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Imagine we started with $100 and were betting on a fair coin with fair odds. There's no edge so Kelly says to bet $0, and hence the Kelly strategy stays at $100
by OscarCunningham 8y ago
Imagine we started with $100 and were betting on a fair coin with fair odds. There's no edge so Kelly says to bet $0, and hence the Kelly strategy stays at $100 forever. You can give yourself a very high chance of beating the Kelly strategy if you use a Martingale strategy. Bet $0.01. If you win then you have $100.01 and you should not bet again. If you lose then bet $0.02 next time, and then $0.04 and then $0.08 doubling each time until you win. When you win you will go up to $100.01, which beats the Kelly strategy. So Martingale beats Kelly unless you go bankrupt by losing too many times in a row before getting a win, which only happens with a very small probability. So there are strategies that have a high probability of beating the Kelly criterion.
For simplicity I gave the above example in the degenerate case where the edge is zero. But I think the analogous strategy works in all cases. If you aim for just a penny over the Kelly strategy then you have a high probability of success.
- imh 8y agoFair point. And while it's too late to edit my post, I think I found the flaw in the math. Just because X converges in probability to x, f(X) doesn't necessarily converge in probability to f(x). If that were so, then logarithmic utility would be sound, but it isn't.