It's not that curious, throw in a little calculus and everything makes sense. The root of the problem is that we need to find the probability that a needle will cross a line. For simplicity, we will assign the length of the needle and the distance between the lines a value of 1. As the needle is dropped, it can land at any angle. Assuming the lines run vertically, the maximum horizontal span of the needle, 1, occurs when the angle between the needle and the vertical is π/2 radians. The minimum span, essentially 0, occurs when the needle falls parallel to the lines, so that the angle between it and the vertical is 0.

(If you're not into geometry, π/2 radians is the same as 90 degrees.)

Applying some trigonometry, we can easily show that the horizontal span of the needle is given by sin(x) where x is the angle the needle makes with the vertical. The next step is to find the average span, that is, the average value of the function sin(x) as x varies between 0 and π/2 degrees. From calculus, this is defined as:



      π/2              /
     ∫    sinx dx     /  
      0              /   π/2 - 0
                    /

(Or if you don't like calculus, it is the area under the curve sin(x) divided by the length of the base of the curve, in this case π/2.)

Evaluating this integral, we get

        |π/2
 -cos(x)|     =   cos(0) - cos(π/2)   =   1
        |0
Dividing by π/2 - 0, we get 2/π. This is the average horizontal span of a needle dropped on the vertical lines. Without going into a proof, we can show by common sense that the probability that a needle will touch a line equals its average span divided by the distance between lines (which we defined as 1, so the probability is just 2/π). (Just think, as span approaches the distance between lines, this ratio approaches one, or maximum probability, so as span approaches the average span, the ratio goes to average probability.)

By definition, probability is the number of needles that cross a line (c) divided by the total number of needles (t) Therefore:

 P =   c/t = 2/π
Which can be rearranged to π = 2t/c
As with all probability problems, the more trials involved, the better the estimate becomes.

Note, I didnt mean to ruin the mystery of it...in my opinion the beautiful simplicity of this makes the world an even more interesting place.