= 27d567bc2d48d9f40e9ccc20ea370b15)
PhilGoetz comments on The Prediction Hierarchy - Less Wrong
You are viewing a comment permalink. View the original post to see all comments and the full post content.
= 783df68a0f980790206b9ea87794c5b6)
Loading…= b715d02e85f7b3b4ae65ca3d3e450868)
Subscribe to RSS Feed
Powered by Reddit
= f037147d6e6c911a85753b9abdedda8d)
You are viewing a comment permalink. View the original post to see all comments and the full post content.
Comments (37)
Why, when you consider the case where you calculated the odds of winning the lottery incorrectly, do you increase rather than decrease the odds?
In any case, with a lottery, you do know the odds of winning; they're stated on the ticket.
Edit: I see I misread the remarks. See downthread.
At the moment, I'm calculating my expected value, not the odds, but there are a number of reasons to think that jackpot / stated-odds is optimistic:
The lottery may be a fraud.
The lottery may go bust.
I may lose the ticket.
I may have to split the pot.
In general, the rigorous approach would be to rewrite everything as probability distributions.
Besides: if you want to assume the average lottery ticket is more valuable that I would - e = -0.5*t, say - that's your right. I make no justification for my priors.
You quoted this from somewhere:
This says that rationally, you should assign a much higher expected value to the ticket. But all 4 factors you just listed are ones which would make the expected value of the ticket lower.
Oh, I see what you mean. That wasn't a quote, actually - it was essentially an articulation of ciphergoth's clever (but incorrect) argument. The purpose of this post was to explain my method for rebutting it.
It's just a restatement of the Pascal's Mugging problem, but with the lottery in place of the mugging.
I'm still ambivalent about Pascal's Mugging, however - my instinct is to refuse to pay, but I don't feel I can sufficiently justify that response.
The lottery, as an ordinary situation, is far more tractable.
Can't you apply a similar argument? Instead of considering P(mugger's statement is true), you consider P(you have the faintest idea what's going on).
My instinctive probability measurement for such a statement is not so small as 1/3^^^^3. My best retort at the moment is purely pragmatic: never accept such a mugging, because otherwise you will be mugged.
Indeed, the probability that we don't know what's going on is non-negligible. What I'm suggesting is that we don't have to assign a non-negligible probability to the specific hypothesis "this mugger is speaking the literal truth" - instead we avoid overconfidence by trying to consider all of the hypotheses that might hide behind the general assertion "our grasp on this situation is much less than we think" and try to use broader reference classes to see what the outcomes of various strategies might be in those instances, using the strategy you outline for the lottery.
Not to engage in needless turnabout, but how does that translate into math?
Agreed--the trick is that being wrong "only once" is deceptive. I may be wrong more than once on a one-in-forty-million chance. But I may also be wrong zero times in 100 million tries, on a problem as frequent and well-understood as the lottery, and I'm hesitant to say that any reading problems I may have would bias the test toward more lucrative mistakes.
Let's rephrase, then. Suppose for a moment that you are 100% confident a lottery ticket costs $1, you can buy it, it pays $10^6 on a win, etc etc and that you are reading the ticket right now and believe it says the probability the ticket will win is 1/(4x10^6). Should you believe the ticket is +EV?
The wrong calculation: Yes, because you estimate you'll misread the ticket (or it's lying, etc etc) 1 in a million times, which makes the EV 10^6 x (10^-6 + (1-10^-6) x 1/(4x10^6)) = 1 + ~0.25.
The right calculation: No, because you'll misread the ticket 1 in a million times, which makes the EV 10^6*(10^-6 x P + (1-10^-6) x 1/(4x10^6)) = P + ~0.25 where P is whatever probability of winning with 1 ticket you assign to an arbitrary lottery that costs $1 and pays $10^6 where you incorrectly read the probability off the back of the ticket as being 10^-6 (or it's lying, etc etc). If your priors say P ~= 1 then they need adjusting; if they say P ~= 10^-7 to 10^-6 then they probably don't need adjusting. And then the EV is ~= 0.25 again.
AFAICT this is the same as in the post, but I'm not certain I understand precisely where your question is.
Edit: ha, I put 10^-5 to 10^-6 (which is of course silly) instead of 10^-7 to 10^-6, but RobinZ put ~0 anyway