[Manifold Plays Chess 3] Move 9: What will white play?
8
21
resolved Mar 25
45%
Bxc6+
42%
Qxf3
1.4%
d3
0.9%
d4
0.1%
g4
0.7%
Qe2
0.4%
Kf1
0.4%
Rh1
0.0%
Rb2
0.4%
Be2
0.4%
a4
0.5%
Rb1

What will white (Manifold) play in move 9?

Check the game here: https://lichess.org/GF9YULQP

The game so far: 1. e4 e5 2. Bc4 Nf6 3. Nc3 Nc6 4. Nf3 Nxe4 5. Nxe4 d5 6. Bd3 dxe4 7. Rg1 Bf5 8. Bb5 exf3

previous move:



For each response, the average probability in the last four hours before close is measured. With 75% probability, two moves will be randomly drawn, with weight proportional to those market probabilities. With 25% probabilities, three moves will be randomly drawn in the same way. Then for each of the two or three candidate moves, a conditional market is created.

The score of each move will be determined by the average probability in the last 4 hours. The move with higher score will be chosen (and the corresponding condditional market will resolve to the score one move later. The other market(s) will resolve N/A).

More details here:

https://manifold.markets/harfe/will-white-win-in-manifold-plays-ch

Any resign moves or draw offeers are not allowed in this move!

Invalid moves or duplicate moves will be removed from consideration.

Get Ṁ200 play money

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@harfe your random number is: 905204120

Salt: B9DM2arq8p33uxnew32X, round: 2806522 (signature a0a514a5344c12498292788d02fd51b4cad07843abc2b83adceb2f96ddb9b8c3c9b784fbfaa9358d1f75e314100dbc8a036b7b2b0f5ad21656079ca613c90ccfbe57a91dc115caed957d97bf8a452ee6c3fb8db3e96e2bc0add7062bd9d991b1)

Bxc6+, Qe2 it is

@FairlyRandom 999999893

@harfe you asked for a random integer between 1 and 999999893, inclusive. Coming up shortly!

Source: GitHub, previous round: 2806520 (latest), offset: 2, selected round: 2806522, salt: B9DM2arq8p33uxnew32X.

Moves by average probability:

0.440305 Bxc6+

0.399040 Qxf3

0.016676 d3

0.010183 d4

0.008857 Qe2

0.005053 a4

0.004731 Rb1

0.004586 Be2

0.004547 Rh1

0.004527 Kf1

0.001758 g4

0.001210 Rb2 (removed)

pick a number between 1 and 999999893 (inclusive)

Outcomes by integer range:

[ 1-610263800] Bxc6+, Qxf3

[610263801-677266096] Bxc6+, Qxf3, d3

[677266097-718066590] Bxc6+, Qxf3, d4

[718066591-753537217] Bxc6+, Qxf3, Qe2

[753537218-773758655] Bxc6+, d3

[773758656-793961210] Bxc6+, Qxf3, a4

[793961211-812875080] Bxc6+, Qxf3, Rb1

[812875081-831209137] Bxc6+, Qxf3, Be2

[831209138-849385353] Bxc6+, Qxf3, Rh1

[849385354-867482651] Bxc6+, Qxf3, Kf1

[867482652-884816703] Qxf3, d3

[884816704-897133754] Bxc6+, d4

[897133755-907842338] Bxc6+, Qe2

[907842338-999999893] other

Hash of the complete table:

94db298bce840016eeb9a724346f9d293328bd42c41fdc3a9ad7e2842419eaef

bought Ṁ10 of Qxf3

I suppose "touch the queen then suddenly have second thoughts" isn't an option even though if I was playing it would definitely be on the cards.

We could split these moves down into two decisions, one being which piece to move and the other being where to move it.

Not sure if better or worse outcomes.

@MartinRandall Ooh, a cool variant on this for the next round of this game could be:

Enumerate all possible legal moves and randomly divide them into N groups (for some predetermined constant N in the range 2 to 5 roughly). Make a conditional market on "will we win if we pick a move in this group" for each of the N groups. For the top predicted group, divide again into subgroups and repeat until only one move remains.

This allows us to sort of have a conditional market for every move and avoids both the "what if the random bot gives us only bad choices" problem and the "too many conditional markets active simultaneously means not enough liquidity for any of them" problem. I'm sure it still has some other problems, notably in exactly how you want to define the outcome of the conditional markets (wait until the end of the game to resolve vs resolve to some intermediate proxy metric).

