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Comment by OscarCunningham | Hacker News Reader

The dynamic programming and multi-round bets are a distraction there. Since it uses a logarithmic utility function the Kelly criterion is optimal for even a single bet.

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No, the two betting strategies diverge. The Kelly criterion assumes an unlimited steam of future bets, while the dynamic programming approach assumes a fixed number of remaining bets.

The logarithmic utility function makes the Kelly criterion optimal even when there is only one remaining bet. For example say you have $x, the probability of winning is p, and you bet $y. Then you want to maximise

    plog(x+y) + (1-p)log(x-y).

Setting the derivative with respect to y equal to 0 yields

    p/(x+y) - (1-p)/(x-y) = 0.

This rearranges to give

    y = (2p - 1)x,

which is precisely the Kelly criterion.

This is precise the argument at the penultimate time-step in the dynamic programming solution of the multi-round case. The other interesting aspect is that the expected returns are logarithmic, i.e, with y = (2p -1) x

    p log(x+y) + (1-p) log(x-y) = log(x) + C

where C is the Shannon capacity of the binary symmetric channel with cross-over probability p.

By the same argument, the expected wealth after T rounds will be

    log(x) + T C

So, in addition to the optimal strategy, we have also derived the rate of growth of wealth. This is also in tune with the motivation of Kelly's paper where he was showing a relationship between Shannon capacity and optimal gambling (without using a dynamic programming argument)

Correction: I should have said that the Kelly criterion and DP-based approaches diverge when there is a maximum amount of wealth that can be attained (e.g., $250 in TFA), not in scenarios where there are a fixed number of betting rounds remaining.

We can consider the limit T -> infinity to recover the setting with infinite number of bets (but need to look at the rate of growth, otherwise multiple strategies can give infinite returns).

In the result for multi-round bets, if we take the betting horizon T=1, we recover the result for a single bit. But the other way round is not obvious.

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