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?

(In reply to easy math but laterally??? by David)

I got a slightly different answer.

Let D1 be your dice and D2, D3 be your friend's dices.  Then Prob(winning)=
P({D1>D2, D1>D3} U {D1<D2, D1<D3})

P({D1>D2, D1>D3}
= sum(P(D2<D1, D3<D1 | D1=i)P(D1=i), i=1..6
= sum(P(D2<i) x P(D3<i) x P(D1=i), i=1..6
= sum([(i-1)/6]² x (1/6), i=1..6
= (1+4+9+16+25)/216
= 55/216

By symmetry, P(D1<D2, D1<D3) also equals to 55/216.

Then Prob(winning)=2 x 55/216 = 110/216>0.5

So it's good odds, although very slightly.

Posted by Bon on 2004-08-05 17:07:06