I've been wanting to do something like this for a while, so it's good to see it properly worked out here.
If you wanted to expand this you could look at games which weren't symmetrical in the players. So you'd have eight variables, W, X, Y and Z, and w, x, y and z. But you'd only have to look at the possible orderings within each set of four, since it's not necessarily valid to compare utilities between people. You'd also be able to reduce the number of games by using the swap-the-players symmetry.
Right. But also we would want to use a prior that favoured biases which were near fair, since we know that Wolf at least thought they were a normal pair of dice.
Suppose I'm trying to infer probabilities about some set of events by looking at betting markets. My idea was to visualise the possible probability assignments as a high-dimensional space, and then for each bet being offered remove the part of that space for which the bet has positive expected value. The region remaining after doing this for all bets on offer should contain the probability assignment representing the "market's beliefs".
My question is about the situation where there is no remaining region. In this situation for every probability assignment there's some bet with a positive expectation. Is it a theorem that there is always an arbitrage in this case? In other words, can one switch the quantifiers from "for all probability assignments there exists a positive expectation bet" to "there exists a bet such that for all probability assignments the bet has positive expectation"?
I believe you missed one of the rules of Gurkenglas' game, which was that there are at most 100 rounds. (Although it's possible I misunderstood what they were trying to say.)
If you assume that play continues until one of the players is bankrupt then in fact there are lots of winning strategies. In particular betting any constant proportion less than 38.9%. The Kelly criterion isn't unique among them.
My program doesn't assume anything about the strategy. It just works backwards from the last round and calculates the optimal bet and expected value for each possible amount of money you could have, on the basis of the expected values in the next round which it has already calculated. (Assuming each bet is a whole number of cents.)
If you wager one buck at a time, you win almost certainly.
But that isn't the Kelly criterion! Kelly would say I should open by betting two bucks.
In games of that form, it seems like you should be more-and-more careful as the amount of bets gets larger. The optimal strategy doesn't tend to Kelly in the limit.
EDIT: In fact my best opening bet is $0.64, leading to expected winnings of $19.561.
EDIT2: I reran my program with higher precision, and got the answer $0.58 instead. This concerned me so I reran again with infinite precision (rational numbers) and got that the best bet is $0.21. The expected utilities were very similar in each case, which explains the precision problems.
EDIT3: If you always use Kelly, the expected utility is only $18.866.
Can you give a concrete example of such a game?
even if your utility outside of the game is linear, inside of the game it is not.
Are there any games where it's a wise idea to use the Kelly criterion even though your utility outside the game is linear?
Marginal utility is decreasing, but in practice falls off far less than geometrically.
I think this is only true if you're planning to give the money to charity or something. If you're just spending the money on yourself then I think marginal utility is literally zero after a certain point.
Yeah, I think that's probably right.
I thought of that before but I was a bit worried about it because Löb's Theorem says that a theory can never prove this axiom schema about itself. But I think we're safe here because we're assuming "If T proves φ, then φ" while not actually working in T.
I'm arguing that, for a theory T and Turing machine P, "T is consistent" and "T proves that P halts" aren't together enough to deduce that P halts. And as I counter example I suggested T = PA + "PA is inconsistent" and P = "search for an inconsistency in PA". This P doesn't halt even though T is consistent and proves it halts.
So if it doesn't work for that T and P, I don't see why it would work for the original T and P.