If each pot’s opportunities don’t change over time, then one of them has a higher log return than the other. (Or they are equal.) That pot is the only pot that matters asymptotically
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If each pot has the same quality of opportunity (coins with same bias, but different coins), my intuition is you want to bet more than Kelly because the risk of a string of bad luck putting you into a deep hole is lower.
The more coins you have the more certain outcomes are.
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The issue with that logic is that you can’t rebalance between pots. So when your pot gets in a hole, it is stuck in it
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My intuition agrees with - note that you can bet *more* than Kelly and still have positive growth rate (but less than Kelly)
So maybe we can just bet Kelly fraction + epsilon in each pot, and while the *individual* pots have lower growth rate, their *sum* might not
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i.e. E[log(X)], E[log(Y)] < E[log(Z)] does not imply that E[log((X+Y)/2)] < E[log(Z)]
Let me see if I can turn this into a proof either way
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OK so I think:
1) is wrong given his assumptions
2) if you *also* assume that there aren't two pots which have coins with *exactly* the same growth rate, is right given his assumptions
3) his assumptions are stupid
4) given good assumptions, he's wrong
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If there are two pots with the same growth rate, and you can’t rebalance between them, you think you can beat the growth rate?
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yup! (Assuming those coins are uncorrelated with each other.)
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eh sorry I might be wrong here, I'm 50/50
er ok another question about your assumptions: are you assuming that the pots can look at each other's performance (but not spend $ they don't have), or not?
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I don't think you can do better asymptotically if the coin remains the same over time, but if the edge goes down over time you can definitely beat Kelly asymptotically.
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If the edge goes down over time then is the asymptotic growth rate of each pot 0?
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