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, there's a catch. The price to play each additional round increases linearly by 1 dollar each round. (Playing the first round costs $0, second round costs 1, third costs 2, etc.) What is a fair value to charge for entering this game?
This problem seems really challenging. I know that if the price of the game were to remain constant (say, $1), then we can solve this using the approach outlined here: fair value of a hat-drawing game
However, the increasing price seems to make it much more challenging. Additionally, how would this change if instead of increasing linearly, it was a geometric progression (price goes from 1, to 3, to 9, to 27, etc.)? This question was inspired by @Heropup's response to a previous post!
The expected value of playing round $50$ is $1.5$ because it costs $49$ to play and the expected draw is $50.5$. You should therefore only play if you have a $1$ so far. The expected value of playing round $49$ is $\frac 1{100}\cdot 1.5+\frac {99}{100}\cdot 51-48=2.505$ so you should only play it if you have $2$ or below. The first term comes from getting a $1$ and playing round $50$, the second from getting $2-100$ and stopping, and the $-48$ is the cost of playing. You can keep stepping back, finding the threshold to play the next round. I made a spreadsheet and find the value to be about $77.984$. If your first draw is $75$ or above you should keep it.
The same approach works for any price increase algorithm. If it rises rapidly, you don't need to compute many rounds. You should never play when it costs $50.5$ or more, so in the tripling case you only compute up to the fourth round if you start at $1$.