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ciphergoth comments on Advancing Certainty - Less Wrong
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Potentially promising approach, but how does that translate into math?
Let A represent the event when the lottery under consideration is profitable (positive expected value from playing); let X represent the event in which your calculation of the lottery's value is correct. What is desired is P(A). Trivially:
From your calculations, you know P(A|X) - this is the arbitrarily-strong confidence komponisto described. What you need to estimate is P(X) and P(A|~X).
P(X) I cannot help you with. From my own experience, depending on whether I checked my work, I'd put it in the range {0.9,0.999}, but that's your business.
P(A|~X) I would put in the range {1e-10, 1e-4}.
In order to conclude that you should always play the lottery, you would have to put P(A|~X) close to unity.
Q.E.D.
Edit: The error I see is supposing that a wrong calculation gives positive information about the correct answer. That's practically false - if your calculation is wrong, the prior should be approximately correct.
I think this doesn't work, or at least is incomplete, because what is needed (under standard decision theory) to decide whether or not to play is not the probability of the lottery having a positive expected value, but the expected utility of the lottery, which I don't see how to compute from your P(A) (assuming that utility is linear in dollars).
ETA: In case the point isn't clear, suppose P(A)=1e-4, but the expected value of the lottery, conditional on A being true, is 1e5, then you should still play, right?
You're right: recalculating...
Let E(A) be the expected value of the lottery that you should use in determining your actions. Let E(a) be the expected value you calculate. Let p be your confidence in your calculation (a probability in the Bayesian sense).
If we want to account for the possibility of calculating wrong, we are tempted to write something like
where x is what you would expect the lottery to be worth if your calculation was wrong.
The naive calculation - the one which says, "play the lottery" - takes x as equal to the jackpot. This is not justified. The correct value for x is closer to your reference-class prediction.
Setting x equal to "negative the cost of the ticket plus epsilon", then, it becomes abundantly clear that your ignorance does not make the lottery a good bet.
Edit: This also explains why you check your math before betting when it looks like a lottery is a good bet, which is nice.
If we follow your suggestion and obtain E(A) < 0, then compute from that the probability of winning the lottery, we end up with P(will win lottery) < 1e-8. But what if we want to compute P(will win lottery) directly? Or, if you think we shouldn't try to compute it directly, but should do it in this roundabout way, then we need a method for deciding when this indirect method is necessary. (Meta point: I think you might be stopping at the first good answer.)
The parallel calculation would be
I don't put P_typical very high.
Okay, I'll grant you that one. I'm still promoting my original idea to a top-level post.
Edit: ...in part because I would like more eyes to see it and provide feedback - I would love to know if it has some interesting faults.
Edit: Here it is.