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# Coins in a row

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On a table is a row of fifty coins, of various denominations. Alice picks a coin from one of the ends and puts it in her pocket; then Bob chooses a coin from one of the (remaining) ends, and the alternation continues until Bob pockets the last coin.

Prove that Alice can play so as to guarantee at least as much money as Bob.

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Posted · Report post

Alice can total the value of all the odd numbered coins and for all of the even numbered coins. If the odd numbered coins are greater, her first pick will be the first coin. If the even numbered coins are greater, her first pick will be the last coin. After that, she will always pick a coin from the side that Bob last picked from.

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Posted · Report post

I'm not sure how to prove it, but I think whenever possible she should choose the coin with the largest value of all that exist. If that isn't possible then she should look at the last two coins on each end and pick her coin so that Bob's choices are equal to or less than the coin she picks. Or pick such that Bob's choice are minimized.

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Posted · Report post

I'm not sure how to prove it, but I think whenever possible she should choose the coin with the largest value of all that exist. If that isn't possible then she should look at the last two coins on each end and pick her coin so that Bob's choices are equal to or less than the coin she picks. Or pick such that Bob's choice are minimized.

Rob, I think this would work, because I haven't quickly found a counterexample, and it seems intuitive.

BobbyGo's answer is a proof, and got there first.

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