TITLE: How much should I pay for a chance to win 100$?
QUESTION [6 upvotes]: There are $4$ closed doors, with $100\$$ behind one of them. You can pay $X$ to open a door. If the money is there, you can keep it. If not you can pay another $X$ to open the next door, and so on. What is the most I can pay and still win on average?
I think that it's $25\$$. Since the money can be behind any door, you have as much of a chance of getting it on the first door as any other door. Since there are $4$ doors, and equal chance of being behind any door, $100 / 4$ = $25$.

REPLY [3 votes]: If you decide to stop when you find the $\$100$ or on the $n$th door, then your return is
$$\sum_{i=1}^n\frac14(100-iX)+\left(1-\frac n4\right)(-nX)=\frac n8(200+X(n-9)).$$
Let this be greater than $0$ to find that $$X<\frac{200}{9-n}$$
which is maximised when $n=4$ so $X<\$40$.