After this market closes, I will add up the numbers of log(shares) held by each YES trader and each NO trader. If the sum is larger for the YES traders, this resolves YES. Otherwise (including a tie), it resolves NO.

This was inspired by Conflux's market:

Now the the benefit to having more traders on your side, as opposed to having larger bets, will be even more pronounced, to the point where it is basically impossible to manipulate the market by making large bets (though other forms of manipulation would still work).

# 🏅 Top traders

# | Name | Total profit |
---|---|---|

1 | Ṁ1,141 | |

2 | Ṁ458 | |

3 | Ṁ372 | |

4 | Ṁ372 | |

5 | Ṁ226 |

Total log(YES) shares: 76.89

I didn't calculate the total log(NO) shares because it just took the first 11 NO holders to exceed that, with a log sum of 79.11. The total sum of log(NO) shares is much larger.

curl+jq script (and the current status):

> curl https://manifold.markets/api/v0/market/9WizBWmiczVYPqDA6LTR/positions | jq '{yes: map(select(.maxSharesOutcome == "YES") | .totalShares.YES | log) | add, no: map(select(.maxSharesOutcome == "NO") | .totalShares.NO | log) | add}'

{

"yes": 65.90765240401448,

"no": 193.96296403660878

}

@Imuli javascript/paste into console version:

let req = new XMLHttpRequest();

req.open("GET", "https://manifold.markets/api/v0/market/9WizBWmiczVYPqDA6LTR/positions", false);

req.send();

let resp = JSON.parse(req.response);

let count = sh => resp.filter(x => x.maxSharesOutcome === sh)

.reduce((total,x) => total + Math.log(x.totalShares[sh]), 0);

({ yes: count("YES"), no: count("NO") })

@JosephNoonan who wants to make "does IsaacKing have a trick up his sleeve on this market"

@Conflux Yeah, with alts being illegal now, I don't see what his plan could be. Unless he actually has amassed an army to snipe this with YES bets at the last second.

@SeamusBronski It actually does not matter: https://manifold.markets/JosephNoonan/will-team-yes-have-more-total-logsh#O62ngjsTvSuUhY7PZO2c

@42irrationalist If you use a negative base and then a total order the complex plane based on some space-filling fractal, that would be amusing but also very much against the spirit of the question :)

@Imuli Well, don't worry. As mentioned below, I will use a base greater than one, so this is equivalent to asking which team will have the largest product of shares. When I actually calculate the sum, I will use natural log.

This is currently my market closest to meeting the YES criteria for this one:

If we get 25 more traders, we can make it happen.

At the moment, NO shares are so cheap, that if 7 people spend 10 mana on NO shares each, team NO would be winning. And I guess even after that NO shares would still cost less than YES.

And the profit for those who buy these 10 mana positions would be like 500%

@Inosen_Infinity The lead has narrowed even more. Now, all we need is for a single new trader to buy Ṁ2 worth of NO (7 shares), and Team NO will be ahead, with NO shares still being almost 3 times cheaper.

There's a hedgehog voting yes I feel bad for screwing over but I just became a turncoat and joined your cause

@Odoacre I don't think the M100 bet was a good idea. It makes NO shares cheaper while only giving about 5 log-shares to Team YES. Similar to what happened in the sqrt market, but this time making large bets is even riskier, since log grows much slower than sqrt.

@Odoacre We'll see. Although my comment was at least partially motivated by a desire to get people to bet NO.

@AlexRockwell It doesn't actually matter what base I use, as long as the base is greater than 1, since log_b(x)=log_a(x)/log_a(b), for any bases a and b, so using a different base is equivalent to multiplying both sums by a constant factor. But when doing the calculation, I will probably use natural log.

Oh yeah, youre right it doesnt matter :D Without thinking about it much, my intuition was suggesting that with a higher log base there would be more advantage to lots of little owners. But I guess its the same.