Probably not transitive: Difference between revisions
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I first heard a similar version of this puzzle on the {{Math | [[File:Probably_not_transitive.png|right|500px]] | ||
I first heard a similar version of this puzzle on the {{Math Factor}}. | |||
==Puzzle== | ==Puzzle== | ||
Consider the following card game: the 2 through 10 of spades are placed in three piles of three cards each. Player 1 picks any pile, then player 2 picks one of the remaining piles. Each player picks one of their three cards at random - high card wins. Now in the piles pictured | Consider the following card game: the 2 through 10 of spades are placed in three piles of three cards each. Player 1 picks any pile, then player 2 picks one of the remaining piles. Each player picks one of their three cards at random - high card wins. Now in the piles pictured, player 1 has a winning strategy: pick pile A, as any card in that pile would be a winner. But if the cards were arranged differently, things might not be so simple. | ||
Find an arrangement of the cards into three piles so that player 2 has a winning strategy. That is, player 2 can always pick a pile that has a greater than 50% chance of beating the pile picked by player 1. Or prove that this is impossible. | Find an arrangement of the cards into three piles so that player 2 has a winning strategy. That is, player 2 can always pick a pile that has a greater than 50% chance of beating the pile picked by player 1. Or prove that this is impossible. | ||
Current revision as of 18:59, 13 July 2013
I first heard a similar version of this puzzle on the The Math Factor.
Puzzle
Consider the following card game: the 2 through 10 of spades are placed in three piles of three cards each. Player 1 picks any pile, then player 2 picks one of the remaining piles. Each player picks one of their three cards at random - high card wins. Now in the piles pictured, player 1 has a winning strategy: pick pile A, as any card in that pile would be a winner. But if the cards were arranged differently, things might not be so simple.
Find an arrangement of the cards into three piles so that player 2 has a winning strategy. That is, player 2 can always pick a pile that has a greater than 50% chance of beating the pile picked by player 1. Or prove that this is impossible.