Will Team YES have more total log(shares) at close than Team NO?
76
76
Ṁ1.5K
resolved Apr 25
Resolved
NO

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).

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predicted YES

Oh whoops. Was gonna try to manipulate this one, but I got... busy.

predicted NO

@IsaacKing really? you seem inactive?

predicted NO

@IsaacKing How exactly were you planning to manipulate it?

bought Ṁ1,000 of NO

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.

bought Ṁ1,000 of NO

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

}

predicted NO

@Imuli Who wants to buy 4.11x10^55 YES shares to snipe this last minute?

bought Ṁ220 of YES
predicted NO

@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") })

predicted NO

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

predicted NO

@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.

log base what?

predicted NO

@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 :)

predicted NO

@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.

predicted NO

I went one hyperoperation up:

bought Ṁ10 of NO

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.

predicted NO

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.

predicted NO

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

predicted NO

@Inosen_Infinity That's why I'm on team no.

sold Ṁ12 of YES

@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.

sold Ṁ90 of YES

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

sold Ṁ6 of YES

@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.

bought Ṁ100 of YES

@JosephNoonan I think it will be fine

predicted NO

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

bought Ṁ10 of NO

Log base 10? Natural log? :D

@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.

predicted NO

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.