Let's say our pin lands such that its center is a distance x from the nearest line, where x goes from 0 to 1/2". The pin is at an angle ?, where 0 means the pin is perpendicular to the lines and ?/2 means the pin is parallel to the lines. Without loss of generality, we can consider ? only in the range 0 to ?/2.
The pin forms a right triangle, where the hypotenuse is 1/2", and the adjacent side is 1/2" cos(?). The adjacent side tells us how far towards the nearest line the pin reaches or the maximum x the pin can land at to intersect the line at a given ?. If we divide this by our maximum x, 1/2", that gives us cos(?), the probability that a pin at an angle ? will land at an x that causes it to intersect a line.
Integrating cos(?) from 0 to ?/2 and dividing by (?/2 - 0) gives the average probability for a pin to intersect a line across all theta. That integral gives us sin(?) evaluated from 0 to ?/2, or sin(?/2) - sin(0), which is 1. Dividing this by the size of the range, ?/2, gives us 2/?, which is about 63.66%.
Interesting problem. I first tried doing x first and then ?, which gave me an arccosine I didn't have the faintest idea how to integrate. Then I realized if I did it in the other order, I'd just get a regular old cosine, giving me the solution above.
Of course now that I've looked up the integral of arccos and found out that it's doable using integration by parts, I feel silly. Calc 2 was my favorite calc class too. :'(
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Comments (2)
Is it Pi? (3.1416etc) 8)
Let's say our pin lands such that its center is a distance x from the nearest line, where x goes from 0 to 1/2". The pin is at an angle ?, where 0 means the pin is perpendicular to the lines and ?/2 means the pin is parallel to the lines. Without loss of generality, we can consider ? only in the range 0 to ?/2.
The pin forms a right triangle, where the hypotenuse is 1/2", and the adjacent side is 1/2" cos(?). The adjacent side tells us how far towards the nearest line the pin reaches or the maximum x the pin can land at to intersect the line at a given ?. If we divide this by our maximum x, 1/2", that gives us cos(?), the probability that a pin at an angle ? will land at an x that causes it to intersect a line.
Integrating cos(?) from 0 to ?/2 and dividing by (?/2 - 0) gives the average probability for a pin to intersect a line across all theta. That integral gives us sin(?) evaluated from 0 to ?/2, or sin(?/2) - sin(0), which is 1. Dividing this by the size of the range, ?/2, gives us 2/?, which is about 63.66%.
Interesting problem. I first tried doing x first and then ?, which gave me an arccosine I didn't have the faintest idea how to integrate. Then I realized if I did it in the other order, I'd just get a regular old cosine, giving me the solution above.
Of course now that I've looked up the integral of arccos and found out that it's doable using integration by parts, I feel silly. Calc 2 was my favorite calc class too. :'(
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