Add responses as you see fit
1,000🏅 Top traders
| # | Name | Total profit |
|---|---|---|
| 1 | Ṁ893 | |
| 2 | Ṁ125 | |
| 3 | Ṁ116 | |
| 4 | Ṁ114 | |
| 5 | Ṁ73 |
@diadematus I guess I misread your comment as asking someone else to make the market so I made one (https://manifold.markets/Weezing/who-will-win-round-2-of-fide-candid). Hopefully that's ok.
As for me I enjoyed it, but it was not possible to add more option during the tournament, which could be fun.
@diadematus I'm sorry but I don't really see what's unclear with them. 2 games started with 1.e4 and 2 games started with 1.e4 so given a first move, there's at most 2 games starting with it, "N*25% where N is the maximum number on boards on which white's first move is the same." indeed resolves 50% because N = 2 for that question.
Among the four games, one started with 1.d4 Nf6 ; one with 1.d4 d5 ; one with 1.e4 e5 and one with 1.e4 c5. So after the first half-move, some games are not distinguishable. But after 2 half-moves, we can distinguish all-games looking only at the moves played : 1.d4 Nf6 only mathches Avsov vs Nepo ; 1.d4 d5 <=> Gukesh vs Vidit ; 1.e4 e5 <=> Firou vs Pragg and 1.e4 c5 <=> Caruana vs Nakamura. So "N*5% where N is the minimum number of half-moves needed to distinguish all games by looking only at the N first half-moves." resolves 2*5% = 10%
But I see you already resolved them with what I had in mind.
@diadematus For example if 1.e4 is played on 3 boards and 1.Nf3 is played on 1 board, then N = 3 so it resolves to 75%. If all four games start with 1.c4 (please no), then N = 4 and it resolves YES ; if all four games start with a different move, then N = 1 and it resolves to 25%, and so on