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Topic: Ask Marilyn wrong again
Message: Posted by: jammen (Jun 28, 2015 05:02PM)
Puzzle: Say you have four specially marked dice. Two players each select one, and the player who rolls higher number wins. The faces are 1-1-1-5-5-5, 2-2-2-2,6-6, 3-3-3-3-3-3, and 4-4-4-4-0-0. 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 2-2-2-2-6-6.
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This is a non-transitive problem. i.e. A beats B, B beats C, and C beats A. So whoever picks first loses. The 3-s 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 non-transitive puzzler in his wonderful "The Colossal Book of Mathematics".
Message: Posted by: stanalger (Jul 1, 2015 05:30PM)
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 1-1-1-5-5-5 die will give you a 1/2 probability of winning,
your choice of the 2-2-2-2-6-6 die will give you a 14/27 prob. of winning,
your choice of the 3-3-3-3-3-3 die will give you a 1/2 prob. of winning,
your choice of the 4-4-4-4-0-0 die will give you a 13/27 prob. of winning.
Thus, if second player chooses randomly, you should choose the 2-2-2-2-6-6 die, since it's the "strongest" die--in the sense defined by Marilyn.

I don't see any mistake in Marilyn's answer.
Message: Posted by: jammen (Jul 2, 2015 08:57AM)
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.
Message: Posted by: LobowolfXXX (Oct 6, 2015 04:13PM)
The mail she got after her analysis of the Monty Hall problem was hilarious.