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 May 10th, 2018, 04:06 AM #11 Member   Joined: Jan 2018 From: Belgrade Posts: 55 Thanks: 2 Yes, I see what you mean. But, how to "expand" that strategy to greater number of cards (general case)?
 May 10th, 2018, 05:21 AM #12 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,117 Thanks: 1003 Well, that's too complicated for li'l old me! Perhaps apply to the whole 2018 numbers, in "sections of 6", the 3 numbers at both ends being the "sections"? That would leave 2 numbers at the finish line (336*6=2016), but A picks 1st so no problems...
 May 10th, 2018, 08:51 AM #13 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,117 Thanks: 1003 Thought about it some more during my walk! I'm now convinced my previous post is the solution. General case: 1: show A always wins a section of 6 numbers 2: treat game in sections of 6 numbers (3 from both ends) 3: if 1 or 2 or 3 or 4 or 5 numbers left, A wins since he goes first Note: if 4 left, there's a tie possibility for those 4, but A wins the game anyway since he's always at least 1 point ahead. EDIT: still easier: no need to worry about "ending" A will be ahead once passed the halfway point! Last edited by Denis; May 10th, 2018 at 08:55 AM.
 June 26th, 2018, 03:39 AM #14 Member   Joined: Jan 2018 From: Belgrade Posts: 55 Thanks: 2 I've been thinking about this problem. I believe we could weaken the condition that player $\displaystyle A$ must always win in every section of $\displaystyle 6$ numbers. He could win the game even if he plays draw some sections. In the end, even if $\displaystyle A$ plays draw with $\displaystyle B$ in every section of $\displaystyle 6$ (or some other number) numbers, he still can win the game because there are $\displaystyle 2$ numbers in the end and he is on the move. Thus, we can show much easier that $\displaystyle A$ wins or plays the draw in every $\displaystyle 4$ number section (then to show $\displaystyle A$ wins every $\displaystyle 6$ number section). And because $\displaystyle 2018=4 \cdot 504 +2$, there are $\displaystyle 2$ numbers left and $\displaystyle A$ is on the move. So $\displaystyle A$ has the winning strategy. Last edited by lua; June 26th, 2018 at 03:43 AM.
 January 4th, 2019, 03:24 AM #15 Member   Joined: Jan 2018 From: Belgrade Posts: 55 Thanks: 2 I've (read) found very elegant solution. First, we see that sum of all card numbers ($\displaystyle S$) is odd (because there is odd number of odd card numbers). Now, we enumerate card positions with numbers from 1 to 2018. Let's sum numbers on cards positioned on even positions ($\displaystyle S_E$), and numbers on cards positioned on odd positions ($\displaystyle S_O$). We have that $\displaystyle S_E+S_O=S$, and $\displaystyle S$ is odd. So, it must be $\displaystyle S_{E}\neq S_{O}$. 1. Let $\displaystyle S_{E} > S_{O}$ Player $\displaystyle A$ will take card from even position and will keep doing so until the end of the game. He can do so, because player $\displaystyle B$ will be forced to take card on odd position in every turn. So, in the end, player $\displaystyle A$ wins, because he takes all cards on even positions. 2. Let $\displaystyle S_{O} > S_{E}$ In this case, player $\displaystyle A$ takes card from odd position and keeps doing so until the end of the game. Doing so he wins in this case too. Last edited by lua; January 4th, 2019 at 03:29 AM.

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