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May 10th, 2018, 04:06 AM   #11
lua
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Yes, I see what you mean.

But, how to "expand" that strategy to greater number of cards (general case)?
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May 10th, 2018, 05:21 AM   #12
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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...
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May 10th, 2018, 08:51 AM   #13
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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.
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June 26th, 2018, 03:39 AM   #14
lua
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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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