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قم بتسجيل الدخول اولاً لكي يتسنى لك الاعجاب والتعليق.

Clean Tile Problem

المؤلف:  Buffon, G.

المصدر:  "Essai d,arithmétique morale." Histoire naturelle, générale er particulière, Supplément 4

الجزء والصفحة:  ...

8-3-2021

2267

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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).

CleanTileSquare1 Clean 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)

CleanTileTriangle Clean 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.


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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