Comments on: Relatively basic semi-theoretical semi-economic type question.

Question: Relatively basic semi-theoretical semi-economic type question.

Kwantsar — Wed, 09 Nov 2005 13:34:59 -0800

Relatively basic semi-theoretical semi-economic type question.

My task (and the task of sixty other individuals) is to write down a number between 0 and 100 (inclusive). Another party will calculate the average answer of all participants. My objective is to guess as close as possible to 2/3 of the average guess of all participants. The other participants are bright, but not really bright.

Any ideas?

By: fvw

fvw — Wed, 09 Nov 2005 13:42:34 -0800

What is the goal of the other participants?

By: Kwantsar

Kwantsar — Wed, 09 Nov 2005 13:45:58 -0800

They have an identical goal.

By: milkrate

milkrate — Wed, 09 Nov 2005 13:46:34 -0800

This is a game theory question. The answer is 0.

By: Kwantsar

Kwantsar — Wed, 09 Nov 2005 13:47:07 -0800

(A goal identical to mine)

By: knave

knave — Wed, 09 Nov 2005 13:48:01 -0800

Hmm.. no one "bright" would guess higher than 67, right? That is, if everyone else picked 100, 2/3 of the average would be 67. But by guessing 67 you will bring the average down a bit, too...

So now the problem is reduced to a number between 0 and 67, inclusive. At this point, repeat the logic from the above paragraph. Personally, I'd write down 0, but I could be totally whacked.

By: googly

googly — Wed, 09 Nov 2005 13:49:31 -0800

Two strategies:

(A) If we assume that the guesses are distributed randomly, then the average would be about 50, and your best guess would be 33 or thereabouts.

(B) If everyone followed this strategy, then the average guess would be about 33, and your best guess would be 22.

Assume that roughly half of the participants will follow strategy A, and half will be smart enough to follow strategy B. Thus, the average will be about 27.5, and the best guess would be 18.

On preview: You can adjust this strategy according to how cynical you think the other participants will be (how many will guess zero).

By: Rothko

Rothko — Wed, 09 Nov 2005 13:49:43 -0800

Are you asking what the probability or likelihood would be of correctly calculating 2/3rds or more of each number selected by all people in the "other" party? I don't understand your question.

By: adamwolf

adamwolf — Wed, 09 Nov 2005 13:49:48 -0800

If you assume everyone will guess 100, your number will be 66. However, knowing this, the other people will also guess a lower number than 100. The new average will be less than 66, and so the guess drifts down to 0.

By: knave

knave — Wed, 09 Nov 2005 13:49:48 -0800

On postview, milkrate agrees. Cool.

By: raedyn

raedyn — Wed, 09 Nov 2005 14:20:05 -0800

What are they trying to teach with this exercise?

By: justkevin

justkevin — Wed, 09 Nov 2005 14:42:22 -0800

I'm guessing their trying to teach the participants about factoring in others' internal thought processes in your decision making (a la the three hats problem).

Zero would be the correct answer if everyone else was perfectly logical. But Kwantsar said they were bright, but not really bright.

Based on this information, I would guess some participants will not fully extrapolate the conclusion to "0". If only one person guesses over 30, then 0 would lose to 1.

Of course, if everyone else realizes this, then they would put down 1...

So I can clearly not choose the wine in front of me!

By: justkevin

justkevin — Wed, 09 Nov 2005 14:43:01 -0800

their = they're

By: j.edwards

j.edwards — Wed, 09 Nov 2005 14:43:40 -0800

In terms of an experiment, this was posited as a question on livejournal and linked to from the mathematics community a couple weeks ago, but it appears to not have been updated. I guessed zero.

By: vacapinta

vacapinta — Wed, 09 Nov 2005 14:47:02 -0800

You also have to factor in the people who understand all of the above but have decided to say 100 just to mess with things.

By: knave

knave — Wed, 09 Nov 2005 14:56:05 -0800

So I can clearly not choose the wine in front of me!

Bwhahahahahaha...

By: Kwantsar

Kwantsar — Wed, 09 Nov 2005 16:19:12 -0800

Are you asking what the probability or likelihood would be of correctly calculating 2/3rds or more of each number selected by all people in the "other" party? I don't understand your question.

No, Rothko. adamwolf, googly, and knave understand the question, and so do I, but I'm soliciting educated guesses for the number that will actually "win."

By: Melinika

Melinika — Wed, 09 Nov 2005 17:25:36 -0800

Last year, a friend of mine posted this question as a Livejournal poll. Here's everyone's answers with a piece explaining the game and a bit of discussion too. It wasn't as many people playing but you might find it useful.

I won it with a guess of 24, but I didn't reason it out much - I just picked my lucky number since it seemed it would be low enough. (I suppose that's "bright, but not really bright"!) The next closest guess was 17.

By: sfenders

sfenders — Wed, 09 Nov 2005 18:11:57 -0800

If only one person guesses over 30, then 0 would lose to 1.

Not only that, but if everyone chooses 0, the perceived value of "winning" by doing the same is probably quite a bit less than the value of doing better than all of them with 1 if it does win.

By: sfenders

sfenders — Wed, 09 Nov 2005 18:17:28 -0800

from Melinika's link:

When the goal was to guess 70% of the average, some of the groups, including those made up of Caltech undergraduates, game theorists, and computer scientists, averaged under 20, while other groups, such as CEOs, 70-year-olds, and Pasadena City College students, averaged over 50.

That's awesome.

By: littleme

littleme — Wed, 09 Nov 2005 18:49:32 -0800

So I can clearly not choose the wine in front of me!

justkevin wins!

By: louigi

louigi — Wed, 09 Nov 2005 22:29:30 -0800

It's more interesting mathematically if you assume all the numbers are distinct. Then it's not immediately obvious what the best strategy is, from a game theoretic point of view.

I'll work on it.... It's a good thing these threads are open for a year now.