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.
I agree. This should be time-symmetrical; your actions should be the same whether you rolled first then saw market prices or vice-versa. If you observe then roll, you have gained information making 6 more probable at least a little bit, and you must trade in the direction of your information (going long 6 or short the others), which is +EV.
The real question is how much should you trade, since you cannot afford to bet indefinite amounts of money on tiny +EV opportunities. For a prediction market that is a bit tricky, since they won’t typically be on problems where you can read off the posterior distribution from a few summary prices (if it’s 50/10/10/10/10/10%, does that mean that the market is highly certain that it’s a problem which simply has those true frequencies, or could it mean that it’s a deterministic problem but that the market is currently highly uncertain? in the former case, you have almost no edge and want to bet little and in the latter you want to bet a lot), and you have to start looking at things like volatility of the prices to gauge how much information you have relative to it and so how much of an edge and how much you can afford to bet.
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 assuming that the market is basically just doing Bayesian updating; it’s possible that there are some more complicated things happening with the market which make it a bad idea to make this assumption.
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.
I agree. This should be time-symmetrical; your actions should be the same whether you rolled first then saw market prices or vice-versa. If you observe then roll, you have gained information making 6 more probable at least a little bit, and you must trade in the direction of your information (going long 6 or short the others), which is +EV.
The real question is how much should you trade, since you cannot afford to bet indefinite amounts of money on tiny +EV opportunities. For a prediction market that is a bit tricky, since they won’t typically be on problems where you can read off the posterior distribution from a few summary prices (if it’s 50/10/10/10/10/10%, does that mean that the market is highly certain that it’s a problem which simply has those true frequencies, or could it mean that it’s a deterministic problem but that the market is currently highly uncertain? in the former case, you have almost no edge and want to bet little and in the latter you want to bet a lot), and you have to start looking at things like volatility of the prices to gauge how much information you have relative to it and so how much of an edge and how much you can afford to bet.
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 assuming that the market is basically just doing Bayesian updating; it’s possible that there are some more complicated things happening with the market which make it a bad idea to make this assumption.