There exists a 6-sided die that is weighted such that one of the 6 numbers has a 50% chance to come up and all the other numbers have a 1 in 10 chance. Nobody knows for certain which number the die is biased in favor of, but some people have had a chance to roll the die and see the result.
You get a chance to roll the die exactly once, with nobody else watching. It comes up 6. Running a quick Bayes's Theorem calculation, you now think there's a 50% chance that the die is biased in favor of 6 and a 10% chance for the numbers 1 through 5.
You then discover that there's a prediction market about the die. The prediction market says there's a 50% chance that "3" is the number the die is biased in favor of, and each other number is given 10% probability.
How do you update based on what you've learned? Do you make any bets?
I think I know the answer for this toy problem, but I'm not sure if I'm right or how it generalizes to real life...
I'd bet on 6. I have information that the market doesn't have, and my information points to 6 as the answer, so the market is underpricing 6 (compared to how it would price 6 if it had all the information).
Another way to think of it: suppose that, instead of a market, there was just a single person looking at all the die rolls and updating using Bayes's Rule. There have been n rolls and that person has assigned the appropriate probabilities to each of the possible die weightings. Then the n+1th roll is a 6. The person then updates their probability assignments to give 6 a higher chance of being the favored side.
If the prediction market is efficient, then it should be analogous to this situation. The market price reflects the first n rolls, and now I know that the n+1th roll was a 6, so I get to profit (in expectation) by updating the market's probabilities to take that new piece of information into account.
It may be possible to give a more precise answer, but this is what I have for now.
AlexMennen and Oscar_Cunningham have run the numbers and gotten that more precise answer. I did some calculations myself and agree with them. If the market has been efficiently incorporating information, then the prior die rolls included k+1 rolls of 3, and k rolls of each of the other numbers (this gives 1:1:5:1:1:1 odds regardless of k). My roll brings it up to k+1 6's, so the odds should now be 1:1:5:1:1:5 (i.e., 1/14 for most numbers and 5/14 for 3 and 6).
This is ass... (read more)