interesting problem
 

Not exactly poker problem but I thought someone might find this interesting...

A game where you must choose a number between 0-100. There is 20 people and the winner is the one who chooses the number that is two thirds of the arithmetic mean of all the chosen numbers (or closest to it). What number should you pick to have highest chance to win? Does it depend on the number of participants?

I don't know the answer and I can't figure out how to calculate it but I found the problem interesting.

Re: interesting problem
 

33

Re: interesting problem
 

wait a second then everyone should chose 33 and thus your answers should be 22. But you then sku hte thing and everyone should pick 22 and 15 becoms correct then ten then 6 or 7, then 4, 2, 1.

So do our participants know how the game will be evaluated?

Re: interesting problem
 

Yes they do know. I really don't understand how one should get a solution because how would it be possible to know how others think. A university professor told us to pick a number what we thought was the best choice in a class today and he is going to tell the right answer thursday and also tell who won. I am just wondering if there is any right answer.

Re: interesting problem
 

Are you sure individuals choose the numbers, not random like a lottery, and you are trying to come within 2/3 of the mean?

Alternatively are the numbers chosen with or without replacement?

Re: interesting problem
 

Individuals choose numbers and everyone of them has a goal to choose a number which is 2/3 of a mean of all chosen numbers. No replacements.

Re: interesting problem
 

if everyone is playing competently, they will all pick the same number, so there is no correct answer unless we assume some of them are not playing ideally.

but how are they going to play incorrectly? we don't know, and thus the problem cannot be answered given the information here.

Re: interesting problem
 

If you are choosing without replacement, then the lowest winning number possible is 2/3 of the lowest mean possible, which is 9. So you would pick 6. Knowing that, you would not want to pick 0 - 5 which then increases the mean. So now, you're going the other way and assume 6 - 26 will be chosen. You could iterate the whole process again ultimately ending up choosing 80 - 100 but that can't be right so you start down again. So I'll take a stab and say 50 is the best guess.

Re: interesting problem
 

0

Re: interesting problem
 

given that there is no replacement there will be a winning strategy for every number of players > n. (Eg with 90 players there is a range from 30-37 for 2/3 of the mean, so picking one of those numbers gives you 8/90 chance of winning.) I dont believe there can be a winning strategy for as few as 20 players, where 2/3 of the mean can range from 7 to 60.

n is around 42 or 43. Of course you have to know you will get to pick your number sometime in the first k players, where k is the spread from min to max values of 2/3 * mean.

Re: interesting problem
 

yep

Zero,
 

of course. This is not a math problem; it's a logic problem. As each player tries to anticipate what the other 19 players will choose, they will constantly adjust their selection downward until reaching the lowest possible number. In this case that is zero.

BR

Re: Zero,
 

not with the stipulation that numbers are chosen without replacement.

replacement/no replacement
 

English is not my first language, I think I understood the word replacement wrong. However I meant that everyone picks their number without knowing what the others pick and so people may pick same numbers.

Re: Zero,
 

but if everyone thinks about what everyone else is thinking before they pick, like all those poker books tell us to, then everyone will pick zero. if they just guess then god knows what will happen. this solution assumes thinking, competent opponents.

The answer is 16
 

Here's what will happen.

You will choose the right answer, 16.
4 dummies will pick random numbers, so they will avg 50.
5 people will decide that the avg is 50 and 2/3 of that is 33, so they will pick 33.
5 people will think that everyone will pick 33 for the reason above, so they will pick 22.
The remaining 5 people will think that everyone will adjust downward and thus all end up at 0. So they will pick 0.

1*16 + 4*50 + 5*33 + 5*22 + 5*0 = 491
491/20 = 24.55
(2/3)*24.55 = 16.4 close enough

Lots of people are dumb. And lots of people who are very smart in a lot of other ways are pretty bad at math and logic problems, thus will not come up with any rational answer here. The key to solving this problem is having a good grasp on exactly how dumb a given sample of people are.

Re: replacement/no replacement
 

"without replacement" means no two can pick the same number.

If the game were "with replacement", ie everyone can pick the same number, then the game is pointless until you get few enough players (n) that one player can influence 2/3 of the mean by an amount greater than n. For example, at the extreme of two players, one player can shift the mean to anywhere from 1 to 50 if the other player picks 0, and from 25 to 75 if he picks 50. There would be no guaranteed +EV strategy though, since you are always on a guess as to what the other player will pick, and zero is not automatic. Zero becomes automatic (and break even only) when n is greater than the critical point, ie when one player can no longer influence 2/3 of the mean by > 1/n.

Re: The answer is 16
 

I chose number 15 and my thinking was pretty much the same than yours. This sample is not very dumb though, all are university economics students. Still you would be surprised how simple some of them are:)

Re: interesting problem
 

If there's no replacement, and two people pick the same number, do they just throw away one person's entry? There doesn't appear to be any arbitration scheme or way for the second person to submit a second guess.

Re: interesting problem
 

please read my post "replacement/no replacement" in this same thread.

Re: The answer is 16
 

[ QUOTE ]
The key to solving this problem is having a good grasp on exactly how dumb a given sample of people are.

[/ QUOTE ]

Heh. Just like so many other things in life.

Re: The answer is 16
 

Sorry, I dont buy it.

Re: interesting problem
 

Since we have no replacement of numbers the number of players should change the number.

So if we have 100 players 33 is correct.

With 20 players the number could float between 13 and 60ish So the greater the number of players the more correct 33 is and the fewer the players the more the range opens.

In a 2 person game it woulds always be correct to choose 0.

Re: interesting problem
 

The answer will be around 16. I would say most of the kids in the class will not care and just write a number down at random getting you a mean of 50. 2/3 of that is between 16-17. Some people will try to figure it out and will pick a lower number which will lower the answer slightly. I might adjust my answer down to 10-15....If the class thinks like me then the answer would be 5 or 6 entire. If you go to one of those smart school the it could be 0 or 1.

Irony
 

[ QUOTE ]
The answer will be around 16. I would say most of the kids in the class will not care and just write a number down at random getting you a mean of 50. 2/3 of that is between 16-17. Some people will try to figure it out and will pick a lower number which will lower the answer slightly. I might adjust my answer down to 10-15....If the class thinks like me then the answer would be 5 or 6 entire. If you go to one of those smart school the it could be 0 or 1.

[/ QUOTE ]

Re: interesting problem
 

Response from an Economist who has done this in a classroom setting:

"I actually have had students do this a few times. Usually the winning
number is close to two-thirds of two-thirds of half the maximum
allowed answer. So when I said they could give me any number between
1 and 100, The average answer was around 50. Two-thirds of 50 is 33.
The last two times I played this game the winner guessed something
close to 22 or thereabouts. The answer reflects what the respondent
thinks other players are thinking. Apparently most people thought the
rest of the group would have no idea and guess 50, right in the
middle; so a lot of people wrote down two-thirds of 50. But you want
to guess two-thirds of the average guess, and that is around 22 if
most people guess 33. Weird. Anyway if everyone is "rational" in the
game-theoretic sense then the nobody really wins, everyone ties for
first-place by guessing zero: 2/3 * 2/3 * .... = 0."

Re: The answer is 16
 

</font><blockquote><font class="small">In risposta di:</font><hr />
The key to solving this problem is having a good grasp on exactly how dumb a given sample of people are.

[/ QUOTE ]


this is the solution. congratulations ulysses!

Re: interesting problem
 

</font><blockquote><font class="small">In risposta di:</font><hr />
...getting you a mean of 50. 2/3 of that is between 16-17.

[/ QUOTE ]

1/3 of 50 is between 16-17. not 2/3