Vaniver, looks like you were thinking about the problem in the same way that I was, getting repeated chances to buy new numbers.
Sort of. The calculations that I ran are all one-step-ahead calculations, starting with different priors. Consider three different cases:
Pricing X is easy; it's £125. Pricing Z is a bit tougher, but still okay. Pricing Y involves coming up with 13 different prices- the twelve possibilities after the first roll, and then Y (which depends on each of those possibilities!). Doing that with arbitrary n is doable but tough! (It's somewhat easier if you have a set price for each successive number, so you can swiftly terminate trees once you've hit the point that it's no longer worth the price.)
And so, even at 9:1 odds, there is some number of numbers he can read off that will have positive VoI. It will be very low- because it's very unlikely you will get that many informative numbers- but it is true that if you aren't perfectly certain, a test that gives you perfect certainty will have positive VoI.
What do you do then? The conclusion 'I literally won't lift a finger to know more numbers' doesn't seem right unless you're certain of the answer already.
The thing to focus on here is both the amount of additional certainty and the effect of additional certainty. The number you get when you're at 9:1 tells you a lot less than the number you get when you're at 1:1. Imagine the next number being a 1- in the first case, it feels like you just got £100, but in the second case it feels like you just got £500. Similarly, when I'm at 1:1, telling me one additional number is expected to change my guess in some cases. When I'm at 9:1, regardless of what he tells me, I still make the same call.
There is such a thing as certain enough when there are tests that aren't informative enough.
(Interestingly, note that you can never reach perfect certainty that it's 2d6, and there will always be a positive VoI for another number because there will always be a positive chance that it's a 1.)
Great post by the way. Thank you. It sounds like your job is to think about this sort of thing!
I think I now believe that the answer to the original question can't be £125, unless you already know what happens next.
Suppose the question is something like: "Every time you give me a penny, I'll give you the next number. At any time you can stop and make your one guess." It seems to me that there has to be a computer program that is best at playing this game. Do you have any idea what its stopping criterion would be? Or what the price would have to ...
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?