interesting problem

limitti · #1 10-07-2003, 11:42 AM

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.

slavic · #2 10-07-2003, 01:03 PM

33

slavic · #3 10-07-2003, 01:07 PM

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?

limitti · #4 10-07-2003, 01:59 PM

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.

Copernicus · #5 10-07-2003, 02:02 PM

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?

limitti · #6 10-07-2003, 03:13 PM

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.

crockpot · #7 10-07-2003, 03:32 PM

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.

Festus22 · #8 10-07-2003, 03:41 PM

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.

Bozeman · #9 10-07-2003, 03:54 PM

0

Copernicus · #10 10-07-2003, 03:56 PM

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.