I decided to try to figure it out from this angle. What we have is a random walk where, with probability 0.51 we go forward, and with probability 0.49 we go ba

I decided to try to figure it out from this angle. What we have is a random walk where, with probability 0.51 we go forward, and with probability 0.49 we go backward. If we ever reach 10, we quit. I wrote a program to compute the cumulative probability that we stop at 10 over time. #include <stdio.h> #include <math.h> #define GOAL 10 #define MAXN 20000 double p = 0.51; double pr[2][MAXN + 1 + GOAL]; #define PR_(t, x) (pr[(t) % 2][MAXN + (x)]) int main() { int n, i; double totalp = 0; PR_(0, 0) = 1; for (n = 1; n <= MAXN; n++) { PR_(n, -n) = PR_(n-1, -n+1) * (1 - p); for (i = -n+1; i < GOAL; i++) PR_(n, i) = PR_(n-1, i-1) * p + PR_(n-1, i+1) * (1 - p); totalp += PR_(n-1, GOAL-1) * p; printf("n=%d p=%g\n", n, totalp); if (totalp >= 0.99) break; } return 0; } The results: n=5488 p=0.989997 n=5489 p=0.989997 n=5490 p=0.990005 You get to quit much sooner if you can walk away as soon as you have $10!