[aqme28]

Working through this myself...

If you place a coin in the center, then due to symmetry there must be an even number of remaining places for a coin (or you didn't put it in the center).

You can work up from remaining=2 to see that player A always wins with an even remainder.

[tzs]

It is not fully correct to think in terms of even numbers or remainders, because there are an uncountably infinite number of places that a coin can be placed.

It is the right general idea, but don't try to express it in numbers. Just think that for each non-center point P there is a unique corresponding point P' such that the center point is exactly midway between P and P'. That gets you your pairs of places to work with without having to bring numbers into it.

For a coin placed at the center it covers any given non-center point Q if and only if at covers Q's corresponding point Q'. After the first player plays the center we then have that for every point that is not yet covered that point's corresponding point is not yet covered.

Then you have to show that once the center point has been covered by a coin any subsequent coin played that covers a point R cannot also cover R's corresponding point R'.

Then you can proceed with your pairing argument that whenever player 2 covers a set of points player 1 can cover the corresponding points on the next turn, and that after this we have restored the condition that for every point still available its corresponding point is also still available.

[aqme28]

Even though it's uncountable and potentially infinite, it's still even, because of exactly the symmetry relation you're saying.

I just thought that the intuition for an even number of places is a lot easier (for me) to grok.