# Relatively basic semi-theoretical semi-economic type question.November 9, 2005 1:34 PM   Subscribe

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?
posted by Kwantsar to Grab Bag (21 answers total)

What is the goal of the other participants?
posted by fvw at 1:42 PM on November 9, 2005

Response by poster: They have an identical goal.
posted by Kwantsar at 1:45 PM on November 9, 2005

This is a game theory question. The answer is 0.
posted by milkrate at 1:46 PM on November 9, 2005

Response by poster: (A goal identical to mine)
posted by Kwantsar at 1:47 PM on November 9, 2005

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.
posted by knave at 1:48 PM on November 9, 2005

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).
posted by googly at 1:49 PM on November 9, 2005

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.
posted by Rothko at 1:49 PM on November 9, 2005

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.
posted by adamwolf at 1:49 PM on November 9, 2005

On postview, milkrate agrees. Cool.
posted by knave at 1:49 PM on November 9, 2005

What are they trying to teach with this exercise?
posted by raedyn at 2:20 PM on November 9, 2005

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!
posted by justkevin at 2:42 PM on November 9, 2005

their = they're
posted by justkevin at 2:43 PM on November 9, 2005

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.
posted by j.edwards at 2:43 PM on November 9, 2005

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.
posted by vacapinta at 2:47 PM on November 9, 2005

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

Bwhahahahahaha...
posted by knave at 2:56 PM on November 9, 2005

Response by poster: 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."
posted by Kwantsar at 4:19 PM on November 9, 2005

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.
posted by Melinika at 5:25 PM on November 9, 2005

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.
posted by sfenders at 6:11 PM on November 9, 2005

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.
posted by sfenders at 6:17 PM on November 9, 2005

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

justkevin wins!
posted by littleme at 6:49 PM on November 9, 2005

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.
posted by louigi at 10:29 PM on November 9, 2005

« Older Shuffle, damn you! Shuffle!   |   Photo Holiday Cards Newer »