Well, in the particular case he posed with a prior of 0.9 on the 1d12, 2 pieces of information are also useless. In fact, you need 4 pieces of paper to have nonzero value of information (and even then, I think the expected value of the 4 is < £1).
Sure, but I don't see where that changes the analysis. The probability of you getting 4 pieces of information, contingent on getting the first one, has got to be larger than the probability of getting 4, contingent on not getting the first one. (In fact the latter seems to be a contradiction, which presumably has probability zero.) So the first one still has some value, even if it's perhaps rather smaller than the value of the time it takes to do the formal calculation of the value.
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?