My card deck has $2n$ cards, $n$ black and $n$ red. I keep drawing cards (without replacing) until I reach the first black card. What is the probability that the next card is again black?
Surprisingly (to me) it seems that for each $n>1$ the answer is $1/2$, at least it is for $n=2,3,4$.
Is there a way of seeing this without going through all the options?
This question is essentially a duplicate, but the claim is so unintuitive (and, I think, unfamiliar) that it may be worth repeating the argument in this context.
Indeed, suppose the cards are numbered $\{b_1, \cdots, b_n\}$ and $\{r_1, \cdots, r_n\}$. Pick any card $X$ and remove it from the deck. Now shuffle the $X-$less deck. There are $(2n-1)!$ ways to do that. Having done so, insert $X$ into the deck behind the first black card (note: this is where we require that $n>1$). In this way, we see that there are $(2n-1)!$ shuffles of the full deck in which $X$ is the card following the first black card. Thus the probability that $X$ is the card following the first black card is $$\frac {(2n-1)!}{(2n)!}=\frac 1{2n}$$ so we see that each card is equally likely to be the card following the first black.
The desired claim follows at once.
