You have $n$ coins arranged in a circle, labeled $1$ to $n$. You start at the first coin and go around. At each coin you flip it - if it lands heads you keep it, if it lands tails you remove it. Which coin is most likely to be the last coin remaining?
The answer is the coin labeled $n$ is the most likely. Indeed, the last coin is twice as likely than the first and the probabilities are strictly increasing from $1$ to $n$.
This is an interview question I had and I came up with a couple intuitive ways to explain it. The clearest one is that when you reach coin $k$ for the $m$th time, $k-1$ coins have had $m$ chances to disappear and $n-k+1$ coins have had $m-1$ chances to disappear (including coin $k$). When $k$ is the last coin, then every other coin has had one more chance to be removed than that coin and when $k$ is the first coin, then every other coin has had the same amount of chances to be removed than it (which is why the last coin is twice as likely).
What other ways, intuitive or rigorous, can you use to explain this phenomenon?
Also, can you find exact evaluations for $P(k, n)$, i.e. the probability that the $k$th coin in a circle of $n$ coins is the last remaining coin?