Evaluate $ \prod_{n=3}^\infty ( 1 - \binom{n}{2}^{-1} )$ 
$$ \prod_{n=3}^\infty \left ( 1 - \frac{1}{\binom{n}{2}} \right)$$

Expanding it out doesn't give any good patterns, I'm totally stuck.
The answer is 1/3 [according to WolframAlpha: 
 https://www.wolframalpha.com/input/?i=product+from+3+to+infinity+(1+-+1%2F((n)+comb+2)+)

Note that we have
$$\begin{align}
\prod_{n=3}^N \left(1-\frac{1}{\binom{n}{2}}\right)&=\prod_{n=3}^N \left(1-\frac{2}{n(n-1)}\right)\\\\
&=\prod_{n=3}^N\frac{n^2-n-2}{n(n-1)}\\\\
&=\prod_{n=3}^N \frac{n+1}{n}\frac{n-2}{n-1}\\\\
&=\prod_{n=3}^N \frac{n+1}{n}\prod_{n=3}^N \frac{n-2}{n-1}\\\\
&=\frac{N+1}{3}\frac{1}{N-1}
\end{align}$$
Now let $N\to \infty$.
