Choices and Coins (June 2022 challenge question)

There is no way to assure a win, the variability of the spacing between coins negates any plan you might have, like placing your coin in the center and mirroring your opponents moves...the table may be symmetrical, and the coins may be symmetrical individually, but the placement of the coins creates a non-determinant variable...

Good answer.

If you go second you have to hope that your opponent will at some point place a coin which partially covers the centre position.

This is the admin's answer as well. There is only on position that cannot be countered with an equivalent move.

Go first and take the very center of the table. Thereafter, mirror your opponent's move on the other side of the table, if axes are applied across the center length and center width of the table. Your opponent will be the first one to run out of unique spaces to place a coin.

I don't believe you can guarantee a win, but here's a strategy that would probably work against most opponents.

Try to create an even number of "holes" big enough for an odd number of coins.

Your "holes" need to be one coin "wide", n coins long and "oblong". What do I mean by oblong: I mean that you could slide a coin from one end to the other: if a "hole" is shaped so that you can't slide a coin past two intrusions, then it is two holes. For example it's easy to see how you could create a hole big enough for 3 coins, but that you cannot fit one so that there is only enough space for one more: this space is really just three one coin holes.

As the table fills look for opportunities to create pairs of odd sized holes.

At the end of the game it's easy to see how this works for an even number of single holes. For triple holes: if he places a coin in one slot, you place another in the same hole; if he places a coin over two slots, you do the same in another hole.

Etc. Etc.

If he figures out your strategy, or, is working on a similar one, before the shapes on the table become "definable" then it's anyones game. The trouble is that it will get to point where placements will become critical to within tiny fractions of a millimetre: the only way to reasonably play would be on a digital table placing coins by the coordinates of their centres.

I cant explain why mathematically,but intuitively I think it must always be an even number of coins,so going 2nd is the winning strategy.Even going down to the molecular size should not matter.Of course,going up in size to fill the sides would limit it to 1,but I am excluding this from my guess.

Just a SWAG on my part without actually proving it with math.

Shippers never ship or store an odd number of round articles without a spacer somewhere,and I am sure they have done the math on this.
Example:6 pack,12 pack,etc.

Oil used to come in round containers till someone realized how much space they were losing.Eventually,all foods may come in square-sided containers with rounded corners,like gallon milk or water bottles.The radius and thickness of the corners will be the minimum possible to endure the stresses involved. I leave it to industry to figure the best way,and I am sure they have.

A wild thought is:Tilt the table,and vibrate it by tapping on the bottom.This will arrange the quarters into the tightest possible arrangement.No math needed.

(Presuming there is an edge to prevent the coins from slipping off of the table.)

Shippers never ship or store an odd number of round articles without a spacer somewhere,and I am sure they have done the math on this.

24 beers in a case. 24 hours in a day. Coincidence? I think not!

The number of quarters on the table could be even or odd. If it is odd, the first player wins, and if it's even the second player wins.

The first player forces it to be odd by placing the first quarter exactly in the center and by countering each quarter placed by his opponent with one directly opposite, thus maintaining symmetry.

This works for a rectangular table, a square table, a regular hexagon table, a round table, etc.

This strategy doesn't work for an equilateral-triangle-shaped table or any regular polygon with an odd number of sides.

My guess is that it is necessary for there to exist 2 perpendicular axes of symmetry to the table for it to work.

In this case, you certainly could have an odd number. The square of an odd number equals an odd number. You could have 9, 25, 49, 81, 121, 169, 225, 289, etc.

I stand corrected.Sometimes the gut feeling is not right ,(except when you are hungry).