How many chips would you risk for a bet

Question

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.

Answer 1

I would play all $c$ chips, but if I am allowed I would e.g., bet each chip one-by-one, assuming that the probability that I win this time is independent of the previous outcomes. Then if $c$ is large then whp I would have something close to my expected winnings.

Answer 2

The mathematical analysis of this is usually done on expected value, where you want the highest average outcome over the long term. In that case, if the bet is in your favor you should take it and bet all your money. If there is a series of bets that will maximize the expected value, even though the probability you go broke is very high.

You may believe there is a declining marginal utility to money, so winning an enormous amount doesn't count as much as it would seem and going broke is terrible. If so, you have to define the utility curve, then the result can be different. You might see the Kelly criterion for a result along this line.