \$4 choose 2\$ methods to choose the red cards and also \$4 choose 2\$ methods to select the black ones. There room \$8 choose 4\$ ways to choose the four cards if friend don"t care around the colors, therefore the opportunity you obtain two reds and two blacks is \$frac 4 choose 2^28 choose 4=frac 3670.\$ for \$13\$ reds the end of \$26\$ cards drawn from a standard deck it is \$frac 26 choose 13^252 choose 26approx 0.218\$

I have simulated this basic problem and my outcomes to me show up logical yet do no reflect the answers offered here. This is reflect in another question if everyone is interested.

You deserve to go to: Probability of equal no. Of red/black cards from choice - simulation vs. Answers discrepancy.

I would certainly love to know how to fix this discrepancy!

So then taking the original question come the general case, the probability of selecting equal reds and blacks in a sample of \$Y\$ cards taken indigenous a deck of \$X\$ cards (assuming both \$X\$ and \$Y\$ room even and also \$X > Y\$) is:

\$frac inomX/2Y/2^2inomXY\$

At least that"s mine summation! Cheers - Dave

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