What is the log of lowest number rolled before AI wins IMO gold
4
104
110
2124
57%
chance

This market resolves if AI wins IMO gold. To be more precise, the grand prize of this competition is awarded: https://aimoprize.com/

If the group that receives the grand prize announces that they have won IMO gold before the prize is awarded, then that date will be used retroactively.

Starting January 2025, each year around Jan 1 I will ask @FairlyRandom for a random number between 1 and 1024. This market resolves PROB to 10*log(lowest number rolled)/log(2).

It will resolve YES if this market resolves before 2025. It will resolve NO early if a 1 is rolled.

I will not bet on this market.

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Betting guide: if you think that it will take roughly 2^n years for AI to win IMO gold, I recommend that you bet this market to roughly 100 - 10*n percent.

Example: if you think it will take roughly 16 = 2^4 years for AI to win IMO gold, then there will be roughly 15 rolls. In 15 rolls, the minimum number will be 1024/16 = 64 on average, and that means the market will resolve to 10*log(64)/log(2) = 60, which is indeed the recommended percentage in this comment.

Disclaimer: Although I am a mathematician, this is not a full mathematical proof. The actual percentage you should bet might be off by a few percentage points.

Is there a specific rationale behind this market formulation? Like the "amplified odds" things for nuclear war. Or is more of a flourish?

@CamillePerrin it is to ensure that plenty of betting space is left even if the market doesn't resolve for 10 years or so. Also, it makes the risk smaller for YES bettors that an early low roll occurs.

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