You are playing the following game. You are offered two closed boxes, each having a random amount of money between $0 and $100 inside. After picking one and noting the amount of money inside, you are asked to state whether the other box contains more or less money. If you are right, you win $1. If wrong, you lose $1. If you play a large number of times, is there a strategy where you can be almost certain to leave with more money than you started with?

Since each box is filled randomly and independently, if you look in a box where the amount is under $50, then the likelihood is that the other box contains more money. If the first box had over $50, the other box likely contains less money.

First box    prob other box correctly predicted

  0              100/101

  1               99/101

  2               98/101

 ...

 49               51/101 

 50               50/101

 51               51/101 

 ...

 98               98/101

 99               99/101

100              100/101

As the numbers for the first box are equally likely, we only need to add up the probabilities in the second column and divide by 101 to get the overall probability of winning, which is 7600/10201 ~= 0.7450249975492598764, so you'd win almost 3/4 of the time. You'd lose 0.2549750024507401235 of the time, for a net expected gain of 0.4900499950985197529 for every game played.

Posted by Charlie on 2007-02-14 10:38:11