# Perfect strategy for the 2nd player in 'The coins on a table game'

I've just seen this and looked at the duplicate post it directs me to. I'm interested in a point raised in one of the comments in the answer, although it wasn't resolved in that question. It is already established that the perfect strategy for the first player is to place a coin in the exact middle before mirroring their opponent's every move.

However, if the first player has no idea of what they are doing, what is the perfect strategy for the player going second if there is one?

If the first player has no idea of what they are doing, does the second player still have a reasonable chance of winning by playing perfectly?

How badly does a player have to blunder until they would 'throw' the game away to a perfect player? I'm not so sure that only one poorly made move could necessarily achieve this.

Summary of the rules for the game:

This is a 2-player game. You have a large (elevated) rectangular surface, and each player takes turns placing identical (circular) coins on the surface. Stacking is not allowed, and coins already placed aren't allowed to be moved. The player who is not able to place a coin on the surface without it falling off on their turn loses the game.

EDIT: And in response to a witty answer given below, no — you're not allowed to stand the coins on their side.

• I don't think there is one – Beastly Gerbil Jun 1 '16 at 17:54
• Why the down votes? And I would like a convincing proof that there isn't one if that truly is the case. Note that there may be strategies depending on specific cases due to the different places the first player places their coin in or the ratio of the radius of the coin to the side lengths of the table (although I hope it doesn't come down to that). – Shuri2060 Jun 1 '16 at 17:55
• This is an interesting question, though I wonder if it might be better served on Math.SE - I suspect the answer in the general case is not at all trivial, if it even has a reasonable closed form. (Also, is the condition for a coin to fall off "its center does not lie in the rectangle", or is it something else). For what it's worth, if your opponent's first move is such that it would intersect a coin placed at the center, but their coin does not contain the center, then you just play symmetrically through the origin and win. Other cases are less obvious. – Milo Brandt Jun 1 '16 at 17:58
• @MiloBrandt Yes - I wondered if that was the case too, although I thought finding strategies for games would be better suited here. I think it is reasonable to assume that if a coin's centre does not lie on the surface it falls off, although it shouldn't make any difference if you decide that it 'falls off' as soon as any part of it goes over the edge — you just have surfaces of different dimensions in these cases. And yes - that was just the case I was thinking of, although you worded it more elegantly that I did. – Shuri2060 Jun 1 '16 at 18:02
• One thing that occurs to me is that this can be formulated in any metric space - in particular, a legal state of the game is a set of points $P$ inside some set $S$ (here a rectangle) such that $d(p_1,p_2)\geq 1$ for any pair of points $p_1$ and $p_2$ in $P$ and $d$ the distance function, where $1$ is the diameter of a coin. It seems to behave best when $S$ is compact (i.e. a closed rectangle). The one dimension case (i.e. playing on a line segment) is possible to completely characterize with a few tricks. I might turn that into a puzzle here if no one else has asked it. – Milo Brandt Jun 1 '16 at 19:19

Obviously player two wants it to be so that after an even number of coins have been played no more can be. To do that, player two should prevent the forcing of an odd coin game and look for a way to force an even game.

The worst move player two could make would be to place a coin in the exact center on his first turn as that would allow player one to resume the mirroring strategy. At the same time, player two does not necessarily have to block the center move because if player one tries to make the center move on turn 2, he cannot guarantee that he can enforce the mirror strategy because there will be 3 coins already on the table without pairs.

If we consider the maximum number of coins which fit into the rectangular area it is $<- \frac{\pi}{2 \sqrt{3}} * \frac{Area of Rectangle}{Area of Circle}$, and the finite size of the rectangle may reduce it. https://en.wikipedia.org/wiki/Circle_packing

Consider this, at every point in time, I examine the board and count the maximum number of possible additional coins which could be added. Then the game you are proposing is in some ways similar to the old game where you can take 1 or 2 coins from the pile, except that you reduce the maximum number of coins that can be added. By placing a coin in one spot, you reduce the maximum by 1, in another spot you reduce the maximum by 2, in another spot you reduce the maximum by 3, etc.

As far as I can tell, this would be a highly chaotic game where the number you can reduce the maximum by is significantly different with each turn, and highly dependent on previous moves. I cannot see anyway to solve this game perfectly aside from a computer which tries find all possibilities.

If Player One puts the first coin with its center located between $1$ and $\sqrt 3$ radii of the exact center, then Player Two can win by always playing the symmetrical move (rotated $180^o$ about the center from Player One's move). Once the first two coins are placed, it will be impossible to ever place a coin overlapping the center. Thus, after each of Player Two's moves, the board will be symmetrical, meaning any move by player one has a symmetrical move available for Player Two. Eventually, Player One will lose.

If Player One's first coin is anywhere else, then each situation is a special case, and there is no general best strategy for Player Two.