Kelly Can't Fail

Very cool to see no variance in the outcome. But that also makes it feel like there should be a strategy with better expected return due to the unique problem structure. Do we know if the Kelly strategy is optimal here?

The book claims it is optimal for a set of strategies they called "sensible." I didn't think the argument flowed as well as the zero variance part of the proof, so I didn't work it in. I think the source also hinted at a game-theory proof as they called the sub-strategies in the portfolio "pure strategies."

  Do we know if the Kelly strategy is optimal here?

What do you mean by optimal? Do you mean you're willing to risk going bankrupt, if it means a higher expected value?

It is optimal for expected returns yes.

Surely there's some space between risking to go bankrupt and risking of getting less than 9.08 return guaranteed by Kelly strategy.

If you are willing to take some risk in exchange for possibility of higher payout just bet a bit more then Kelly recommends. That's your "optimal" strategy for the amount of risk you are willing to take. I imagine it's expected return is the same as Kelly and calculating it's variance is left as the exercise for the reader.

I have a feeling it’s the highest EV. I tried a strategy of flipping all the cards until there’s only one color left and then betting it all every time. Ran a million trials and got 9.08.

I was thinking these are very different strategies, but they’re not exactly. The Kelly strategy does the same thing when there’s only one color left. The difference is this strategy does nothing before that point.

Still, they feel like limit cases. Betting it all with only one color left is the only right move, so it’s what you do before that. Nothing and Kelly seem like the only good strategies.

The Kelly criterion is the strategy with better return due to the uniquely problem structure.

Ah, but these aren't the same. The Kelly strategy has zero variance, whereas this strategy likely has very high variance.

It would be interesting to do the math and show why they're equal. It seems like you should be able to make the same sort of portfolio probability argument.

To start, your minimum return is 2x, and depending on how many cards of a single color are left at the end, you get a return of 2^N. You could take the summation of those N card returns, times the probability of each, and that must come out to 9.08 on average.

I guess the number of possible arrangements of cards with N of one color remaining is... The number of permutations of N times 2 times the number of permutations of 52 minus N times 26 choose N?

Ah, yes this works, you can see it here: https://www.wolframalpha.com/input?i=%28summation+of+N%21+*+....

That is: (summation of N! * (52 - N)!* (26 choose N) * 2^N/52! from N=0 to 26 (for some reason the * 2 for different suits was over counting, so I removed it. Not sure why? Also it seems like it should be from 1 to 26, but that also doesn't give the right answer, so something is whack)

  I imagine it's expected return is the same as Kelly

Given two options with the same expected return, most people would prefer the lower variance.

Accepting higher variance with no increase in expected return has a name: gambling.

In this game, all strategies have the same expected value, so long as they follow the rule 'if the remaining deck is all the same colour, then you should bet everything you have on that colour'.

Of course they're not the same. They have the same EV and the strategies do the same thing in a condition that always happens: there's only one color left. The variance is wildly different.