Probability of equal no. of red/black cards from selection - simulation vs. answers discrepancy Following reading this thread: "Probability of drawing exactly 13 black & 13 red cards from deck of 52", I created a simple simulation using Excel/VBA to help my son grasp the concept - he's only 7 but wanted to know more...
In this simulation, I chose 2, 4, 6...50 cards from a deck of 52 over 10,000 iterations each and counted how many events of equal red/black cards occurred for each round. The chance of equal red/black cards were then recorded and graphed - results as follows:


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*Excel graph of simulated probabilities
These results are in no way remotely close to the solutions given to the original problem. However, to me the simulated results are logical (2 should be the same as 50, 4 the same as 48 etc.). So where have I/we gone wrong?


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*Numeric comparison of sim results - P(Sim) - and solutions given to the original post - P(Ans).
Unfortunately my high school days were too long ago to address the deterministic answer. However, I am (supposed to be) an expert at simulations and this problem is a very basic one to perform.
I guess it boils down to my disputing the provided answers: ie. for 10 selected cards, the solution of (10C5)^2/(52C10) ~ 0.0004%, nothing like 27.82% from the sim. Logically this should also equal 42 selected cards (10 left over) but the provided answer calculates as (42C21)^2/(52C42) ~  1.83E+13.
So if anyone can resolve the differences or point out to me where I've missed something, that would be wonderful.
Many thanks and regards,
David E

The chance of getting five red and five black is $\frac {{26 \choose 5}^2}{52 \choose 10} \approx 0.2735$, very close to your simulation.  The $26 \choose 5$s are the number of ways to choose five of the $26$ red (black) cards.
