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Oscar_Cunningham
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Celtic Knots on a hex lattice
Oscar_Cunningham5mo85

Celtic knot enjoyers might also like the daily game Celtix where the objective is to separate five coloured strands of a Celtic knot.

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An anti-inductive sequence
Oscar_Cunningham1y110

A simple version of this is the sequence 1,2,4,8,16,.... Each term is 1 greater than what you would predict by fitting a polynomial to the preceding terms.

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Translations Should Invert
Oscar_Cunningham2y192

Yup, you want a bi-interpretation:

Two models or theories are mutually interpretable, when merely each is interpreted in the other, whereas bi-interpretation requires that the interpretations are invertible in a sense after iteration, so that if one should interpret one model or theory in the other and then re-interpret the first theory inside that, then the resulting model should be definably isomorphic to the original universe

Bi-interpretation in weak set theories

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How tall is the Shard, really?
Oscar_Cunningham2y60

They will in fact stop you from taking fancy laser measuring kit to the top of the Shard: https://www.youtube.com/watch?v=ckcdqlo3pYc&t=792s.

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Kelly betting vs expectation maximization
Oscar_Cunningham2y10

I don't think that's the only reason - if I value something linearly, I still don't want to play a game that almost certainly bankrupts me.

I still think that's because you intuitively know that bankruptcy is worse-than-linearly bad for you. If your utility function were truly linear then it's true by definition that you would trade an arbitrary chance of going bankrupt for a tiny chance of a sufficiently large reward.

I mean, that's not obvious - the Kelly criterion gives you, in the example with the game, E(money) = $240, compared to $246.61 with the optimal strategy. That's really close.

Yes, but the game is very easy, so a lot of different strategies get you close to the cap.

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Kelly betting vs expectation maximization
Oscar_Cunningham2y10

It bankrupts you with probability 1 - 0.6^300, but in the other 0.6^300 of cases you get a sweet sweet $25 × 2^300. This nets you an expected $1.42 × 10^25.

Whereas Kelly betting only has an expected value of $25 × (0.6×1.2 + 0.4×0.8)^300 = $3220637.15.

Obviously humans don't have linear utility functions, but my point is that the Kelly criterion still isn't the right answer when you make the assumptions more realistic. You actually have to do the calculation with the actual utility function.

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Kelly betting vs expectation maximization
Oscar_Cunningham2y10

The answer is that you bet approximately Kelly.

No, it isn't. Gwern never says that anywhere, and it's not true. This is a good example of what I'm saying.

For clarity the game is this. You start with $25 and you can bet any multiple of $0.01 up to the amount you have. A coin is flipped with a 60/40 bias in your favour. If you win you double the amount you bet, otherwise you lose it. There is a cap of $250, so after each bet you lose any money over this amount (so in fact you should never make a bet that could take you over). This continues for 300 rounds.

Bob's edge is 20%, so the Kelly criterion would recommend that he bets $5. If he continues to use the Kelly criterion in every round (except if this would take him over the cap, in which case he bets to take him to the cap) he ends with an average of $238.04.

As explained on the page you link to, the optimal strategy and expected value can be calculated inductively based on the number of bets remaining. The optimal starting bet is $1.99, and if you continue to bet optimally your average amount of money is $246.61.

So in this game the optimal starting bet is only 20% of the Kelly bet. The Kelly strategy bets too riskily, and leaves $8.57 on the table compared to the optimal strategy.

Kelly isn't optimal in any limit either. As the number of rounds goes to infinity, the optimal strategy is to bet just $0.01, since this maximises the likelihood of never going bankrupt. If instead the cap goes to infinity then the optimal strategy is to bet everything on every round. Of course you could tune the cap and the number of rounds together so that Kelly was optimal on the first bet, but then it still wouldn't be optimal for subsequent bets.

(EDIT: It's actually not certain that the optimal strategy in the first round is $1.99, since floating point accuracy in the computations becomes relevant and many starting bets give the same result. But $5 is so far from optimum that it genuinely did give a lower expected value, so we can say for certain that Kelly is not optimal.)

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Kelly betting vs expectation maximization
Oscar_Cunningham2y72

If Bob wants to maximise his money at the end, then he really should bet it all every round. I don't see why you would want to use Kelly rather than maximising expected utility. Not maximising expected utility means that you expect to get less utility.

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Kelly betting vs expectation maximization
Oscar_Cunningham2y40

Can you be more precise about the exact situation Bob is in? How many rounds will he get to play? Is he trying to maximise money, or trying to beat Alice? I doubt the Kelly criterion will actually be his optimal strategy.

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A hundredth of a bit of extra entropy
Oscar_Cunningham3y70

I tend to view the golden ratio as the least irrational irrational number. It fills in the next gap after all the rational numbers. In the same way, 1/2 is the noninteger which shares the most algebraic properties with the integers, even though it's furthest from them in a metric sense.

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