I thought of a problem which was related, but not the same, and it seems much harder. I don't know how to start solving it.
The sheet may(let's say 50%) have been switched by Oswald's son Norbert, except he doesn't know his numbers that well, so he just wrote "2d6" on the top and then filled it with nothing but the number 4, because 6-2 is 4. Oswald doesn't notice this, since this is a bar and well, he's drunk since Norbert's been a handful lately.
Presumably, if this happens, at some point you will realize that buying information is not helping you in the slightest and stop. You are probably not going to pay 10,000 pounds for information 1 pound at a time on the 10,001st 4, because you will have probably considered the possibility that while it has an equal chance of being either, this doesn't seem likely to be a distribution of a 2d6 or a 1d12.
How would you calculate a maximum amount in total you should be willing to pay for information on a potentially corrupted bet like this?
The only answer I can think of here is to take a prior over all possible sequence generating computer programs, weighting them for length, and then to jack up the ones that simulate 2d6 or 1d12, and then to use the numbers to update on that.
I don't know if I've just described Solomonoff Induction or similar, but it sounds complicated, and yet I notice that if I'd just seen 10000 consecutive 4s I'd be pretty hot for the 'always gives 4' theory, and I wonder how I'd be doing that with my limited supply of slow neurons.
I posted this problem to my own blog the other day. When I posted it, I thought it looked very easy, more fiddly than difficult:
I reasoned thus:
There's no reason that you should have any opinion on which piece of paper he's brought. So you start off thinking 50:50, and that leads you to believe that he's effectively just given you £500.
If he tells you a number, then your belief will change. Say he tells you 1, then you know that he's brought the 1D12 results, and so you're now able to tell him that, and collect your £1000.
If he tells you 7, then that's twice as likely to be the 2D6 talking as the D12, and you should shift your prior to 1:3.
If you've got a prior of 1:3, then your guess (that it's the 2D6) is now worth £750, on average.
So when you get a new number, your prior shifts, the bet changes value. Average over all the cases and that's what you'll pay to know the first number.
Using this reckoning, I thought the answer to the puzzle was £125.
But now I'm not so sure, because the same reasoning tells you that if, for whatever reason, you start out 9:1 in favour of the 1D12, then the value of the new information is zero. (Because whatever the new information is, it won't be enough to change your mind).
But can that really be true? Because that implies that if Omega keeps making you the same offer for £1, then you should keep turning it down.
But if he told you a hundred numbers, you'd be damned sure which piece of paper he'd brought. So surely they have some value over £1?
But maybe you say: "Well, you can't put a value on the information unless you know how many extra opportunities you'll get."
Really? I'm sure that I'd pay £1 for the number in the original problem, and sure that I wouldn't pay £1000.
Where am I mis-thinking, and how should I calculate the answer to my puzzle?
Edit:
Just to clarify, if you buy the first number and it's a 2, and then you buy the second number and it's a 12, then I think you're now back in the same situation with a prior of 9:1 and an expected gain of £900.
I think you'd be mad to stop buying numbers at this point, since there's £100 you're not certain of yet. But if I don't believe that the price is £0, why do I believe that the price for the first one is £125?
Edit II:
It seems that the opinion of most people is that the problem is under-determined, in the sense that you don't know what options are coming. Fair enough.
In which case, what's wrong with the intuition that your beliefs alone determine the worth of your option to guess?
And in the more specific version where Oswald charges a price of one penny for every result, and you can keep buying them one-by-one until you decide you're certain enough and guess, what criterion do you use to stop guessing?