×   Home   Blog   Newsletter   Privacy   Contact Us   About

Coin on an infinite Chessboard

There’s a classic puzzle (I think I first read about it in Professor Povey’s Perplexing Problems), about throwing a circular coin onto an infinite chessboard.
The question asks, if you throw the coin, at random, what is the probability that the coin will land inside a single square (not span over any of the lines on the board).
In the examples below, all the coins in green fulfill this requirement, those in red fail.
Clearly it depends of the diameter of the coin, and the spacing of the grid. We’ll define the radius of the coin to be r and the spacing between the lines of the square grid to be s.
If the coin has a diameter larger than the grid (r > s/2), then it’s never possible: However the coin lands, it’s going to span at least one square. (If the diameter were exactly equal to spacing, then it would require infinite precision to not touch a side).
With a little thinking we can see that there is a square in the middle of each grid, such that, if the coin were centered inside this square, it would not touch the edge. This square is r away from the edge of larger square. (We can describe the inner square as the locus of points).
The inner square has a length of edge t = s – 2r.
If we assume that throwing the coin at random results in a point anywhere in the large square being equally likely, then the probability of if landing with its centroid inside the smaller square is simply the ratio of the area of the smaller square over the larger square.
This is another one of those carnival games that is not a good bet. If the square has a spacing twice the diameter coin (seems fair?), and you win by tossing your coin so that it does not cross an edge, what are the chances of you winning?
Setting r = 1 (unit width coin), and s = 4 (a diameter is twice the radius), then we see the chance is just 25%. Not a good bet if they pay even money on success.
What ratio would be needed to make it an even game? For a coin of unit radius r = 1, then to get Pr to 0.5 we need to have a grid with spacing s = 2(2 + √2) ≈ 6.828 - almost seven times! (a spacing of 3.414× the diameter).
This is what this ratio looks like:
To make the game fair (a 50% chance), the spacing of the grid needs to be almost seven times the radius of the coin!
I don’t know about you, but if we didn’t trust the math, this grid seems far larger than appears necessary. You can see how the carnival games, which use larger relative sized coins, scam money off innocent players.
Here is a graph of the percentage probability of successfully landing inside a square plotted against the ratio of the grid spacing to coin radius:


Carrying on this theme, what is the probability that the coin will land over a corner?
This is a little more complicated. What we’re looking for is the chance that center of the coin lands close to the corner of the grid. Through symmetery, just looking at one grid square, we can see there are four corner sectors that enable success.
There are a couple of cases to consider. The first is when the coin is relatively small compared to the grid. If the coin has a radius less than half the grid spacing (r < s/2) then there are no overlapping regions.
There are four quarter circles, each of area (πr2/4), and the area of the square is just s2, so the probability is:
That’s right, you can estimate Pi by throwing coins over a grid! For more information about this concept, read this article about Buffon’s Needle.
In the example where the grid spacing gives a 50% chance of landing in the middle s = 2(2 + √2), the probability of landing over a corner comes out to be approximately 6.7%
Things get a little more complicated as the radius of the coin increases. When the coin is greater than half the grid spacing (s/2 < s < √2/s), the loci overlap, and we have to do a little Boolean algebra to make sure we don’t double count.
The circles overlap and we don’t want to overcount the lenses We can simplify the math realizing that, through symmetry, there is reflection line midway through the grid, and this cancels out half the lense, so all we need to really calculate is the area of the right triangle.
Each triangle has an area = ½ base × height. We can work out the height from Pythagoras. Each individual triangle has an area:
Next there is a sector in the middle (with angle α). The area of a sector = ½r2α. (We can determine this angle by starting with the right angle (π/2) of the corner and subtracting two lots of the inside angles from the triangles).
In total we need eight of the triangles, and four of the sectors, to calculate the non-overlapping regions of the circles, and to get the probability we divide this by the area of the grid square.
The final answer comes out to:
This gives us the answer for when the coin is larger the s/2, but less than s/√2. When the coin gets larger than this, it’s a sure thing that it will clip at least one corner!
Merging these two together, here is the graph of the probability of landing over a corner based on the ratio of the coin radius to the size of the grid.