A semi-technical question about prediction markets and private info

post by CronoDAS · 2017-02-20T02:20:40.146Z · score: 6 (7 votes) · LW · GW · Legacy · 17 comments

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...

comment by Dagon · 2017-02-20T05:25:39.299Z · score: 5 (5 votes) · LW · GW

Missing information - how many people have rolled the die how many times before participating in the market? If you don't expect that there's private information that the market can make public, you shouldn't expect it to indicate truth.

Also missing - what are the actual contracts? Is this a wager on the next roll of the die? Meaning 50% is paying 1:1, 10% paying 9:1 (\$100 bet on 3 pays \$100, \$100 bet on 6 pays \$900 if it wins)?

In that case, if the market is one person who saw one roll, I bet 5 contracts on 6, and one on each of 1,2,4,5. If the market is hundreds of people, even if they've each only seen it once (independently; hundreds of rolls with one observer each, not 100s of observers of one roll), then the market has likely already worked out the correct odds, so my observation doesn't add much.

comment by philh · 2017-02-20T10:36:04.024Z · score: 3 (3 votes) · LW · GW

Note: if the market is hundreds of people, and it's a market on which face comes up 50% of the time (not on which face will come up on a specific roll), and it only gets 50% odds on that being a particular number, then something unusual is happening. An efficient market under these circumstances should be very confident.

(I haven't done any explicit calculations, but I'm reasonably confident.)

comment by Good_Burning_Plastic · 2017-02-20T11:02:59.880Z · score: 1 (1 votes) · LW · GW

Unless only one roll of the die was seen by the hundreds of people, and it came up "3".

comment by philh · 2017-02-20T12:37:57.580Z · score: 1 (1 votes) · LW · GW

Ah, sure. It would have been better of me to say "if the market has collectively observed hundreds of rolls".

comment by Unnamed · 2017-02-20T07:38:33.422Z · score: 4 (4 votes) · LW · GW

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.

comment by gwern · 2017-02-20T20:52:26.912Z · score: 3 (3 votes) · LW · GW

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.

comment by Unnamed · 2017-02-20T21:47:54.645Z · score: 1 (1 votes) · LW · GW

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.

comment by AlexMennen · 2017-02-20T07:37:08.202Z · score: 4 (4 votes) · LW · GW

Let's assume prediction markets are efficient and you didn't already possess any relevant information that you weren't trading on beforehand. Then you should treat the market odds as a prior and your die roll as evidence, in exactly the way you always do Bayesian updates. In this case, that means it looks like that gives you a posterior probability of 5/14 each for the die being weighted in favor of 3 or 6, and 1/14 for each of the other possibilities. Contrary to what other commenters were saying, it doesn't matter what information led to the market odds under these assumptions.

comment by Oscar_Cunningham · 2017-02-20T10:33:20.392Z · score: 3 (3 votes) · LW · GW

I don't know if there are any economic theorems about how markets incorporate information from their participants, but if we assume that they incorporate all of the information then it must be that the market participants had seen every number equally often except for seeing one extra three. So you should update as though you had seen an extra three, to get odds of 1:1:5:1:1:5. Then it would be worth placing a bet on six and against each other number.

comment by Zvi · 2017-02-20T17:16:32.382Z · score: 2 (2 votes) · LW · GW

The market prices reflect what one would see if one person had rolled the die once and had the result be a 3. Getting the result any other way seems pretty hard.

I would therefore assume that the market probably got that way because one person rolled the die once, and got a 3, so now I have two rolls, 3 and 6, and I should update accordingly.

If the prediction market let me bet non-trivial amounts without moving the prices, then something strange is going on, and "it's time for some game theory."

comment by CronoDAS · 2017-02-20T19:11:39.086Z · score: 0 (0 votes) · LW · GW

The numbers I picked are kind of arbitrary - yes, they are what you would get from someone rolling the die once and getting a 3. The basic concern is that prediction markets might be vulnerable to information cascades - you think the prediction market knows better than you, so you don't go and contradict it even though your personal analysis of the situation gives a different answer.

comment by Oscar_Cunningham · 2017-02-21T12:08:29.567Z · score: 0 (0 votes) · LW · GW

I think what Unnamed says is the most important observation:

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).

Any time you have private information that the market doesn't have you should bet to move the market in the direction of your information. The difficult question is how much you should bet.

What information do they have?

This is the general problem of a mixture of experts when all you have are the predictions but not the information on which the predictions are based (at least for the market). I don't think there is a real answer to that until you input more information into the system.

We want P(side | I_them, I_me)

We have P(side|I_them), P(side|I_me)

The latter don't give the former.

comment by ChristianKl · 2017-02-21T12:49:18.855Z · score: 0 (0 votes) · LW · GW

Any time you have private information that the market doesn't have you should bet to move the market in the direction of your information. The difficult question is how much you should bet.

I don't think that's true. If it would be true I don't think most mutual fund managers would underperform the SAP 500. A mutual manger might get some superficial information about stocks he invests in that take actual research.

Mutual fund managers have incentives other than maximizing expectation of price. OPM.

Also, they likely overestimate the value of their privileged information.

comment by 9eB1 · 2017-02-20T04:11:11.090Z · score: 0 (0 votes) · LW · GW

This is an interesting puzzle. I catch myself fighting the hypothetical a lot.

I think it hinges on what would be the right move if you saw a six, and the market also had six as the favored option. In that situation, it would be appropriate to bet on the six which would move it past the 50% equilibrium, because you have the information from the market and the information from the die. I think maybe your equilibrium price can only exist if there is only one participant currently offering all of those bets, and they saw a six (so it's not really a true market yet, or there is only one informed participant and many uninformed). In that case, you having seen a six would imply a probability of higher than 50% that it is the weighted side. Given that thinking, if you see that prediction market favoring a different number ("3"), you should indeed bet against it, because very little information is contained in the market (one die throw worth).

The market showing a 50% price for one number and 10% for the rest is an unstable equilibrium. If you started out with a market-maker with no information offering 1/6 for all sides and there were many participants who only saw a single die, the betting would increase on the correct side. At each price that favors that side, every person who again sees that side would have both the guess from the market and the information from their roll, then they would use that information to estimate a slightly greater probability and the price would shift further in the correct direction. It would blow past 50% probability without even pausing.

Those don't seem like very satisfactory answers though.

comment by ike · 2017-02-20T03:49:32.033Z · score: 0 (0 votes) · LW · GW

Has anyone rolled the die more than once? If not, it's hard to see how it could converge on that outcome unless everybody that's betting saw a 3 (even a single person seeing differently should drive the price downward). Therefore, it depends on how many people saw rolls, and you should update as if you've seen as many 3s as other people have bet.

You should bet on six if your probability is still higher than 10%.

If the prediction market caused others to update previously then it's more complicated. Probably you should assume it reflects all available information, and therefore exactly one 3 was seen. Ultimately there's no good answer because there's Knightian uncertainty in markets.