Tuesday, August 14, 2012

A Slightly Harder Problem in Decision Theory

I'd be surprised if a cat could do this. But it think it uses the exact same tools as the easy one. 

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 number on the sheet?"

And actually there might be some feline subsystem that solves a problem a bit like this.


  1. Aye, that's a good lolly of dosh. I'm too faced to bet any folding, but I might lay down a few bob or even a nugget 'cause I am feeling at the moment like a jammy trainspotter.

    If the number is a one, I win two monkeys. That's mint.

    If the number is a two or a twelve, I'd be a plank to think I knew the answer.

    If the number falls between two and twelve, I'd be nutter to bet on anything but 2D6.

  2. You've got a monkey in the bag! Getting a 1 would lock in another. That makes the gen worth an apple on its own!

    Question is, will you go as far as a spit roast?

  3. seems like it'd be equivalent to ask about two lists of flips from a fair coin and a loaded coin.

  4. That would be a good start! This problem turns out to be much harder than I thought it was.

    Almost everyone including me initially thinks the answer is £125. But the same reasoning means that if you've already bought several numbers, the next one is absolutely worthless. Which is just silly.

    The simplest version I can think of is where you've got a 2:1 heads-biased coin and a 2:1 tails biased coin. But I can't work out what the answer is for that either. Or even work out if there has to be an answer.

  5. the two biased coins thing seems good, too.

    i went ahead and simulated the 1d12 vs. 2d6 version,
    using the strategy that seeing 1,2,3,11,12 imply 1d12
    and seeing 5,6,7,8,9 imply 2d6,
    and got 62.5% correct guesses (out of 10 million trials),
    which leads me to agree that it's worth £125.