@A @MartinRandall I feel such over-complexification makes it worse rather than solve any problems. The current systems is already quite a lot for most, as shown by the few of us really participating.

I don’t think the problem with this system is picking move candidates to get its conditional, but the way those conditionals are resolved (quasi self-resolved) solely based on the probability on the next winner and not on some external real-world, market-independent metric.

Sure, there is a bit of the real-world chess game factor here, which is why picking terrible blunders like Rg1 before didn’t quite work that well for the manipulator (they offloaded a big loss onto their alt) or Qe2 in this decision nearly impossible to prop. And of course, for the last move, it’s probably 95% determined externally. But the back propagation is weak beyond a couple of moves.

@deagol I agree that the conditional markets are the weakest link here.

Qe2 only cost me 1 mana and I already lost m800 on this game so it's not my main error.

@MartinRandall ha don’t get me wrong I’m not worried about you losing 1 mana, but some manipulator might give it a shot with M$500 and perhaps guarantee it gets a conditional, and still fail. Just pointing out this is not the right time to manipulate through blunders (need to prop them too high next) but through good moves. If @jack is able to unwind after this move, then yes, blunders may be back in play.

Rg1 before didn’t quite work that well for the manipulator (they offloaded a big loss onto their alt)

This is not actually the case. As far as I can tell Rg1 has been very profitable for me - there is what appears to be a large loss only because you sold the market down right before close, but that doesn't actually matter. I count the profits of both this account and Jack2 as the same bucket and every so often transfer them all to my main account, I don't offload losses. The not-so-secret reason I'm using an alt is because the two accounts can have separate pots of mana and separate limit orders - e.g. I can ensure that if my manipulation mana all gets invested I still have non-manipulation mana for my regular trading.

Also I haven't run the numbers properly but I think I have just about made back the M5000 I dumped into that first free answer market lol.

@jack Yeah, of course you made a killing on that one, sorry I misremembered. I meant an earlier move where you fought really hard I think against Martin, maybe one of the knight moves. I specifically remember I couldn’t understand why you forced it to resolve like that, letting the alt take a big loss while you took a big profit. I think it was almost a wash, so I thought it was a sort of transfer thing, as you say. I also remember you didn’t squeeze all the profit you could have out of Martin’s bets, and I commented about it, only to later see the resolution you allowed in which he didn’t lose that much. Anyway, sorry for my confusion with the old moves.

@jack Yep, it was move 4.Nf3 with almost 10k shares on both sides between you, your alt, and Martin, only to resolve almost painless.

@deagol Yeah, the reason was that when I started doing heavier manipulation trading on that market, I used the alt account to partition and protect my limit orders. Naturally, the shares bought while manipulating were pricier so those took a slight loss while the shares I bought earlier on Jack took a slight gain. No particular reason for forcing it to resolve like that, that's just the outcome that I thought best optimized my overall profit, including on future moves.

answered
Rb2

@A Illegal move. Thanks for the subsidy! ;)

sold Ṁ0 of Rb2

@deagol Oops, I meant Rb1 lol. Consider the 1 mana as a gift in the spirit of cooperation :)

answered
Bxc6+
bought Ṁ7

The best chance for white to get some compensation for the piece, other than SF3 blunders, is to create these doubled and isolated c pawns for black, and then go after them.

answered
Qe2

@MartinRandall Don’t waste your lotto ticket on this (gives up the queen), because even if the randombot picks it, there’s no way anyone will be able to prop its conditional anywhere close to 70% (that’s what it will take)

@deagol currently a 4% chance the choice will be between two blunders, I think? But I think this would even lose to other blunders.

bought Ṁ38 of Qxf3

@MartinRandall I’m prety sure the chance of picking two blunders is way lower, I think roughly like the product of their probabilities. 4% may be closer to the chance of picking a single blunder among the 2-3 picks.

@deagol at time of writing the two top options have 80%, so picking two blunders is about 20% x 20% = 4%.

@MartinRandall that’s not how it works

@deagol

two moves will be randomly drawn, with weight proportional to those market probabilities

Can't draw the same move twice, so I guess I'm making a wrong assumption about how that's excluded.

@MartinRandall Not all the others are blunders, and the majority is remaining subsidy. The real blunders I see (currently 3) total less than 2%, and yes I agree even if two get picked including Qe2, that would still lose.

now I count 5 blunders total 2.6%

I estimate roughly 1 in 1500 chances of picking two

@MartinRandall specifically, I don’t consider any of d3, g4, d4, or a4 as blunders. Maybe the last 2 as inaccuracies but d3 or g4 quite decent.