Let X be a random variable indicating the number of heads that have come up. After n flips, E[x] = pn, where p = 0.51 in our case. When X >= n/2 + 5, we're up

Let X be a random variable indicating the number of heads that have come up. After n flips, E[x] = pn, where p = 0.51 in our case. When X >= n/2 + 5, we're up 10 dollars. In other words, we want to find n such that Pr[X < n/2 + 5] < .01 If we express that in terms of the mean, we can apply a Chernoff bound: Pr[X <= E[X] - a] <= Exp[-2a^2/n], for 0 < a < E[X] If E[x] = pn, then a = pn - (n/2 + 5), which for p = 0.51 is a = .01n - 5. So, solving Exp[-2a^2/n] = .01 (with help from Wolfram alpha) gives a quadratic equation with two solutions: n = 348.257 and n = 651.743 The first is too small, a would be negative (n needs to be at least 500). The second is good enough. 652 does the trick.