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Who wins? (Posted on 2004-07-08) Difficulty: 3 of 5
A "friend" offers to play the following game: you throw a die, and he throws two dice. If both his dice are either higher or lower than yours, he wins; otherwise, you win.

First, you reason: out of three dice, one will always be the "middle" one, and only one out of three times it will be mine, so my odds are just 1/3 -- I shouldn't play.

After a while, you realize that you forgot about duplicate numbers. About 50% of the time, all three dice will be different, and then you have 1/3 chance of winning. But on the other 50%, you assuredly win, so the game stands 2/3 in your favor.

It's clear that BOTH lines of reasoning cannot be right, if any. Should you play, or shouldn't you?

Note: you can solve this mathemathically, or you can use "lateral thinking"; can you find both ways?

See The Solution Submitted by Federico Kereki    
Rating: 4.0000 (2 votes)

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Solution Mathematically | Comment 2 of 9 |

The second line of reasoning merely needs mathematical justification and greater accuracy than the numbers being different "about 50% of the time".

All three dice will be different (5/6)*(4/6) = 5/9 of the time, so your probability of winning is (5/9)/3 + 4/9 = 17/27 = 62.96296296296296296... %.

  Posted by Charlie on 2004-07-08 08:41:18
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