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?


I tried focusing instead on the number of times he tries a key and fails.  So if he gets it on the 3rd try, he misses $2x$.  The probability of doing this, given that he's drunk, is $(4/5) * (4/5) = 16/25$.  On the other hand, the probability of him missing twice in a row given that he's sober is $(4/5) * (3/4) = 3/5$.  Applying Baye's rule, I get 
$$Pr(\text{drunk}\mid \text{missed twice}) = (16/25)/((16/25 + (3/4)(4/5)) = 0.51$$
Given that he misses $3x$, I get
$$Pr(\text{drunk}\mid \text{missed  }3x) = ((4/5)^3)/((4/5)^3 + (2/3)(3/4)(4/5)) = 0.62$$ 
$$Pr(\text{drunk}\mid \text{missed }4x) = ((4/5)^4)/((4/5)^4 + (1/2)(2/3)(3/4)(4/5)) = 67.2$$
$$Pr(\text{drunk}\mid \text{missed } 5x) = ((4/5)^5)/((4/5)^5 + 0) = 1$$
The result has the desirable property that the probability starts at $0.5$ and gets higher the more we observe he starts missing the lock.  I'm thinking the success on the $x$ attempt should not enter the calculation.  I justify this because, we're given the observation that he finally opens the door, so that's not part of our probability calculation.  What's really uncertain is the number of times he has to try before he opens it.  
