Oddly Enough

First we show that the value of $S = \displaystyle \sum_{r=0}^{\infty} \frac{(-1)^r}{2r+1}$ is $\frac{\pi}{4}$ . I have two methods here for the proof:

Method-1:

Clearly $S = \displaystyle \sum_{r=0}^{\infty}(-1)^r \int_{0}^{1} x^{2r} dx$

= $\displaystyle \int_{0}^{1} \bigg(\sum_{r=0}^{\infty} (-1)^r x^{2r} \bigg)dx$

= $\displaystyle \int_{0}^{1} \frac{1}{1+x^2} dx = \boxed{\frac{\pi}{4}}$

Method-2:

By series expansion of $\ln(1+i)$ , we see that $S$ is its imaginary part. Hence, $S = \text{Im}\bigg(\ln\bigg(\sqrt{2} e^{i \frac{\pi}{4}}\bigg)\bigg) = \text{Im}\big(\ln(\sqrt{2}) + i \frac{\pi}{4}\big)= \boxed{\frac{\pi}{4}}$

Now, let us come to the problem. Let me use $y$ in place of $a$ and $x$ in place of $b$ .

Clearly, as the integer nearest to $\frac{y}{x}$ is odd, hence,

$\frac{y}{x} \in \big(1/2,3/2\big) \cup \big(5/2,7/2\big) \cup \big(9/2,11/2\big) \cup \dots$

Now, as $x,y \in (0,1)$ , area of sample space is $1$ .

Given below is a graph, showing triangles $1,2,3,4,5,\dots$

Clearly, area of favorable region = $\text{Ar(1)+Ar(2)+Ar(4)+Ar(6)+Ar(8)+}\dots$

Now, altitude of each of these triangles is $1$

Hence, area of favorable region = $\frac{1}{2} \times 1 \times \bigg(1 - 1/2 + 1-2/3 + 2/5-2/7 + 2/9-2/11 + 2/13-2/15 \dots\bigg)$

= $\frac{1}{2}\big(2S - \frac{1}{2}\big)$

= $\frac{\pi - 1}{4} \approx 0.5354$

This is the required probability as area of sample space is $1$ . Hence, answer is $\boxed{535}$

One can use Monte Carlo method with computer with this code:

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var n=10000000;
var c=0;
for(var i=0;i<n;i++)
{
    var a=Math.random();
    var b=Math.random();
    if(Math.round(a/b)%2)
        c++;
}

var prob=c/n

Although Monte-Carlo would have been easy I decided against it after I made this diagram, on the grounds that such results might be subject to considerable variation. On the other hand, I have a congenital disinclination for reducing infinite series.

I decided to proceed by summing areas. Since the region corresponding to the first odd integer is miserable to work with I decided to sum the even-numbered regions instead. The boundaries of these are obviously equations like y=x/1.5 and y=x/2.5, y=x/3.5 and y=x/4.5, and so on. The probability of the region where x/y rounds to 0 is 0.25. To avoid round-off errors I made the summation using fractions.

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from fractions import Fraction

upper=Fraction(5,2)
lower=Fraction(3,2)

total=0
for i in range(5000):
    term=1/lower-1/upper
    total+=term
    if i%100==0:
        print 0.5*total
    upper+=2
    lower+=2

Convergence was slow (reminding me of one of Euler's calculatiosn). The final value here was 0.214576334205 giving a probability of 1-(.25+0.214576334205) or 0.535423665795.

Consider the problem geometrically. We have points $(a,b)$ confined within a unit square with its bottom left vertex at the origin.

Let $\left [\dfrac{a}{b}\right ]$ be $\dfrac{a}{b}$ rounded to the nearest integer.

We have

$\left [\dfrac{a}{b}\right ]=1\iff \dfrac{1}{2}\leq \dfrac{a}{b}<\dfrac{3}{2}\iff b\leq 2a<3b$

$\left [\dfrac{a}{b}\right ]=3\iff \dfrac{5}{2}\leq \dfrac{a}{b}<\dfrac{7}{2}\iff 5b\leq 2a<7b$

And so on. We can interpret these regions in the following diagram.

https://www.desmos.com/calculator/p1nf8wndrs

Where the shaded regions are those where $\left [\dfrac{a}{b}\right ]$ is odd.

To avoid dealing with the larger, non-triangular region differently, just subtract the non-shaded area from 1.

$\begin{aligned}\textnormal{Pr}\left (\left [\dfrac{a}{b}\right ]\textnormal{ is even}\right )&=\frac{1}{2}\cdot 1\cdot \left (\frac{1}{2}+\left (\frac{2}{3}-\frac{2}{5}\right )+\left (\frac{2}{7}-\frac{2}{9}\right )+\left (\frac{2}{11}-\frac{2}{13}\right )+...\right )\\&=\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+\frac{1}{7}-\frac{1}{9}+...\\&=\frac{1}{4}+1-\left (1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...\right )\\&=\frac{5}{4}-\frac{\pi}{4}\href{https://brilliant.org/wiki/pi/#exactly-defining-pi-as-an-infinite-series}{\textnormal{ by the Leibniz formula for }\pi}\\&=\frac{5-\pi}{4}\end{aligned}$

We then have that $\textnormal{Pr}\left (\left [\dfrac{a}{b}\right ]\textnormal{ is odd}\right )=1-\dfrac{5-\pi}{4}=\dfrac{\pi -1}{4}$ .

$\left \lfloor 1000\cdot\dfrac{\pi-1}{4}\right \rfloor=\color{#20A900}{\boxed{535}}$

Extra Information

• To prove that $1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+...$ , we have that $\arctan \theta=\theta-\dfrac{\theta^3}{3}+\dfrac{\theta^5}{5}$ from its Maclaurin series. Substituting $\theta=1$ gives the result.
• A similar problem was covered by 3Blue1Brown, from where I learnt this method.

Wow, this was fun and kind of unexpected. What's interesting is that if the Floor of a/b was used instead of Round of a/b, we'd end up with (1/2)Log(2) instead of something involving π. How about that?