North Korea were caught cheating in 1991 and given a 15 year ban until 2007. They were also disqualified from the 2010 IMO because of weaker evidence of cheating. Given this, an alternative hypothesis is that they have also been cheating in other years and weren’t caught. The adult team leaders at the IMO do know the problems in advance, so cheating is not too hard.
One other argument I’ve seen for Kelly is that it’s optimal if you start with $a and you want to get to $b as quickly as possible, in the limit of b >> a. (And your utility function is linear in time, i.e. -t.)
You can see why this would lead to Kelly. All good strategies in this game will have somewhat exponential growth of money, so the time taken will be proportional to the logarithm of b/a.
So this is a way in which a logarithmic utility might arise as an instrumental value while optimising for some other goal, albeit not a particularly realistic one.
Why teach about these concepts in terms of the Kelly criterion, if the Kelly criterion isn’t optimal? You could just teach about repeated bets directly.
Passive investors own the same proportion of each stock (to a first approximation). Therefore the remaining stocks, which are held by active investors, also consist of the same proportion of every stock. So if stocks go down then this will reduce the value of the stocks held by the average active investor by the same amount as those of passive investors.
If you think stocks will go down across the market then the only way to avoid your investments going down is to not own stocks.
I just did the calculations. Using the interactive forecast from 538 gives them a score of −9.027; using the electoral_college_simulations.csv data from The Economist gives them a score of −7.841. So The Economist still wins!
Does it make sense to calculate the score like this for events that aren’t independent? You no longer have the cool property that it doesn’t matter how you chop up your observations.
I think the correct thing to do would be to score the single probability that each model gave to this exact outcome. Equivalently you could add the scores for each state, but for each use the probability conditional on the states you’ve already scored. For 538 these probabilities are available via their interactive forecast.
Otherwise you’re counting the correlated part of the outcomes multiple times. So it’s not surprising that The Economist does best overall, because they had the highest probability for a Biden win and that did in fact occur.
My suggested method has the nice property that if you score two perfectly correlated events then the second one always gives exactly 0 points.
I think the correct thing to do would be to score the single probability that each model gave to this exact outcome. Equivalently you could add the scores for each state, but for each use the probabilities conditional on the states you’ve already scored. For 538 these probabilities are available via their interactive forecast.
EDIT: My suggested method has the nice property that if you score two perfectly correlated events then the second one always gives exactly 0 points.
Maybe there’s just some new information in Trump’s favour that you don’t know about yet?
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.
Consistency of T isn’t enough, is it? For example the theory (PA + “The program that searches for a contradiction in PA halts”) is consistent, even though that program doesn’t halt.