We ought not be able to predict a value in a random sequence, but it seems that in this case (for width equal to needle length, say) we can predict that a rational expression involving pi will occur.
Added by OP Mike Jones on 8.May.2011 (Beijing time):
I believe the answer is affirmative. Below is my proposed proof which, even if it turns out to be defective, will at least clarify what I meant by the question:)
Theorem: pi is irrational. Proof: By the well-known solution to Buffon’s Needle Problem, the sequence f(n)/n converges to 1/pi, where n is the number of tosses of the needle, and f(n) is the number of line-crossings of the needle, where the needle has unit length, and the parallel lines are unit distance apart. Notice that it should be impossible to predict with certainty any value of the ratio f(n)/n. However, because of the discrete (as opposed to continuous), and physical, nature of this process, if this sequence of rational numbers converges to a rational number, it is certain that it will sooner or later take on this rational number as one of its terms. This contradiction establishes that 1/pi is irrational. Therefore pi is irrational. Q.E.D.