Drunk man with a set of keys. I found this problem in a contest of years ago, but I'm not very good at probability, so I prefer to see how you do it:

A man gets drunk half of the days of a month. To open his house, he has a set of keys with $5$ keys that are all very similar, and only one key lets him enter his house. Even when he arrives sober he doesn't know which key is the correct one, so he tries them one by one until he chooses the correct key. When he's drunk, he also tries the keys one by one, but he can't distinguish which keys he has tried before, so he may repeat the same key.
One day we saw that he opened the door on his third try.
What is the probability that he was drunk that day?


Suppose you considered $1000$ instances where the drunk man came to the door. On average, $500$ of them will be when he is drunk, and $500$ when he is sober. If he is sober, then the chance of him taking $3$ tries to open the door is $\frac{1}{5}$ by symmetry. If he is drunk, then the chance is $\frac{4}{5}\cdot\frac{4}{5}\cdot\frac{1}{5}=\frac{16}{125}$. This means that on average there will be $100$ instances of him taking $3$ tries whilst sober, and $64$ whilst drunk. Given that we know he has taken $3$ tries, the probability of him being drunk is therefore
$$
\frac{64}{64+100}=\frac{16}{41} \, .
$$
