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May 7th, 2018, 11:26 AM   #1
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Players with enumerated cards

Quote:
 On 2018 cards, the whole numbers from 0 to 2017 were printed. Then the cards are placed on the table in an arbitrary order. Players A and B alternately take one card, but at the same time they can only take one of the two end cards. The game begins player A. The game ends when all the cards are taken, and the winner is the player with greater sum of the numbers on the cards taken. Prove that one of the players has a winning strategy. Which one ?
I tried the strategy of taking always maximumum of end numbers but didn't manage to solve the problem.

 May 7th, 2018, 03:09 PM #2 Senior Member     Joined: Sep 2015 From: USA Posts: 2,165 Thanks: 1139 If say player A forces both end cards to be taken is it then B's turn to choose?
 May 8th, 2018, 02:49 AM #3 Member   Joined: Jan 2018 From: Belgrade Posts: 54 Thanks: 2 I'm not sure I understand what you mean. I'll try to make it clear. Cards are placed in front of the players in one line from left to right (random order). Player A starts. In one move, he can take only one card: the one from the left end OR the one from the right end of the line. When he takes his one card, it is player's B turn. He can do the same as player A: take only one card: the one from the left end OR the one from the right end of the line. When he takes his card, it is player's A turn, and so on. The game is over when all the cards are taken. I hope it's clearer now. Last edited by skipjack; May 10th, 2018 at 01:01 AM.
 May 8th, 2018, 05:14 AM #4 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,451 Thanks: 946 Is this example with 6 cards correct? 3 6 1 4 2 5 : cards shuffled, then placed in a row face down; A starts: 3 6 1 4 2 A : A = 5 3 6 1 4 B _ : B = 2 A 6 1 4 _ _ : A = 5+3 = 8 _ 6 1 B _ _ : B = 2+4 = 6 _ 6 A _ _ _ : A = 8+1 = 9 B is left with last card: B = 6+6 = 12 : the winner! Or are the cards face up????? If not, makes no sense! Last edited by Denis; May 8th, 2018 at 05:21 AM.
 May 8th, 2018, 05:25 AM #5 Member   Joined: Jan 2018 From: Belgrade Posts: 54 Thanks: 2 Yes, that's it. Only one thing. In the text of the problem it is not said that cards are faced down (so that players can not see number on cards), but it is not said the opposite either. So I suppose that players can see numbers on cards.
 May 8th, 2018, 05:31 AM #6 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,451 Thanks: 946 Well, they MUST be able to see the numbers... else problem makes no sense. Can you get a better/clearer translation of the problem?
May 8th, 2018, 06:07 AM   #7
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I'll try to translate it better:

Quote:
 The whole numbers from 0 to 2017 were printed on 2018 cards. Then, the cards are placed on the table in one line in an arbitrary order. Players A and B alternately take one card, but at one time they can only take one of the two end cards. The game starts with player A. The game ends when all the cards are taken, and the winner is the player with bigger sum of the numbers on the cards he has taken. Prove that one of the players has a winning strategy. Which one?
I believe it's not much better.
But with your example, you've got the point.

 May 9th, 2018, 04:45 PM #8 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,451 Thanks: 946 Been playing around with this... Sticking to my example: these 2 appear to be the "worse case" for A: 1 5 3 4 6 2 and 1 5 4 3 6 2 However, A will win both with score 11 to 10: do you agree?
 May 9th, 2018, 11:21 PM #9 Member   Joined: Jan 2018 From: Belgrade Posts: 54 Thanks: 2 Yes, but strategies of A are different in these two cases. In the first case, A takes 2 in his first move, and in the second case, he must take 1 to have a winning stategy.
 May 10th, 2018, 03:57 AM #10 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 13,451 Thanks: 946 Agree. BUT can we not say that the "strategy" is the SAME: A plays such that he gets the 4 (higher of 2 last cards) in both cases? I guess it depends on what "strategy" means.

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