I have set up a stall where you may play a fabulous game.
 
Presented before you are 4 unusual 6-sided dice:
 

A Big Red die:
has the numbers 5, 5, 5, 5, 1 and 1 on its sides

A Large Yellow die:
has the numbers 6, 6, 2, 2, 2 and 2 on its sides

A Medium Green die:
has the numbers 6, 4, 4, 2, 2 and 1 on its sides

A Small Blue die:
has the numbers 3, 3, 3, 3, 3 and 1 on its sides
 
I inform those that are unaware that the average value they would roll with each of the 4 dice are (roughly) 3.66, 3.33, 3.17 and 2.67 respectively. All dice are fair and players find it impossible to cheat when rolling them.
 
I request a $1 payment from you to play. You may choose any one die. Then I may choose any of the remaining dice. We then roll. If you roll more than or the same as me, I return your original $1 stake and a bonus $1 prize. If I score more than you, I keep your stake and you win nothing.
 
What would be your strategy if you wanted to walk away from my stall with the most amount of money possible?
 
(Thanks go to an old university professor would showed us something similar, which instantly intrigued me, in a Probability lecture)

(In reply to Oops by TomM)

Well, seeing as I didn't say you had to play at all, there's obviously one better strategy for ending up with more money in the long run... ;-)

Actually, I didn't notice the 24 either... *embarassed look*
I just found it really interesting at the time that such a combination could exist where, no matter on the die chosen, there was always another that could beat it - such a situation had never occurred to me before I was shown that...

Posted by Nick Reed on 2002-07-24 03:59:33