There are $6$ cups and under one of the cups, there are $500$ dollars. In each round, by paying a fixed cost $c$, you can either open a cup or stop playing. What's the maximum amount you are willing to pay to play this game?
In the first round, I pay c dollars and the chances of opening the correct cup is $1/6$ and the chances of opening the wrong one is $5/6$. So my expected payoff would be $1/6*500 + 5/6*0 - c$. However, if I want to continue, for the second round, my expected payoff would be $1/6*500 + 5/6(1/5*500 + 4/5*0) - 2c$ and so on. Assuming I want to continue until the end the final expression would be $$E = \frac{1}{6}\times500 + \frac{5}{6}(\frac{1}{5}\times500 + \frac{4}{5}(\frac{1}{4}\times500 + \frac{3}{4}(\frac{1}{3}\times500 + \frac{2}{3}(\frac{1}{2}\times500 +
\frac{1}{2}\times0)))) - 5c$$.
So setting $E=0$ and finding $c$ would be the fair price for this game.
Does this make sense? What would be the correct way of solving this problem?
What is the cost structure?
As the question formulated now, I'll pay at most 500, open all the cups and get my 500 back. Probably there should be some costs for opening each cup or something similar. Suppose that the cost of opening each cup is $c$ and that you are risk-neutral:
The number of cups to lift until getting the 500 is uniform $\{1,\ldots,6\}$ so the expected number of cups to lift is $3.5$ and the expected cost is $3.5c$. The maximum $c$ is the one that makes me indifferent between playing or not playing, so $3.5c=500$ or $c=142.8$.