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Topic: Ask Marilyn wrong again 


Puzzle: Say you have four specially marked dice. Two players each select one, and the player who rolls higher number wins. The faces are 111555, 2222,66, 333333, and 444400. You're given the first choice. Which die should you choose? Marilyn says "I'll tell you which die is the strongest (meaning that it will score more wins when played against all other dice, and you figure out why it doesn't matter which die you choose!) The answer will appear June 29 at parade.com. Ok, ready? The strongest die is 222266.  This is a nontransitive problem. i.e. A beats B, B beats C, and C beats A. So whoever picks first loses. The 3s beat the 2's. The 4's beat the 3's. The 5's beat the 4's. And the 2's beat the 5's. Martin Gardner has a beautiful nontransitive puzzler in his wonderful "The Colossal Book of Mathematics". 


Yes, if second player correctly analyzes game, that player can always pick a die that will beat your die 2/3 of the time. So it doesn't matter which die you choose. However, if the second player simply chooses a die at random, your (first) choice of the 111555 die will give you a 1/2 probability of winning, your choice of the 222266 die will give you a 14/27 prob. of winning, your choice of the 333333 die will give you a 1/2 prob. of winning, your choice of the 444400 die will give you a 13/27 prob. of winning. Thus, if second player chooses randomly, you should choose the 222266 die, since it's the "strongest" diein the sense defined by Marilyn. I don't see any mistake in Marilyn's answer. 


Well, the problem is clumsily worded as there is no "strongest die". If random play is assumed then it must be stated in the conditions of contest. 


The mail she got after her analysis of the Monty Hall problem was hilarious. 