The question here is whether a random walk simulated by iterated equal chance steps in either direction is always its own best predictor.
In other words if we start a market at 50%, and then iteratively flip a coin, and on each coin flip we say the underlying true probability has gone up or down by some step size, and if that number ever hits 0 or 100, it will immediately resolve, is that number in fact always the best possible predictor for the eventual resolution of the market?
You can see a live example of such a market here - https://manifold.markets/YaakovSaxon/random-walk-market-free-money-see-d-a8967bd564e6?r=WWFha292U2F4b24.
It seems to me intuitively that this should be the case on the following argument: suppose we use a step size of 5, and the market is presently at 95. So we have a half chance of going to 100 and resolving YES, and a half of chance of going down to 90. Assuming 90 is a good predictor, averaging those two chances gives you 95 so 95 is its own best predictor. Now look at 90. If you are on 90 you have a half chance of getting to 95 (already "proven" accurate) and a half chance of 85. So the average is 90, so 90 is also provisionally proven. And so on, until you get to 50 which obviously must be correct as there's no reason for either resolution to be more likely starting from 50.
I would think that this should then be true even if each coin flip has its own arbitrary step size, so long as the step size is equal in each direction. And it should even be true if it's a probability distribution of different step sizes (eg roll a die, 1:-8,2:-5,3:-1,4:+1,5:+5,6:+8) so long as it's equal in both directions and so long as the maximum step size would never "clip" beyond 0 or 100, which would waste part of the step and make it effectively unequal.
All of this is included in the question: Would a market under such conditions (every person gets to select a probability distribution for their roll each time subject only to both-directions-equality and no-clipping conditions) always be it's own best predictor of resolution to 0 or 100.
I think the answer to this is YES, but I'm not much of a mathematician and none of this is very rigorous. So I'm using this as a test of a new quasi-self-resolving mechanism.
The basic idea is that I will tend to prefer to resolve to Market (if only to avoid the work of determining the answer myself), but that I retain the right to resolve to my own best-effort-judgement instead.
In general my guiding criteria for doing so will be if it seems to me that the market answer has been manipulated in bad faith, or is otherwise just way off from what seems seems reasonable to me. In that case I will put in the hard work to find the correct answer as best I can.
In this case that would probably involve discussing it with proper mathematicians.
It is totally ok and encouraged to post arguments and proofs in the comments.
If there is ambiguity in the question, let me know and I will try to clarify it.
If for some reason even after clarifying my intention the question is PROVEN to be inherently uncomputable or unsolvable, then it would resolve NO.
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@YaakovSaxon Looking at the comments section in the Random Walk market it is an absolute massive wall of text, impossible to decipher what the market rate in a reasonable time frame with out a programmatic tool. If you run that sort of market again that is dependent upon random number generators and user participation, I suggest maybe using some sort of webform that accept and validate roles and can be linked to from the market description although it may open up your process to some sort of scripting attack if you cannot validate the submitter's identity.
@YaakovSaxon regularly @'ing people eligible to re-roll seems to have helped prompt more rolling activity on the market.
@YaakovSaxon I mean on the other market. This just seems to be the place people are chatting about it since it’s a less cluttered chat
@YaakovSaxon I think it might be best to keep the rules till resolution and start a new market if you want to update the ruleset.
Removing the wait will let someone roll until resolved overnight to 0 or 100 with out necessarily giving anyone an opportunity to respond, although it would relieve the bookkeeping issue and needing audits.
A wait of 5 rolls might encourage more activity but making it not retroactive makes for a bookkeeping hardship differentiating between the old regime and the new. But making it retroactive violates the positive expectation of traders trading under the old rules although it may be marginal, like a 5 or 10 percent rug pull.
Note: In your sandbox example market, you can't resolve to Market if the market has not corrected to the underlying probability set forth in your criteria. You would need to resolve to the underlying probability to enforce the theoretical impetus to correct the market after each roll/step. One issue is that the market as designed is a moving target underlying probability so it requires many trades to keep the market correct and there is a Flat Transaction Fee for each market order placed which cuts into the positive expected return. With out the flat market fee it would always make sense to chase the underlying probability, however the "friction" introduced into the system by fees means as the market runs long and takes on many many transactions the overall expected return is negated and then overwhelmed by thr fees. The market the grinds to a halt as existing market participants need to just hold and only new market participants would have a net positive expectancy for their participation provided they do not over participate. The last time someone rolled a 0,1 random number in the example market was 2 days ago. I think the existing market participants either have a net negative expectation or have lost interest since the market has not corrected to the underlying probability in the last 2 days either.
@ShitakiIntaki tldr, The hypothesis is true in a frictionless market, the introduction of fees moves market participants away from the theoretical ideal behavior.
@ShitakiIntaki Yeah that’s what I meant to say, was resolving to the “true probability”
Though I’d prefer to let it run until it goes to 100 or 0 if we can get some more rolls going again
Yeah, this is true. A random walk is a proven example of a "martingale", which is a stochastic process that is its own expected value. See https://en.wikipedia.org/wiki/Martingale_(probability_theory)
Can also play around with Wolfram Alpha such as in the case of 25% steps starting from 50% https://www.wolframalpha.com/input?i=%7B%7B1%2C+0.5%2C+0%2C+0%2C+0%7D%2C+%7B0%2C+0%2C+0.5%2C+0%2C+0%7D%2C+%7B0%2C+0.5%2C+0%2C+0.5%2C+0%7D%2C+%7B0%2C+0%2C+0.5%2C+0%2C+0%7D%2C+%7B0%2C+0%2C+0%2C+0.5%2C+1%7D%7D%5Ex+limit+as+x+approaches+infinity
Yes; see page 4 of https://math.dartmouth.edu/~doyle/docs/walks/walks.pdf
The probability of ending at 100 must be a linear function with f(0) = 0 and f(100) = 1, and the only such function is f(x) = x/100
This is true (if I've read the problem statement correctly). The math to derive it isn't too hard (it can be written in a page of elementary math) but I'm too lazy to write it out and I wasn't able to find a reference for it in 1 minute of google searching (something like 1d random walk with sinks).
Your proof seems to be circular, your statement asserting that 90% is the correct probability assumes both that 95% is correct and that 85% is correct. It does turn out to be true, but the derivation needed for it is different.