Don't Cross A Corner!

Relevant wiki: Basel Problem

For the segment not to cross any other vertices it means that the difference in $x$ and $y$ coordinates of the two vertices are coprime. Therefore, we are looking for the probability that two natural numbers chosen at random are coprime, i.e. they don't have prime factors in common. By notting that 1 in 2 numbers has 2 as a factor, 1 in 3 numbers has 3 as a factor, 1 in 5 numbers has 5 as a factor, etc... we can write the probability as

$\displaystyle P\{a,b \text{ coprime} \}= \left(1-\frac{1}{2^2}\right)\left(1-\frac{1}{3^2}\right)\left(1-\frac{1}{5^2}\right) \cdot \ldots = \prod_{p \text{ prime}} \left(1 - \frac{1}{p^2} \right)$

But the right hand size is just $\frac{1}{\zeta(2)} = \frac{6}{\pi^2} \approx 0.60792$ by the Euler product formula for the Riemann zeta function.

Relevant wiki: Euler's Totient Function

Since $\begin{aligned} \sum_{m \le x} \phi(m) & = \; \sum_{m \le x} \sum_{d|m} \mu(d) \tfrac{m}{d} \; = \; \sum_{d \le x} \mu(d)\sum_{m \le \frac{x}{d}} m \; = \; \sum_{d \le x} \mu(d)\left(\tfrac{x^2}{2d^2} + O\big(\tfrac{x}{d}\big) \right) \\ & = \; \tfrac12x^2 \sum_{d \le x} \frac{\mu(d)}{d^2} + O(x \ln x) \; = \; \tfrac12x^2 \sum_{d=1}^\infty \frac{\mu(d)}{d^2} - \tfrac12x^2 \sum_{d > x}\frac{\mu(d)}{d^2} + O(x\ln x) \\ & = \; \frac{x^2}{2\zeta(2)} + O(x\ln x) \end{aligned}$ as $x \to \infty$ , the probability that two randomly chosen integers in $1,2,3,...,x$ are coprime is $\tfrac{2}{x^2}\sum_{m \le x}\phi(x) \; = \; \tfrac{1}{\zeta(2)} + O\big(\tfrac{\ln x}{x}\big) \hspace{2cm} x \to \infty$ Since nonzero $a,b$ in $-x,-x+1,...,x-1,x$ are coprime if and only if $|a|, |b|$ in $1,2,...,x$ are coprime, and $a,b$ are never coprime if either $a$ or $b$ is zero, this is also (at least asymptotically as $x \to \infty$ ) the probability that two randomly chosen integers in $-x,-x+1,...,x-1,x$ are coprime. Thus the desired limit is $\zeta(2)^{-1} = \boxed{\tfrac{6}{\pi^2}}$ .

We need to find 2 integers which are relatively prime.

Relevant wiki: Euler's Totient Function

I didn't know anything about how to solve Euler function's sum, so I just

and divided by $\dfrac{100000^2}{2}$

:D