# How many chips would you risk for a bet

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.

• There is a vast literature on this kind of topic. The most successful practitioner was probably Edward Thorp It is well worth reading some of his stuff. – almagest Sep 24 at 21:23

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.