Say you have a total of c chips, and I present to you a game with a probability $p$ of winning, and payout $k:1$ (where $p > 1/(k+1)$, so this is a +EV bet). How many chips would you risk to take this bet?
For a more concrete example, say you have 100 chips, and I present to you a game with probability of winning, $p=0.66$. If you win, I will pay 1:1 odds, how many chips would you bet?
I kind of synthesized this question from poker theory. I'm curious to know if there's an optimal solution. My thinking is that this just measures your risk adverseness and there's no mathematical way of finding a bound on c. Not entirely sure.