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 solution by Charlie)

I had said that the overall probability distributions for the needle crossing a 10-cm-spaced line and for crossing a 30-cm-spaced line are independent, but this is not the case.  A needle is less likely to cross both at the same time than if they were independent events, as when a needle is oriented nearly parallel to the widely spaced lines, is is more likely than ordinarily to cross the narrow spaced lines and less likely than ordinarily to cross the widely spaced lines, and vice versa.

It is only for a given orientation of the needle that the two crossing events are independent.  Thus the combinatorial analysis should take place within the integral, rather than after the integration.  The correct answer is:

Integ{0 to pi/2}Sin[x]+Cos[x]/3-Sin[x] Cos[x]/3 dx / (pi/2),

which, by using Wolfram's Integrator for the integral, is

[-cos(x)+((cos(x))^2)/6 + sin(x)/3]{0 to pi/2} / (pi/2),

which evaluates to (1/3 + 1 - 1/6) / (pi/2)

or about 0.742723067762179. So there'd be about 743 needles crossing at least one line.  (Std. dev. sqrt(743*.257) or about 14).

Edited on January 21, 2005, 4:19 am

Posted by Charlie on 2005-01-20 13:51:09