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Clean Tile Problem


Clean tile is a game investigated by Buffon (1777) in which players bet on the number of different tiles a thrown coin will partially cover on a floor that is regularly tiled. Buffon investigated the probabilities on a triangular grid, square grid, hexagonal grid, and grid composed of rhombi. Assume that the side length of the tile l is greater than the coin diameter d. Then, on a square grid, it is possible for a coin to land so that it partially covers 1, 2, 3, or 4 tiles. On a triangular grid, it can land on 1, 2, 3, 4, or 6 tiles. On a hexagonal grid, it can land on 1, 2, or 3 tiles.

Special cases of this game give the Buffon-Laplace needle problem (for a square grid) and Buffon's needle problem (for infinite equally spaced parallel lines).

CleanTileSquare1Clean tile on a square grid

As shown in the figure above, on a square grid with tile edge length l, the probability that a coin of diameter d will lie entirely on a single tile (indicated by yellow disks in the figure) is given by

 P_1=((l-d)^2)/(l^2)=(1-d/l)^2,
(1)

since the shortening of the side of a square obtained by insetting from a square of side length l by the radius of the coin d/2 is given by

 Deltal=2(1/2d)=d.
(2)

The probability that it will lie on two or more (indicated by red disks) is just

 P_(>=2)=1-P_1=1-(1-d/l)^2.
(3)

For the game to be fair with two players betting on (1) a single tile or (2) two or more tiles, these quantities must be equal, which gives

 d=1/2(2-sqrt(2))l=0.29289...l.
(4)
CleanTileSquare2

The probability of landing on exactly two tiles is the ratio of shaded area in the above figure to the tile size, namely

P_2=4d/(2l)(1-d/l)
(5)
=2(1-d/l)d/l.
(6)
CleanTileSquare3

On a square grid, the probability of a coin landing on exactly three tiles is the fraction of a tile covered by the region illustrated in the figure above,

 P_3=(d^2-pi(1/2d)^2)/(l^2)=(1-1/4pi)(d^2)/(l^2).
(7)
CleanTileSquare4

Similarly, the probability of a coin landing on four tiles is the fraction of a tile covered by a disk, as illustrated in the figure above,

 P_4=(pi(1/2d)^2)/(l^2)=1/4pi(d^2)/(l^2).
(8)
CleanTileTriangleClean tile on a triangular grid

As shown in the figure above, on a triangular grid with tile edge length l, the probability that a coin of diameter d will lie entirely on a single tile is given by

 P_1=((l-sqrt(3)d)^2)/(l^2)=(1-sqrt(3)d/l)^2,
(9)

since the shortening of the side of an equilateral triangle obtained by insetting from a triangle of side length l by the radius of the coin d/2 is

 Deltal=2(1/2dcot30 degrees)=sqrt(3)d.
(10)

The probability that it will lie on two or more is just

 P_(>=2)=1-P_1=1-(1-sqrt(3)d/l)^2.
(11)

For the game to be fair with two players betting on (1) a single tile or (2) two or more tiles, these quantities must be equal, which gives

 d=1/6(2sqrt(3)-sqrt(6))l=0.16910...l.
(12)
CleanTileHexagon

As shown in the figure above, on a hexagonal grid with tile edge length l, the probability that a coin of diameter d will lie entirely on a single tile is given by

 P_1=((l-1/3sqrt(3)d)^2)/(l^2)=(1-1/3sqrt(3)d/l)^2,
(13)

since the shortening of the side of a regular hexagon obtained by insetting from a triangle of side length l by the radius of the coin d/2 is

 Deltal=2(1/2d)sec30 degrees=1/3sqrt(3)d.
(14)

The probability that it will lie on two or more is just

 P_(>=2)=1-P_1=1-(1-1/3sqrt(3)d/l)^2.
(15)

For the game to be fair with two players betting on (1) a single tile or (2) two or more tiles, these quantities must be equal, which gives

 d=1/2(2sqrt(3)-sqrt(6))l=0.50730...l.
(16)
CleanTileRhombus

In a quadrilateral tiling formed by rhombi with opening angle theta, insetting from a rhombus of side length l gives

Deltal_1=1/2dcottheta
(17)
Deltal_2=1/2dtantheta,
(18)

so

 Deltal=Deltal_1+Deltal_2=1/2d(cottheta+tantheta)=1/2dcscthetasectheta.
(19)

Therefore, the probability that a coin will lie on a single tile is

P_1=((l-1/2dcscthetasectheta)^2)/(l^2)
(20)
=(1-d/(2l)cscthetasectheta)^2.
(21)

The probability that it will lie on two or more is just

 P_(>=2)=1-P_1=1-(1-d/(2l)cscthetasectheta)^2.
(22)

For the game to be fair with two players betting on (1) a single tile or (2) two or more tiles, these quantities must be equal, which gives

 d=(2-sqrt(2))lcosthetasintheta.
(23)

As expected, this reduces to the square case for theta=pi/4.


See also

Buffon-Laplace Needle Problem, Buffon's Needle Problem

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References

Buffon, G. "Essai d'arithmétique morale." Histoire naturelle, générale er particulière, Supplément 4, 46-123, 1777.Mathai, A. M. "The Clean Tile Problem." §1.1.1 in An Introduction to Geometrical Probability: Distributional Aspects with Applications. Taylor & Francis: pp. 2-5, 1999.Solomon, H. Geometric Probability. Philadelphia, PA: SIAM, 1978.

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Clean Tile Problem

Cite this as:

Weisstein, Eric W. "Clean Tile Problem." From MathWorld--A Wolfram Web Resource. https://mathworld.wolfram.com/CleanTileProblem.html

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