In the martingale doubling system the player doubles his bet each time he loses. Suppose that you are playing roulette in a fair casino where there are no 0's, and you bet on red each time. You then win with probability 1/2 each time. Assume that you enter the casino with 100 dollars, start with a 1-dollar bet and employ the martingale system. You stop as soon as you have won one bet, or in the unlikely event that black turns up six times in a row so that you are down 63 dollars and cannot make the required 64-dollar bet. Find your expected winnings under this system of play.
If a random variable $X$ is defined for as the amount won / lost, $X = -63$ if the outcome is six blacks in a row and $X = 1$ otherwise. $P(X= 1) = 63/64$ and $P(X = -63) = 1/64$, so I think the expected value is $1*63/64-63*1/64 = 0$, so the game is fair. Is this correct?