讓你任選５０個不重複的數字，分別寫在５０張紙條上，折好放在盆子了。

我則從盆子裡任意拿出一紙條，打開。這時我可以有兩個選擇：１。停下，２。繼續從盆子裡拿紙條。這樣重複，直到我停下（拿出最後一張紙條後，就算停下）。

現在我和你打賭，如我停下時，我手中紙條上的數字是５０個數字中最大的，我就算贏。否則就算輸。

我贏的機率有多大？得到這個機率的方法是什麼？
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.