Elementary Math Fractions, Percentages, Word Problems, Equations, Inequations, Factorization, Expansion

 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: 14,587 Thanks: 1038 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: 14,587 Thanks: 1038 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. Tags cards, enumerated, players Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post Aavar Probability and Statistics 0 April 20th, 2017 04:38 AM rexden1 Advanced Statistics 2 October 2nd, 2011 12:01 AM johnmath Applied Math 2 June 15th, 2011 04:40 PM fathwad New Users 2 May 8th, 2007 07:55 PM

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