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"Newspaper Beauty Contest"

Joined
3/5/11
Messages
34
Points
18
Saw an interesting study quoted in a paper where they asked this question to about 1000 subjects:

"Pick a number from 0 to 100. The winner is the person who chooses the number closest to 2/3rds of the group's average response."

There is a "rational" answer.
 
Each number (between 0 and 100) has a prob of 1/101 of getting picked. Expectation of this game is for each number to be picked 1000/101 times. E(x)=sum (0*1/101+1*1/101+2*1/101+....+100*1/101)=(0+1+2+...+100)/101=((100+0)/2 * 101)/101=50.
Assuming all subjects want to win, they will all pick 2/3 of 50 (because each assumes 50 to be the average). but that will change the answer to 2/3 of 50, or 33. Then all "rational" subjects should pick 2/3 of 33, or 22....
This loop will not end until we reach a number smaller then 1. At then end, Assuming all subjects are "rational" and all subjects assume all subjects to be "rational" too and to want to win (as opposed to subjects picking numbers randomly), all should pick "0", and all will win.
 
I like Alexandre's answer.

It's worth adding the fact that people don't choose numbers randomly. A particular issue is the way they choose birthdays in these, so that 0 is rare, but then 1..28 is more common, 29-31 a but less likely then 32 to 100 less likely still.

This helps one in games such as lotteries, since although you can't alter your chances of winning, you can reduce the number of people you share it with.
 
By the way, for my analysis to work in the case above, the basic assumption is for ALL subjects to be thinking alike, which is never the case.
I read somewhere that 7 is the most common number chosen between 1 and 10 which is weird. I wonder if 7 has a higher probability than most numbers in the 0-100 game.
 
We don't always think rationally... 7 is considered lucky by many so when they have a choice to select between 0-10 ,7 is selected with highest frequency.....
 
Alexandre is right, the rational answer is 0. It may seem like an aggressive bet because your assumption is that the other players are rational too, but it is a less risky assumption than it may seem.

Bonus question: assume that a portion of the players see the rational solution and choose zero. The rest of the players are randomly distributed among the remaining numbers, 1 to 100 (we are assuming nothing about the rationality of the remaining players, so they are free to do something ignorant like pick 100). How big does the zero-choosing portion have to be in order for zero to be the correct answer with P = 0.5?
 
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