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Bux in a Box (Posted on 2007-02-14) Difficulty: 2 of 5
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?

  Submitted by Kenny M    
Rating: 3.0000 (3 votes)
Solution: (Hide)
Yes - the following is a winning strategy - there may be others. Pick a “target number”, such as $50. Pick the first box. If the amount of money in this box is less than the “target”, declare the second box to contain more money than the first. If the first box contains more money than the target, then declare the second box to contain less money than the first. Use this same target repeatedly.

Proof:

Let the boxes contain S (small) and L (large) amounts of money, and your Target=T.

1) If T is less than S and
a) You pick S, you will lose
b) You pick L, you will win
2) If T is between S & L and
a) You pick S, you will win
b) You pick L, you will win
3) If T is greater than L and
a) You pick S, you will win
b) You pick L, you will lose

For the first and third cases, you win 50% of the time. For the second case, you will win all the time. Believe it or not, this works as a winning strategy even if you pick a target quite far from the mean, such as 10 or 90. Only for very extreme values of the target, such as 5 or 95 do you sometimes lose.

Comments: ( You must be logged in to post comments.)
  Subject Author Date
Some ThoughtsWhat about pennies?Steve Herman2007-02-19 12:02:09
re: solution -- overall chances of coming out aheadCharlie2007-02-14 11:23:53
SolutionsolutionCharlie2007-02-14 10:38:11
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