The Prediction Hierarchy

I haven't read the paper through, but the similarity in algebra cannot be denied. I have added a reference to the post.

Thanks, interesting read. Could you expand more on the points of similarity and difference between your argument and RobinZ's? They currently seem very disparate approaches to me.

Thank you!

A lot of credit has to go to ciphergoth and Wei_Dai, actually - I was just trying to run a stack trace on my instantaneous rejection of the original lottery argument; they're the ones who made me quantify it.

Seconded. In fact I'm almost tempted to declare that this supersedes my post

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:

1. The lottery may be a fraud.

2. The lottery may go bust.

3. I may lose the ticket.

4. 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.

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

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 me be more precise: before you see anything wrong with your calculations, you have no real reason to expect locating an error in them to give you evidence of anything specific. Therefore, when doing your initial post-calculations, the prior probability is appropriate.

After you find an error in your calculations, you can usually fix the error in your calculations.

Not quite. It depends on your beliefs about how the calculation could go wrong and how much this would change the result. If you are very confident in all parts except a minor correcting term, and are simply told that there is an error in the calculation, then you can still have some kind of rough confidence in the result (you can see how to spell this out in maths). If you know the exact part of the calculation that was mistaken, then the situation is slightly different, but still not identical to reverting to your prior.

That's a good point - standard cautions against overconfidence should reduce p(C), much like time pressure, exhaustion, and complexity of argument.

The second assumption, however, is harder to justify. There are many ways that a calculation of odds could go wrong (putting a decimal point in the wrong place, making a multiplication error, unknowingly misunderstanding the laws of probability, actually being insane, etc.) If we could really enumerate all of them, understand how they effect our computed payout probability, and estimate the probability of each occurring, then we could compute this missing factor exactly. As things stand though, it is probably untenable. It should not be expected though that errors that make the payout probability artificially larger will balance those that make it artificially smaller. Misplacing a decimal point, for example, will almost certainly be noticed if it leads to a percentage greater than 100%, but not if it leads to one that is less than that (creating an asymmetry).

This is a valid point, and one I missed in my writeup. (Toby_Ord said something similar, but that was in response to a specific question.)

It is probably a useful skill to recognize asymmetries in the possible direction of error, such as that which you pointed out. I can see two ways to handle this:

a. Additional terms in the derivation, such as P(decimal-point error) and P(sign error), with the e term restricted to the unanticipated-error case.
b. Modification of e.

Don't be so overconfident, apparently in 1997 New York ran a promotion that doubled the payouts of a single game on exactly 4 days in exactly 1 month, and payed out $1.40 to $1 on average.

If by 'do the figures change' you mean 'does it ever become a good bet' then no.

Your odds of winning once go up as you increase the number of tickets you buy (# of tickets purchased * Chance of winning per ticket). The expected value of a given ticket remains the same. All you are doing is focusing more money away from other possibilities. If you buy 5 tickets a week for your entire life, and the odds of winning are 1 in 100 million, then you have a 0.000169 chance of winning the lottery, but you could have spent your 16 thousand on a new TV or a vacation.

As mattnewport and LucasSloan point out, it doesn't change the actual numbers - a bad bet multiplied a thousandfold is still a bad bet - but it does change the wrong numbers: buying a thousand tickets for a 0.01% chance of a million dollars is a losing bet again.* More evidence that the ignorance argument fails.

* How I calculate this (changes in italics):

According to your calculations, "none of these thousand tickets will win the lottery" is true with probability 99.9975000312185%. But can you really be sure that you can calculate anything to that good odds? Surely you couldn't expect to make forty thousand predictions of which you were that confident and only be wrong once. Rationally, you ought to ascribe a lower confidence to the statement: 99.99%, for example. But this means a 0.01% chance of winning the lottery, corresponding to an expected value of a hundred dollars. Therefore, ... these thousand tickets still lose, because you spend a thousand to win a hundred.

I make no such point. If you read the post, nowhere do I assume any specific relation between [p(L)*j - t] and e - my point is specifically that you should use something with no dependence on your calculation (and strictly more reliable) to draw conclusions from when you're not sure.