# How much should I pay for a chance to win 100$? 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, $$100 / 4$$ = $$25$$. • What do you mean "how much should you pay"? I guess you are not the one who chooses$X$, otherwise why not choose$X=0$? – Mark Apr 14, 2019 at 21:32 • yes good catch, "How much should I be willing to pay?" is more appropriate. Apr 14, 2019 at 21:34 • What are you after? Do you want to be guaranteed not to lose, or do you just not want to lose on average over many games? Either way, I would be willing to go above$\$25$. Apr 14, 2019 at 21:37
• On average over many games. Apr 14, 2019 at 21:40
• Hint: Try to compute the expected value of the cost you pay. What is the chance that you get it on the first try? What's the chance that you get it on the second try? Apr 14, 2019 at 21:44

You can find the money behind door 1,2,3 or 4 with equal probability and if the price is $$X$$ to open a door, your winnings are $$(100-X)$$, $$(100-2X)$$, $$(100-3X)$$ or $$(100-4X)$$ depending what door you find the money behind.
Therefore your expected return is $$\frac14\bigl((100-X)+(100-2X)+(100-3X)+(100-4X)\bigr) = 100 -2.5X$$
For you to expect to make a profit then $$X$$ must be less than $$\40$$. So you would make a profit over many games if $$X$$ is less than $$\40$$.
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 $$X<\40$$.