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simple game theory question

  • Aug. 10th, 2004 at 2:20 PM
side-beard-flip
(from Division of Labor)
My good friend, Neil Emerick in South Africa who runs Global Economic Software, sent me the following question. I pass it along to you here. If you'd like to add to his database of answers, e-mail him at
neilemerick[at]wol[dot]co[dot]za.

Imagine a company is running a competition for a prize of 10,000 dollars.
They challenge people to send in a number between 1 and 100. The prize will
go to the person(s) who picked the number that was 2/3rds of the average
number sent in.

What number would you send in?

With splitting the pot on ties, the only Nash equilibrium is for everyone to send in 0. Any other result is unstable, as everyone has incentive to move to the 2/3 point, which will tend to decrease the average towards 0. And once it is there, no one profits by moving. It also matters whether the metric is "closest to 2/3", or "closest to 2/3 from above". With the latter metric, all 0 is not stable, because if one person moves to 1, they win. It is unclear to me what the answer is in this case, its certainly much harder.

Another interesting point is that this game is ripe for collusion. Suppose everyone knows to pick 0. If my partners and I have a rough idea of how many people there are, some of us pick 100, and one of us picks the 2/3 point.

Comments

[info]zuleikhajami wrote:
Aug. 10th, 2004 02:28 pm (UTC)
Wouldn't 0 not be a number between 1 and 100?
[info]patrissimo wrote:
Aug. 10th, 2004 02:30 pm (UTC)
oops
you're right, replace 0 with 1 everywhere I said it...and replace 1 with 2 in the "closest to 2/3 from above" case.
[info]alexx_kay wrote:
Aug. 10th, 2004 02:55 pm (UTC)
Re: oops
But 2 doesn't actually trump (every one else chooses 1) under any definition of "average" that I know. So everyone choosing 1 is actually a stable solution. It's what I'd choose.
[info]patrissimo wrote:
Aug. 10th, 2004 03:00 pm (UTC)
Re: oops
I proposed 2 as the solution only if the rule is "closest to 2/3 FROM ABOVE". In that case, the defector to 2 would be the sole winner. Which means no one should choose 1, since it can never win.
[info]randallsquared wrote:
Aug. 10th, 2004 03:38 pm (UTC)
Re: oops
No, the average of all ones is one, and two thirds of one still leaves one as the closest from above.
[info]patrissimo wrote:
Aug. 10th, 2004 03:40 pm (UTC)
Re: oops
You're right, thanks.
[info]patrissimo wrote:
Aug. 10th, 2004 03:42 pm (UTC)
Re: oops
Ok, so only if there are few enough people is it right for an individual to defect. ie if there are few enough people that defecting upwards can drag the 2/3 avg. above 1, then 1 is not an equilibrium under "closest from above". If there are many people, then 1 is an equilibrium unless you have collusion, and the size of the collusive group necessary depends on the number of participants.
[info]candid wrote:
Aug. 10th, 2004 04:13 pm (UTC)
You have hit one of my pet peeves.

I agree that the only NE is for everyone to pick 1. (modulo some modifications of the game that you keep making in the comments)

To which I say, SO WHAT?

Nash Equilibrium carries zero predictive value in one-shot games.

A bunch of people are going to send in 66, and a bunch more people are going to send in 22, and a bunch more people are going to ...

In a game like this, unless you're playing with only game theorists (and maybe even then), picking as low as possible is a sure way to lose.

If I were playing (and I consider myself a game theorist), I'd put in a guess in the 20s.

But don't take my word for it. I'll do a poll.
[info]patrissimo wrote:
Aug. 10th, 2004 04:17 pm (UTC)
yeah
Oh, I agree with you. The title of the original post was "Game Theory Question", hence my focus on the game theoretic answer. In the real world, the correct submission depends on the audience, and what numbers they are likely to send in. I agree that 1 is a terrible submission.
[info]kirinn wrote:
Aug. 11th, 2004 07:35 am (UTC)
Indeed, as several people have said now, the real-world answer totally depends on the audience. My initial instinct given a large audience was to pick something around 22, being 2/3 * 2/3 * 50 -- figuring a lot of people would know 50 is the random average, multiply it by 2/3 once, and stop there. But on further thought even that is probably assuming too much math literacy for a random audience. Maustin's answer below might be closer. Hard to say.
[info]contrariandoer wrote:
Aug. 10th, 2004 04:18 pm (UTC)
I'll get a team of 100 people and each one will cover one number and then
split the winning among us.
[info]neonelephant wrote:
Aug. 10th, 2004 06:45 pm (UTC)
We actually did this one in the game theory class I took, except that the payoff was a bag of M&Ms. I don't recall what number wound up winning. I do know that not everyone picked 0 (I think somebody picked 100 just to throw the results off, though)
[info]maustin wrote:
Aug. 10th, 2004 08:56 pm (UTC)
I'd predict that about 80% would pick randomly, 12% would pick 66, 8% would pick 33, and I'd pick 34 :) (as the best answer if those percentages did pick those numbers).

It's an interesting problem though. You can't apply math to psychology very effectively. It reminds me of a game I was showing [info]sunshine__girl that I call bid. The two (or three) people involved get a full suit (2-A) of cards, and you have a face down pile that is another suit shuffled. Your goal is to blind bid for each card as you flip it over, the person bidding the most winning. Once you use a card it's out for the game.

It's easy to beat random, and it's easy to beat any strategy that you know, but it's hard to figure out what portion of people will go meta to what extent :)

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