What is the intuition for $P_{\text{even}}$ being $< \frac{1}{2}$? For $c \in \mathbb{R}$, $[c]$ denotes the nearest integer function, i.e., $[c] = \text{argmin}_{n \in \mathbb{Z}} |n-c|$.
Suppose we pick two random $x,y \in (0,1)$ with a uniform distribution. What is the probability, $P_{\text{even}}$, that $\left[\frac{x}{y}\right]$ is even?  
It turns out that the answer is $P_{\text{even}} = \frac{1}{4}(5 - \pi) \approx 0.465 < \frac{1}{2}$. 
Although the argument is straightforward enough to understand, I'm not quite sure of the intuition behind this. Naively, one would think it is exactly $\frac{1}{2}$. 

What is the intuition for this probability being less than $\frac{1}{2}$? 

 A: The easiest intuitive proof for $P_{even}<0.5$ is the following: The probability of $1$ is $\frac5{12}=\frac{25}{60}$. The probabilities of $0$ and $2$ add to $\frac{23}{60}$. Then $3$ is more likely than $4$, $5$ is more likely than $6$, and so on.
A: Draw a picture of the unit square with the lines $y=(k+{1\over2})x$ for $k=0,1,2,\ldots$, and add the line $y=x$, which separates the region between $k=0$ and $k=1$ (for which the nearest integer to $y/x$ is $1$) into two pieces, of area $A_{u}$ and $A_{\ell}$ (for upper and lower, respectively).  The line for $k=0$ exits the unit square at $y={1\over2}$, while the lines for $k=1,2,3,\ldots$ exit at $x={2\over3},{2\over5},{2\over7},\ldots$.  With obvious labeling of the regions, it's easy to see that
$$A_0=A_\ell\qquad\text{and}\qquad A_u\gt A_2\gt A_3\gt A_4\gt\cdots$$
It follows that
$$P_{\text{even}}=A_0+A_2+A_4\ldots\lt A_\ell+A_u+A_3+\cdots=A_1+A_3+\cdots=P_{\text{odd}}$$
Note, this is essentially Arthur's answer, without any explicit calculation of any areas.
