Negative "eeny meeny miny moe"

post by jkaufman · 2019-08-20T02:48:41.509Z · score: 22 (8 votes) · LW · GW · 4 comments

As a kid, I learned the rhyme as:

Eeny, meeny, miny, moe,

Catch a tiger by the toe.

If he hollers, let him go,

Out goes Y, O, U!

Since kids can't predict where it will end, and adults are not supposed to try, it's a reasonably fair way of drawing lots.

At times I've heard versions where the selected person wins instead of loses, and while with two kids it doesn't matter, with three or more it matters a lot!

Let's model each kid having a choice at each stage between "accept" and "protest". While protesting probably doesn't work, if enough of you protest it might. If you do the positive version, where the selected kid wins, the winner accepts but the others may choose to protest. This isn't good: everyone has reason to protest except the single winner.

On the other hand, with the negative version, where one kid is eliminated at once, it's the other way around. When the first kid is eliminated they may protest, but the other kids all accept because then they retain their chance to win. With each successive round the dynamic is the same, plus the already-eliminated kids all choose accept out of a desire for fairness. Even with the last elimination there's still only one person choosing protest.

The iterative process is `O(n)` instead of `O(1)`, but it also works much better because it keeps a majority for "accept" at each stage.

(If you have a very large group of kids, then I could imagine a `O(log(n))` version being worth the added complexity. Divide the kids into three groups, and do negative eeny meeny miny moe on the groups. A third of the kids may protest, but you've still got two thirds accepting. Then redivide those remaining two thirds into three groups, and keep going.)

cross-posted from https://www.jefftk.com/p/negative-eeny-meeny-miny-moe

comment by Dagon · 2019-08-20T16:14:48.591Z · score: 2 (4 votes) · LW · GW

Ehn. For kids who will EVER accept this as fair, you're putting too much thought into politics. If the kids are this manipulable, they'll probably accept your authority in the one-shot case as well.

Also, more iterations gives them more time to realize that you're cheating (by shifts in how to count syllables) or that the game is fully deterministic (and you're cheating by deciding who to start with).

This is only usable for such low-stakes cases where the participants don't mind that it's not fair. And in those cases, don't waste time on pointless complexity. Of course, if this is part of the entertainment, I reverse that advice - choose the single-elimination method to extend the duration of the tension of not knowing.

comment by rossry · 2019-08-20T10:49:46.809Z · score: 1 (1 votes) · LW · GW

Another (related?) advantage is that the incentives to manipulate and catch manipulation are much better balanced with the negative ("you're out") version. Consider:

• Perfectly cheating in the positive version improves your chances of winning by (n-1)/n, and stopping you from doing so improves each other person's chances by 1/n.
• Perfectly cheating in a round of the negative version improves your chances of winning by 1/(k(k-1)), where k is the number of people in to start the round. Stopping you from doing so improves each other person's chances by the same amount.
• The total (summed) incentive to manipulate in the negative version is (n-1)/n, the same as in the positive case.
comment by maximkazhenkov · 2019-08-21T08:44:42.613Z · score: 2 (2 votes) · LW · GW
Perfectly cheating in a round of the negative version improves your chances of winning by 1/(k(k-1)), where k is the number of people in to start the round. Stopping you from doing so improves each other person's chances by the same amount.

I think 1/(k(k-1)) is the improvement in each other person's chance of getting into the next round, not the improvement in chance of winning the whole thing. The point still holds though since the absolute numbers are so much smaller in any single round.

comment by rossry · 2019-08-21T14:55:24.967Z · score: 1 (1 votes) · LW · GW

Oh, you're right. The net incentive to catch cheaters is actually... 1/(k(k-1)^2), then? The relative incentive story is worse, though still better in total than the positive version, and better still if you assume a constant-size disincentive to be caught cheating.