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?

 A: The key thing here is this: let $T$ be the number of tries it takes him to open the door.  Let $D$ be the event that the man is drunk.  Then
$$
P(D\mid T=3)=\frac{P(T=3, D)}{P(T=3)}.
$$
Now, the event that it takes three tries to open the door can be decomposed as
$$
P(T=3)=P(T=3\mid D)\cdot P(D)+P(T=3\mid \neg D)\cdot P(\neg D).
$$
By assumption, $P(D)=P(\neg D)=\frac{1}{2}$.  So, we just need to compute the probability of requiring three attempts when drunk and when sober.
When he's sober, it takes three tries precisely when he chooses a wrong key, followed by a different wrong key, followed by the right key; the probability of doing this is
$$
P(T=3\mid \neg D)=\frac{4}{5}\cdot\frac{3}{4}\cdot\frac{1}{3}=\frac{1}{5}.
$$
When he's drunk, it is
$$
P(T=3\mid D)=\frac{4}{5}\cdot\frac{4}{5}\cdot\frac{1}{5}=\frac{16}{125}.
$$
So, all told,
$$
P(T=3)=\frac{16}{125}\cdot\frac{1}{2}+\frac{1}{5}\cdot\frac{1}{2}=\frac{41}{250}.
$$
Finally, 
$$
P(T=3, D)=P(T=3\mid D)\cdot P(D)=\frac{16}{125}\cdot\frac{1}{2}=\frac{16}{250}
$$
(intentionally left unsimplified). So, we get
$$
P(D\mid T=3)=\frac{\frac{16}{250}}{\frac{41}{250}}=\frac{16}{41}.
$$
A: Let's first compute the probability that he wins on the third try in each of  the two cases:
Sober:  The key has to be one of the (ordered) five, with equal probability for each, so $p_{sober}=p_s=\frac 15$.
Drunk:  Success on any trial has probability $\frac 15$.  To win on the third means he fails twice then succeeds, so $p_{drunk}=p_d=\frac 45\times \frac 45 \times \frac 15 = \frac {16}{125}$
Since our prior was $\frac 12$ the new estimate for the probability is $$\frac {.5\times p_d}{.5p_d+.5p_s}=\frac {16}{41}=.\overline {39024}$$
A: 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.  
A: I'll do it for the case where the man has $k$ keys and gets the correct key on the $j$th try. Essentially, we're trying to find
$$P(\text{Drunk|$j ^{th}$ attempt successful})$$
By Bayes' Theorem,
$$P(\text{Drunk|$j ^{th}$ attempt successful}) = \frac{P(\text{Drunk $\cap$ $j ^{th}$ attempt successful})}{P(\text{$j ^{th}$ attempt successful})}$$
Let's find the denominator first. We can do this with the Law of Total Probability
$$P(\text{$j ^{th}$ attempt successful}) = P(\text{$j ^{th}$ attempt successful|Drunk})P(\text{Drunk}) + P(\text{$j ^{th}$ attempt successful|Sober})P(\text{Sober})$$
We are told that he comes home drunk about half the time, so $P(\text{Drunk}) = \frac{1}{2}$ and $P(\text{Sober}) = \frac{1}{2}$.
We also know that when drunk he has the chance to repeat keys and when sober he tries the keys without repetition, so
$$P(\text{$j ^{th}$ attempt successful|Drunk}) = P(\text{$j$ attempts with repetition}) \\ = \left( \frac{k-1}{k}\right)^{j-1}\frac{1}{k}  \\ (\text{Because  we get the wrong key $j-1$ times and then the right key}) $$
$$P(\text{$j ^{th}$ attempt successful|Sober}) = P(\text{$j$ attempts without repetition}) = \frac{1}{k}$$
Putting this together:$$P(\text{$j ^{th}$ attempt successful}) = \left( \left( \frac{k-1}{k}\right)^{j-1}\frac{1}{k}   \right)\left( \frac{1}{2} \right) + \left( \frac{1}{k} \right) \left( \frac{1}{2} \right) \\ = \left( \frac{1}{2} \right)\left[ \left( \frac{k-1}{k}\right)^{j-1}\frac{1}{k}    + \left( \frac{1}{k} \right)\right]$$
Now all that's left is the numerator which we've honestly already done a lot of the work for $$P(\text{Drunk $\cap$ $j ^{th}$ attempt successful}) = P(\text{$j ^{th}$ attempt successful|Drunk})P(\text{Drunk}) \\ \text{(This is the first term in our law of total probability equation)} \\ = \left( \frac{1}{2} \right) \left( \frac{k-1}{k} \right)^{j-1}\frac{1}{k}  $$
Altogether now we have:
$$P(\text{Drunk|$j ^{th}$ attempt successful}) = \frac {{\left( \frac{1}{2} \right) \left( \frac{k-1}{k} \right)^{j-1}\frac{1}{k}}}{\left( \frac{1}{2} \right)\left[ \left( \frac{k-1}{k}\right)^{j-1}\frac{1}{k}    + \left( \frac{1}{k} \right)\right]} \\ = \frac{\left( \frac{k -1 }{k} \right)^{j-1}}{1 + \left( \frac{k -1 }{k} \right)^{j-1} }$$
If you substitute in the numbers you have, you should end up with $\frac{16}{41} = \overline{.39024}$
A: 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
$$
\frac{64}{64+100}=\frac{16}{41} \, .
$$
