Fair Value Hat Draw Game At each round, draw a number $1-100$ out of a hat (and replace the number after you draw). You can play as many rounds as you want, and the last number you draw is the number of dollars you win, but each round costs $1/x$ (where $x$ is the amount drawn- ex. draw $2$ pay $50$ cents). What is a fair value to charge for entering this game?
 A: At each round, you start the game afresh (having lost some amount). So any sensible strategy must be independent of the previous moves. In this case, it will be of the form "If I draw at most $k$, then continue; otherwise stop". Now fix $k$, and suppose the expectation of the game is $E$. I assume, contrary to Hagen von Eitzen, that you also have to pay for your last draw. Then we get:
$$E = \frac1{100}\sum_{i \le k}\left(E - \frac1{i}\right) + \frac1{100}\sum_{i > k}\left(i - \frac1{i}\right)$$
Rearranging,
$$E=\frac{101}2 + \frac12k - \frac1{100-k}\sigma$$
where $\sigma = \sum_1^{100}\frac1i = 5.1873775\ldots$  
We want to find the $k$ that maximises this expectation. Solving for real $k$ gives $k = 100 - \sqrt{2\sigma} = 96.779\ldots$ So the optimal integer solution is $k = 96$ or $97$. Checking, we find that $k=97$ is optimal, with
$$E = 99 - \frac{\sigma}3 = 97.27087\ldots$$
So you should keep drawing until you see 98 or higher. And a fair price for this game is \$97.28 (because the house can't be expected to run at a loss).
