让你任选５０个不重复的数字，分别写在５０张纸条上，折好放在盆子了。

我则从盆子里任意拿出一纸条，打开。这时我可以有两个选择：１。停下，２。继续从盆子里拿纸条。这样重复，直到我停下（拿出最后一张纸条后，就算停下）。

现在我和你打赌，如我停下时，我手中纸条上的数字是５０个数字中最大的，我就算赢。否则就算输。

我赢的机率有多大？得到这个机率的方法是什么？
解：

37.43%. Stop if and only if the last card is largest and at least 19 cards have been taken out.

We should stop only if the last number is the biggest of all numbers taken out. At this point, all numbers before the last number does not matter, because we only care about the maximum number. (i.e., our decision should be the same to sequence {2, 3, 1, 5} and {3, 1, 2 5}). So the only question is how many numbers we need to see to decide to stop. So the optimal strategy must be: stop if and only if the last number is the biggest so far and the total number of cards we have seen is at least n, where the optimal n is to be decided.

This means we could stop at n-th card, n+1-th card, …, or N-th card. The probabilities of these events are: 1/n, (n-1)/n/(n+1), …, (n-1)/(N-1)/N respectively. (If we stop at the n-th card, then the n-th card is the largest among the first n cards, so prob = 1/ n. If we stop at the n+1-th card, then that card is the largest of the first n+1 cards, so prob = prob(we didn’t stop at the n-th card) * 1/ (n+1) = (n-1)/n/(n+1). We can find other probabilities in a similar way.) The probability of winning in these events are n/N, n+1/N, …, N/N respectively. So the total winning probability is 1/N + (n-1)/N * (1/n + 1/(n+1) + … + 1/(N-1)), for n = 1, …, 50. Therefore we need to find out at which n this probability is maximized.

I found when n = 19, the probability is maximized at 37.43%. Therefore, the best strategy is stopping if and only if the last card is largest and at least 19 cards have been taken out.