If I throw 1000 10-centimeter-long needles on to a tiled floor, where each tile is a 10 cm x 30 cm rectangle, approximately how many needles will end up lying across a crack?

You may assume that the widths of both the needles and the cracks are negligible.

(In reply to re: Simulation supports by Hugo)

Yes, that site gives the formula

P(l;a,b) = (2*l*(a+b) - l^2) / (pi*a*b)

where l is the length of the needle and a and b are the spacings on the lattice.

In this problem, it comes out to (2*10*(10+30) - 10^2) / (pi*10*30) = 7 / (3*pi) = .742723067762179.

The site notes that this is valid for l < a,b, but I think it's apparent that l = a satisfies, though it will cease to hold for l > a or l > b.

 

Edited on January 21, 2005, 4:40 pm

Posted by Charlie on 2005-01-21 16:38:07