# What's the Value of Information?

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:

> The eccentric millionaire Oswald Mega walks into a bar and he says:
> 
> "This morning, I was showing my newborn about Dungeons and Dragons. We took a couple of six sided dice and rolled them, and wrote the results, which are just numbers from 2 to 12, on a piece of paper with 2D6 written at the top.
> 
> Then we took a twelve sided dice, and we wrote 1D12 at the top of a piece of paper, and then we rolled it lots and wrote down the results, numbers between 1 and 12, on the paper.
> 
> How she laughed at the difference in the patterns! Truly fatherhood is a joy.
> 
> Now, I've brought one of the pieces of paper with me, and if you can tell me which one it is, I'll give you £1000.
> 
> How much would you be willing to pay me to know the value of the first result on the sheet?"

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?