If given a standard deck of 52 cards, what is the probability of drawing a black card on the 1st, 2nd, 3rd, 4th … $n^{th}$ draw?
I understand that the first is $26/52$, but the second gets a bit more complicated because there is two scenarios:
a) The first draw is black so the probability is $26/52$ * $25/51$
b) The first draw is non-black, so the probability is $26/52$ * $26/51$
I'm not sure what the formulaic expression of this is and how to consider both of these options in one calculation. I know it uses combinatorics, but I'm lost.
Thanks!
Best Answer
Say you shuffle the cards randomly, and that you are picking from the top. Then you want the probability of the first n cards being all black. There are $n! {26 \choose n}$ possible permutation of black cards being at the top. Now, you may choose any permutation for the remaining cards, so there are $(52-n)!$ choices left. Hence, the probability of getting permutation that gives us our desired situation is
$$\frac{n! {26 \choose n} (52-n)!}{52!}$$