# Probability in the Wild

This isn’t really “statistics” or “in the wild” since it’s a problem from an advanced probability class I am taking, but it’s a neat problem so you’ll just have to live with.

Here is the game:
I get a die and the opponent gets a die. (These dice are not necessarily numbered 1,2,3,4,5,6. The sides could be numbered 1,1,1,3,3,3, for example, but the dice has six sides.) We both roll. If you’re number is higher than my number, you win a dollar. If my number is higher than your number, I win a dollar.

Can you design three six sided dice in such a way, that if the opponent chooses his die first, I can always choose a die from that remaining two that makes my expected winnings positive?

STOP HERE IF YOU WANT TO TRY THIS YOURSELF!

You probably didn’t even try. So here’s an answer.
Here are three possible dice:
A: 0 0 6 6 6 6
B: 4 4 4 4 4 4
C: 2 2 2 2 8 8

Dice A will beat Dice B with probability 2/3.
Dice B will beat Dice C with probabilty 2/3.
Dice C will beat Dice A with probabilty 5/9.
(Pr(C>A)=P(C>A|A=0)*P(A=0)+P(C>A|A=6)*P(A=6)=1*1/3+1/3*2/3=5/9)

So if I let my opponent choose first, I can always make a choice that will make my expectation positive.

The moral:
Make these dice, head to a bar, and fleece some drunks ( Just don’t tell them I sent you). I guess that’s the “in the wild” part.

Cheers.

Posted on March 11, 2009, in Uncategorized. Bookmark the permalink. 1 Comment.