An Example from my Probability Class I Need Help w/

[irchans]

I agree with emp1346's answer. I thought it might be interesting to compare this question to a similar coin toss probability question.


The probability of getting n red cards when drawing 2 n cards from a size 4 n deck is

C(2*n, n)^2/C(4*n, 2*n) ~= Sqrt(2/Pi/n).

The probability of getting n heads when flipping a coin 2 n times is

C(2*n, n)/2^n ~= Sqrt(1/Pi/n).


So the probability of getting n heads is about the probability of getting n red cards divided by Sqrt(2). Hmmm.

[AaronBrown]

That makes some sense. Let's say you put the red cards in one pile and the black cards in another. You flip a coin 26 times, if it's heads you take a card from the red pile, if it's tails you take a card from the black pile. In that case, the chance of having 13 red cards is the same as flipping 13 heads in 26 attempts.

But when you shuffle, the probability of getting a red card is the number of remaining red cards divided by the total number of remaining cards. That makes it more likely to get exactly 13 because if you already have a lot of red cards the next card is more likely to be black; and if you don't have many red cards, the next card is likely to be red.

It's interesting that the difference approaches the square root of two. I'm sure there's a good reason for it, but I don't see it immediately.