[travisjungroth]

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.

[foota]

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.

[foota]

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)

[travisjungroth]

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.