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First or second? (Posted on 2017-05-23) Difficulty: 4 of 5
Two players alternatively erase some 9 numbers from the sequence 1,2,...,101 until only two remain. The player that starts wins x−54 dollars from the player that plays second, x being the absolute value of the difference between the remaining two numbers.

Would you rather be the first or the second player?
Explain your decision by providing your strategy.

See The Solution Submitted by Ady TZIDON    
Rating: 3.0000 (1 votes)

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Solution Winning Strategy (spoiler) | Comment 1 of 6
The last player to go is the one who is in position to maximize the difference between the two remaining numbers.  In this case, the 1st player has 6 turns and crosses out 54 numbers, while the 2nd player has 5 turns and crosses out 45 numbers.  So the 1st player is also the last player, and is therefore in control.  

Strategy:

Player 1, Turn 1:  Player 1 makes a neutral move, crossing out the central numbers from 47 to 55.  Note that this leaves 46 numbers on each side of the crossed out center. Player 2, who only crosses out 45 numbers, cannot cross out all of them in his 5 turns.

Player 2, Turn 1: Player 2 can do no better than crossing out numbers on one or both ends.

Player 1, Turn 2: Player 1 crosses out numbers adjacent to the central block, but on the side opposite the side that player 2 has played.  For instance, if Player 2 on his first turn crosses out 6 numbers on the LEFT and 3 on the RIGHT, then player 1 responds by crossing out 3 on the LEFT (44-46) and 6 on the RIGHT (56-61).  The central block is now 18 wide, at least.  If player 2 made the mistake of playing, for instance, on 60, then player 1 plays on 62 instead of 60, and the block which crosses the center is 19 wide.  Player 2 will regret his mistake.

etc.

At the end, the central block will be at least 54 numbers wide. There will be 1 number remaining on each side of the central block, so they are at least 55 apart.  So Player 2 owes at least 1 to Player 1.  If Player 2 has erred and crossed out n numbers between the 2 remaining numbers, then he owes (1+n) to Player 1.

So I would prefer to be Player 1.

Note that if the payoff were x-56, then Player 1 would still be in control, but the best that he could do would be to limit his losses to 1.  With that payoff, I would rather be Player 2.  So, control is not everything.

Edited on May 23, 2017, 8:32 pm
  Posted by Steve Herman on 2017-05-23 20:16:43

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