probability
I have been having problems solving the following problem.
Two real numbers $x$ and $y$ are chosen uniformly at random in the interval $(0,1)$. What is the probability that the closest integer to $x/y$ is even?
I wrote a computer program to estimate the answer but I can't see how to solve it without one.
Hint: to get an even sum, you need two odds or two evens.
Added: to get two evens is $\frac 59 \cdot \frac 48$. Can you get the chance of two odds? As they are disjoint, you can add them.
Selecting two points $x$ and $y$ from the interval $[0, 1]$ is identical to selecting a single point from the unit square. For the sake of clarity, let us calculate the probability that $y/x$ is closest to an even integer, because that's just the slope. By symmetry, that is equal to the corresponding probability for the ratio $x/y$.
Divide the unit square into two portions, one below the line $y = x$, and one above it. In the bottom portion, points that qualify are below the line $y = x/2$; this section has area $A_l = 1/4$.
In the upper portion, points that qualify are in successively smaller (inverted) triangles with apex at the origin, and bases along the segment from $(0, 1)$ to $(1, 1)$. These bases run from $2/3$ down to $2/5$, then $2/7$ down to $2/9$, then $2/11$ down to $2/13$, etc. Their collective area is therefore
$$ A_u = \frac{1}{2} \Bigl( \frac{2}{3}-\frac{2}{5}+\frac{2}{7}-\frac{2}{9}+\cdots \Bigr) = \frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\frac{1}{9}+\cdots = 1-\frac{\pi}{4} $$
using this well-known series.
The combined area is therefore $A_l+A_u = (5-\pi)/4 \doteq 0.46460$. Here's a diagram of the configuration:
Best Answer
The integer closest to $\frac xy$ is $n>1$ if $n-\frac12<\frac xy<n+\frac12$ or equivalently $(2n-1)y<2x<(2n+1)y$ (it doesn't really matter what you decide for the boundary cases or if you take $\le$ instead of $<$). This describes a triangle with vertices $(0,0)$, $(\frac2{2n-1},1)$ and $(\frac2{2n+1},1)$ and that has area $A_n=\frac12(\frac{2}{2n-1}-\frac2{2n+1})=\frac2{4n^2-1}$. In the special case $n=0$, we obtain a triangle with vertices $(0,0)$, $(1,0)$, and $(1,\frac12)$ and area $A_0=\frac14$. The sought probability equals $$ p =\sum_{k=0}^\infty A_{2k}\approx 0.4646.$$
This sum is also essentially the imaginary part of $\sum_{n=1}^\infty \frac{i^n}n$. More precisely (with the $\frac54$ coming from special treatment for $A_0$ and $\frac in$), $$ p = \frac54-\Im\left(\sum_{n=1}^\infty \frac{i^n}n\right)=\frac54-\Im(\ln i)=\frac{5-\pi}4.$$