You are given a 100*50 snooker table (felt area) and two balls of diameter 1. One ball is placed in the center of the table and the other ball is randomly positioned. What is the probability that I will be able to shoot this second ball directly into the top left pocket without touching the central ball? (Assume pocket has radius 1)
95.2%
Instead of placing the pocket at the top left, let's put it at the top right, because it was easier for me to think this through. The answer will be the same.
Let's put the upper left hand pocket at position (50,100). It has a radius of 1, so the pocket intersects the table edges at (49,100) and (50,99). The opening is sqrt(2) units wide. Each ball has a radius of 0.5 units, so the center of the ball must be at least 0.5 units distance away from the hole for a direct shot in without touching the sides. This means the center of the ball must pass between points (49.13, 99.52) and (49.51, 99.13) lying on the rim of the pocket. The distance between these two points is approximately 0.58 units.
In order to miss the center ball, the center of the second ball must not pass within 0.5 units of the center ball, which means it cannot pass within the boundaries of a circle of radius 1 located at position (25,50).
If we draw a line from point (49.13, 99.52) tangent to the circle at (25,50), touching on the left side, and draw another line from point (49.51, 99.13) to a point tangent to the right side of the circle, we get an approximation of a rhombus with a top of 0.58 units, a base of 2 units and a height of approximately 55.9 units. If we extend the tangent lines past the center towards the lower right hand of the table, we can form another approximate rhombus with a top of 2 units, a base of 6.9 units and a height, again of 55.9 units. The base width was determined by noting that the ration of the top of the first rhombus to the base of the first rhombus will be proportional to the ration of the top of the second rhombus to the base of the second rhombus.
The area of the second rhombus is approximately:
(2+(6.92)/2)*55.9 = 248.8 sq. units, but overlaps the edges of the table a bit, so we must subtract another 4 sq. units (approx.) to compensate. Finally, we subtract half of the area around the central circle and we find the area of the "forbidden zone" is about 248.8  4  3.2 = 241.6 sq. units.
The second ball, though randomly placed on the table, cannot be placed on top of the center ball, nor within the pockets, so the effective area of the table is 50*100  2(3.2) = 4993.6 sq. units. (The first 3.2 sq. units subtracted is for the area around the center ball, the second is for the four quarter areas formed by the corner pockets).
The probability of randomly placing the second ball outside of the "forbidden zone" is then approximately given by:
P = (4993.6  241.6)/4993.6) = .952, or 95.2%

Posted by Mindrod
on 20060213 00:02:44 