Question of Buffon's Needle I looked at the gif on wikipedia that explains Buffon's needle, but I have two questions.
First, why do you only consider $x$ as the distance from the center of the needle to the closest line, so it varies from $0$ to $\frac{d}{2}$ (where $d$ is the distance between the two lines)? Why can't you just say it is the distance from the center of the needle to the top line, in which it would vary from $0$ to $d$.
Second, why do you have to use $x$ and $\theta$ for the region of probabilities? Why can't you directly use $x$ and $y$ and the relation $x \leq y$?
 A: 
First, why do you only consider $x$ as...

Since there are repeated parallel lines ("rulings"), defining "top line" requires at least as much extra work as defining the distance from the needle's center to the nearest line.  

Second, why do you have to use $x$ and $\theta$...

I'm not sure the distribution of $y$ would be easily understood without reference to the uniform distribution of angle $\theta$.

The assumption that the needle's center position and its angle are uniformly and independently distributed seems natural.  Intuitively the needle comes down at a position and angle that have no preferred values.  By symmetry of the rulings (and of the needle), we may assume (taking reflections if necessary) that the center of the needle is within $[0,d/2]$ where $d$ is the distance between parallel rulings (lines of the grid), and that the angle formed by the needle with another parallel line through the center is within $[0,\pi/2]$, i.e. parallel or at right angles.

If the length of the needle is $L$ (easier to distinguish than the lowercase letter), then one end of the needle extends a "vertical" distance $y = \frac12 L \sin \theta$ towards the nearest "horizontal" ruling.  In the simplest case, where $L \lt d$ (so at most one crossing is possible), a crossing occurs exactly when $y \gt x$.
Since the function that relates $\theta$ to $y$ is nonlinear, the uniform distribution of $\theta$ becomes a nonuniform distribution of $y$.  It is intuitively clear, I take it, that the sine curve is not flat.  In any case it suffices to notice that although the $\theta$ subintervals $[0,\pi/4]$ and $[\pi/4,\pi/2]$ are of equal length (and therefore of equal probability), the images by $\sin \theta$ are $[0,\sqrt{2}/2]$ and $[\sqrt{2}/2,1]$, which are not of equal length (but do inherit equal probabilities).
