Will the total number of digits in the positions in this market be evenly divisible by the total number of participants?
Jul 1

At the close of this market, I will look at each position in the market, take the number of digits in it (so, a $1,000 mana position is four digits) and add up all the digit counts.

If that number is divisible evenly, without remainder, by the number of participants - the market resolves YES. Otherwise NO. I will place an initial stake in the market because everything divides by one and no one wants a divide by zero error!


A has Ṁ 327,789 in YES, B has Ṁ 2,476 in YES, C has Ṁ 5 in NO.

6 + 4 + 1 = 11, 11 / 3 = 3 remainder 2, market resolves NO.

Get Ṁ600 play money
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opened a Ṁ30 NO at 22% order

this market is basically if the number of single digit positions is equal to the number of double digit positions, since the average is never going to be 3. and betting a single digit amount has little benefit to the better- unless you use alts or something. so it's probably NO

@ii you don't think we'll see 4, 5, or 6 digit positions?

@MattCWilson If we do, it's not going to make the average 3 digits. the market is mathematically equivalent to if the average number of digits is an integer, i should have made that clear in the previous post

bought Ṁ100 YES

Just to help demonstrate how this market will evaluate: at the present moment, YES has it with 30 digits of positions divided by 15 positions.

(A couple single digit NO positions couldn’t fit in my screenshot sorry 😞)

Steal of a buy for new YESes right now (provided you buy in for 2 or 18 digits worth of mana!)