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Coin contest (Posted on 2015-11-27) Difficulty: 3 of 5
The coins of various denominations are arranged in a chain (e.g., a row, a semicircle) with two "end points" and the others each having two neighbors.
A move consists in removing an end coin, whereby the coin's immediate (and the only) neighbor becomes naturally an end coin available for the removal on one of the successive moves.
The players take turns, 1st defined randomly.
When all the coins have been removed, the game ends, and the players count their bounties. The player with the larger amount wins.

Devise a winning strategy for one of the players.

No Solution Yet Submitted by Ady TZIDON    
Rating: 5.0000 (1 votes)

Comments: ( Back to comment list | You must be logged in to post comments.)
Simple case | Comment 1 of 5
If one coin is worth more than all the other coins combined then there are two possibilities.  
1.  The high value coin in on one end.  Player 1 wins. 

2.  The bush value coin is not on one end.  Player one will win if there are an even number of coins.  Player 2 will win if the number of coins is odd.  Recursive proof: consider 3 coins. Player 2 wins.  Adding a coin adds a move to bring it to this state. 

  Posted by Jer on 2015-11-27 16:19:13
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