
Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion 
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May 10th, 2018, 04:06 AM  #11 
Member Joined: Jan 2018 From: Belgrade Posts: 46 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: 12,771 Thanks: 862 
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: 12,771 Thanks: 862 
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: 46 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. 

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