That's odd

Let $x$ be chosen, uniformly and at random, from the interval $(0,1)$ on the $x$ -axis, and $y$ from the interval $(0,1)$ along the $y$ -axis. So the sample space is a unit square with lower-left corner $(0,0)$ and upper-right corner $(1,1)$ , (the sides of which will be represented by dotted lines, so as to avoid division by zero).

Then $P$ will be the combined areas of the regions within this square where the closest integer to $\dfrac{x}{y}$ is odd. This condition can be quantified as saying that

$| \dfrac{x}{y} - (2n + 1)| \lt \dfrac{1}{2} \Longrightarrow (2n + \dfrac{1}{2}) \lt \dfrac{x}{y} \lt (2n + \dfrac{3}{2})$

for some non-negative integer $n$ . For a given value $n$ the corresponding region within the unit square will be that lying between the lines

$y = \dfrac{2}{4n + 1}*x$ and $y = \dfrac{2}{4n + 3}*x$ .

For $n = 0$ the area of the corresponding region is $\dfrac{1}{4} + \dfrac{1}{6}$ , and for $n \ge 1$ the area is $\dfrac{1}{4n + 1} - \dfrac{1}{4n + 3}$ . we thus have that

$P = \dfrac{1}{4} + \dfrac{1}{6} + \dfrac{1}{5} - \dfrac{1}{7} + \dfrac{1}{9} - \dfrac{1}{11} + ..... =$

$\dfrac{5}{12} - 1 + \dfrac{1}{3} + \displaystyle\sum_{n=0}^{\infty}\dfrac{(-1)^{n}}{2n + 1} = -\dfrac{1}{4} + \dfrac{\pi}{4} = 0.53539816.....$ .

Thus $\lceil 10000*P \rceil = \boxed{5354}$ .

This problem came in Putnam 1993 . And can be done once we know that $pi/4$ =1-1/3+1/5-1/7+1/9-....