Kelly Betting for a Repeated Asymmetric Coin Game
Summary
The document considers how much of a $1 million bankroll to wager on a coin game that pays a stated amount on tails and loses the stake on heads. It compares a single play with repeated plays and discusses Kelly sizing, which maximizes expected logarithmic wealth over repeated bets. One response argues that repeated opportunities make growth-based sizing more meaningful, while a single-play objective depends on what the bettor is trying to maximize.
The answers expose an ambiguity in how the payout is described. The proposed one-third fraction uses a win payoff of three times the stake in addition to the original stake convention, while the log-growth calculation models a win as a net gain of two times the stake and a loss of one stake, yielding one quarter. Under the latter payoff interpretation, that calculation is the exact repeated-play log-growth optimum. The document does not settle which payout convention applies, and its recommendation changes with that convention and with the objective for a one-time bet.
Key ideas
- Kelly sizing maximizes expected logarithmic wealth for repeated bets under a specified payoff model.
- The answer says repeated opportunities make growth-based sizing more relevant than a single play.
- The one-third and one-quarter fractions arise from different interpretations of the stated winning payout.
- A single-play choice depends on the objective, such as maximizing expected payoff rather than long-run growth.
- The payout convention and the possibility of ruin must be defined before selecting a stake fraction.
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# How much shall we bet on head/tail with $1m bankroll? # How much shall we bet on head/tail with $1m bankroll? I was asked this question in a trading interview: how much would you bet in a game where you win 300 on tail and loses your 100 on heads? how much will you bet if you can play game once or multiple times with a $1m bankroll? Here is my thoughts and was wondering if this is correct or there was a better way to answer it: On a 1 bet, our expected gain is 1 and standard deviation is 2, thus sharpe is 0.5. Since we need to risk 1 to win 3, we have 3 to 1 odds and thus need to win only 25% of time to break-even. This is a really a high EV game for us and we should bet "a high amount". Using Kelly Criterion, f= (bp - q) / b yields f= (3*0.5 - 0.5) / 3 = 1/3. This is the theoretical bet size to maximize the expected growth rate of your wealth. So we should bet 1/3rd here of our bankroll if we can play the game once. I would bet less if we can play it multiple times since we'd lose a lot of EV in the future should we go bankrupt. I said 1/10th if we can't change bet sizing, does this make sense and how can we quantify the bet sizing here for the multiple game scenario. Theoretically, Kelly says we should bet the same here 1/3rd? There was a similar question here but it doesn't address the multiple game scenario. Thanks! ## Answer by roz (score 3) https://quant.stackexchange.com/a/51371 I think you have it backwards regarding how much to bet if you play once vs. many times. The optimal amount to bet is given by the Kelly criterion as you said. But you should be MORE inclined to bet closer to the optimal Kelly fraction if you get to play many times. The more you play the more your outcome will approach the expected outcome (which is where your edge is). If you get to play only once Kelly has nothing to tell you really. ## Answer by safetyduck (score 1) https://quant.stackexchange.com/a/51416 For continuous play you generally want to maximize the expected log return (time average rate of growth under geometric wealth dynamics) conditional on not going bust. This is basically Kelly setup without the approximation. Assume you can bet fractional amounts to simplify: ``` In [62]: import sympy as s In [63]: J = s.log(1 + a * (3 - 1)) * 0.5 + s.log(1 + a * (0 - 1)) * 0.5 In [64]: dJ = s.diff(J, a) In [65]: a_star = s.solve(dJ, a)[0] In [66]: a_star Out[66]: 0.250000000000000 In [67]: J.replace(a, a_star - 0.1) Out[67]: 0.0499226674848581 In [68]: J.replace(a, a_star + 0.1) Out[68]: 0.0499226674848581 In [69]: J.replace(a, a_star) Out[69]: 0.0588915178281917 ``` So that is 0.25 of your wealth at each bet time. For the single bet case, it is a bit of a weird mandate, but if your mandate is literally just maximize expected payoff, lose up to all the bankroll then, since the payoff is just $1 + 0.5 a$ you should just bet everything.
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