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.