Suppose we're in the middle of the flipping. Let X be the number of heads since the last tail, and T(X) be the probability of Andy losing given X. We know:T(4) = T(0)/2, with four heads either the next flip will be a head, and Andy wins, or a tail, in which case he has T(0)T(3) = [T(4) + T(0)]/2, if the next flip is heads, Andy has T(4), if it's tails he has T(0)T(2) = [T(3) + T(0)]/2T(1) = [T(2) + T(0)]/2T(0) = [T(1) + 1]/2, if the next flip is heads, Andy has T(1), if tails he loses for sure.That's easily solved to:T(4) = 8/17T(3) = 12/17T(2) = 14/17T(1) = 15/17T(0) = 16/17You can do this by hand easily enough. I set it up as a matrix in Excel and inverted it.The first flip can be either heads, in which case we get T(1) or tails, in which case we get T(0). So Andy's probability of losing is (15/17 + 16/17)/2 = 31/34. Since he loses $3 31 times out of 34, that's a total of -$93, and wins $24 3 times out of 34, that's a total of $72, he loses $21 in every 34 games.
Last edited by Aaron
on June 10th, 2005, 10:00 pm, edited 1 time in total.