Expected value of game involving 100-sided die The following question is from a Jane Street interview. 
You are given a 100-sided die. After you roll once, you can choose to either get paid the dollar amount of that roll OR pay one dollar for one more roll. What is the expected value of the game? (There is no limit on the number of rolls.)
P.S. I think the question assumes that we are rational.

Here’s an approach like Uwe’s that avoids solving a recurrence and with some other things explained.
First note that there is no reason to change your strategy at any point. Dice have no memory, so the value of the “rest of the game” is the same as the value of the complete game. So the best strategy is to play until the result is at least $b$ for some fixed $b$.
For a single roll result $X$, the probability of $X\geq b$ is $\dfrac{100-b+1}{100}$, and (intuitively and actually) the expected length of the game is the reciprocal of that probability, or $\dfrac{100}{100-b+1}$ rolls. You paid \$1 per roll, so you paid a total of $\dfrac{100}{100-b+1}$ dollars. You won between \$$b$ and \$$100$, so the expected payout is the average of the integers from $b$ to $100$, or $50+\dfrac b2$, dollars. (The average of a sequence of consecutive integers is always the average of the smallest and largest ones.) 
So the expected value of the game is $50+\dfrac b2-\dfrac{100}{100-b+1}$. This is sort of a tilted branch of a hyperbola, and it should be pretty clear without graphing that it has a unique maximum, so you can compute values one by one and zero in on the maximum of \$$86\frac5{14}$ when $b=87$.
