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!
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
$$\frac{n! {26 \choose n} (52-n)!}{52!}$$