Find the infinite product
$$\prod_{n=3}^\infty 1- \dfrac{1}{n\choose 2}$$
It is quite simple to expand and cancel alternating terms to obtain $\frac13$ as the answer. However, notice how the expression looks the product of "not" probabilities of some event iterated over different sample spaces. All I want is a simple probabilistic proof of this evaluation.
One can do this by finding a bijection between this expression and that event. For example, I thought of the situation as having bags having $2,3,4,\dots$ balls respectively. Now we select $2$ balls from each bag one by one, and write the probability of not coming up with a specific pair of balls from each bag. This is exactly equal to the given expression.
Update: Check @Ian's answer for such a bijection. Now the problem remains to connect it to 1/3.
A simple interpretation: you have a bag containing two red balls and a blue ball. You draw two balls, at least one is blue. You replace the balls you drew and throw in another blue ball. You draw two balls, at least one is blue. Your quantity is the probability of this chain continuing forever, instead of being terminated by drawing both of the red balls at some point.