4 ms·

Solving this numerically, I get 14414. I used the following line of Mathematica code: For[n=14400, Sum[PDF[BinomialDistribution[n,.51],k],{k,n/2+5,n}] < .99,

• madcaptenor 17y ago
  I'm glad to see an answer from this approach. (Maple wasn't cooperating for me.) And notice that it compares well with the 14507 from the normal approximation to the binomial.
• tspiteri 17y ago
  I did it in C, my code was a bit longer. (And it has the advantage that you do not need an initial estimate for the problem to be solved in a very short time.) #include <math.h> #include <stdio.h> double p(int throws, double pwin, int req_wins) { int win, i; double log_dp = log(pwin) * throws; double log_loss_win = log((1 - pwin) / pwin); double ret = 0; for (win = throws, i = 0; win >= req_wins; win -= 2, ++i) { ret += exp(log_dp); log_dp += log_loss_win + log((throws - i) / (i + 1.0)); } return ret; } int main() { double pwin = 0.51; int req_wins = 10; double conf = 0.99; int min, max = 1; while (p(max, pwin, req_wins) < conf) max *= 2; min = max / 2; while (min < max) { int mid = (min + max) / 2; if (p(mid, pwin, req_wins) < conf) min = mid + 1; else max = mid; } printf("For %d throws, P(win $10 or more) = %f\n" "For %d throws, P(win $10 or more) = %f\n", min - 1, p(min - 1, pwin, req_wins), min, p(min, pwin, req_wins)); } The output from this program: For 14413 throws, P(win $10 or more) = 0.989777 For 14414 throws, P(win $10 or more) = 0.990004