This is an interesting problem from one of my friends’ oral exam.
A and B play a game of drawing cards without replacement from 4 cards, of which 2 are red and 2 are black. They draw one card and bet a dollar each time. If the drawn card is red, then A wins a dollar from B. Otherwise, A loses a dollar to B. Suppose that now A is allowed to stop the game whenever he wants. How much should A pay B before the game starts to make it a fair game?
Solution (CLICK ME):
Note that A would keep flipping cards until he has seen two red cards. Next, we list all the possible outcomes (R: red, B: black) and corresponding probabilities: RR(BB), RBR(B), BRR(B), BRBR, BBRR, RBBR, each with probability $\frac{1}{6}$. (Why?) Now, we can compute the expected outcome of player A, $EX$: $$ \mbox{E}X = \frac{1}{6}\times(2+1+1+0+0) = \frac{2}{3}, $$ which is the desired quantity.