Non Random Distribution for "Random" Number Selection
May 20th, 2010 at 5:52:06 PM
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I saw a question from an old mailbag (#205) where it poses the following question:
The answer is 1/3; however, this assumes that people are likely to pick numbers randomly. It is well known that people are more likely to pick numbers 7-9 than 1-3 thereby introducing a distribution instead of randomness. (Incidentally, Nate Silver of FiveThirtyEight used this notion to accuse a pollster of fabricating data http://www.fivethirtyeight.com/search/label/strategic%20vision)
How would this affect probability and how would you go about solving for the optimum number?
Quote:
You are playing a game that involves three people: (a) yourself, (b) your opponent, and (c) a referee. Each of you picks a real number between 0 and 1 in secret. Once all the numbers have been selected, they are revealed. The player who guessed closest to the referee's number, without going over, wins. If you are closer, you win $1. If your opponent is closer, you lose $1. If both players go over, or there is a tie, the game is a tie.
Is there a number you can pick that will maximize your expected return, if the other player picks randomly? What if the other player has a strategy too?
The answer is 1/3; however, this assumes that people are likely to pick numbers randomly. It is well known that people are more likely to pick numbers 7-9 than 1-3 thereby introducing a distribution instead of randomness. (Incidentally, Nate Silver of FiveThirtyEight used this notion to accuse a pollster of fabricating data http://www.fivethirtyeight.com/search/label/strategic%20vision)
How would this affect probability and how would you go about solving for the optimum number?
