4 ms·

I think he was asking for a numeric answer.

• tome 17y ago
  If he wants one he can calculate it using the formula I've provided. A normal approximation to the binomial distribution might help too.
  • deleted 17y ago
    [deleted]
    • madcaptenor 17y ago
      The amount you win on a single flip has mean 0.02 and variance 0.9996. So the amount won after n flips, for large n, is approximately normally distributed with mean 0.02n and 0.9996n. (This uses the Central Limit theorem.) The question, then, is to find n large enough that this has probability 0.99 of being at least 10. A normally distributed random variable has probability 0.99 of being at least 2.33 standard deviations below its mean, so we want to find n such that 0.02n - 2.33*sqrt(0.9996n) = 10 and solving for n (this is a quadratic equation in sqrt(n)) gives 14507. (Your 10576 is what I get from my method if I look for 97.5% certainty.)
      • tome 17y ago
        Yes, we used the wrong CDF. 14507 is what we get as well with the correct one.