What will be the first published counterexample to Collatz conjecture?
9
67
2050
84%
Collatz conjecture is true
0.3%
42
3%
I predict Yev will sell his stake in #1.
0.7%
63728127
0.3%
No loans :-(

If someone publishes a counterexample Collatz conjecture and the mathematical community accepts it, this market resolves to that number. If someone publishes a proof of Collatz conjecture and the mathematical community accepts it, this market resolves #1. If this market doesn't resolve by the end of 2049, it resolves N/A.

Apr 14, 12:08am: This was inspired by my Discord conversation with Gurkenglas about free-response market mechanisms. The point was to test / demonstrate a specific type of market failure with the current mechanism.


Note: When I created this market the answers were numbered. #1 refers to the first answer submitted, in this case "Collatz conjecture is true".

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answered
I predict Yev will sell his stake in #1.

@Gurkenglas This is not a valid answer so it will not be chosen. You should make a derived market if you predict that.

What type of market failure is this demonstrating?

@IsaacKing I don't remember the details, but under DPM, #1 is going to be close to 100% regardless of the actual probability.

@Yev Is that just because it's not feasible to bet M$10^-1000 on each integer up to 10^1000? (Or whatever the best distribution would be.)

@IsaacKing It's a bit weirder than that.

Consider a different market where you have to guess a random 1000-digit number. If you submit a random answer and bet it up to 10^-1000, you expect to make a (very tiny) profit. It's not feasible to bet on all possible answers, but you can still maximize your profit by betting on as many answers as is feasible.

But in this market if you believe #1 is overpriced, the only way to make expected profit is to actually solve Collatz conjecture.

@Yev Why? You just choose a probability distribution over the integers where you'd most expect to find a counterexample and bet the appropriate amount on a finite number of those.

@IsaacKing I mean with the current system, where you need to submit each answer individually.

I guess there was some ambiguity in your question. I assumed that the counterfactual is that you send 10^1000 individual requests to manifold servers. But if the counterfactual is that Manifold lets you submit a succinct description of a probability distribution, then the answer is yes.

answered
63728127

@Adam converges to 1 after 949 steps

bought Ṁ2 of 63728127
this one seems pretty promising!
bought Ṁ1 of 42
This does not work: 42 -> 21 -> 64 -> 32 -> 16 -> 8 -> 4 -> 2 -> 1