Drunk men finding their tents Problem:
There are four couples camping at the lakeside. They start drinking and as the women get bored and tired, they all go to sleep into their tents. The men continue and get very drunk. In the morning they all go randomly into a tent (but each to a separate one).  What is the probability that 


*

*P (all men go to their own tents)

*P (3 go their own tent, and 1 to a foreign one)

*P (2 go their own tent, and 2 to a foreign one)

*P (1 go their own tent, and 3 to a foreign one)

*P (all men are mistaken)


My attempt:


*

*P (all men go to their own tents) = 1/4 * 1/3 * 1/2 * 1 = 1/24 = 0.04167 


because the P (1 man finds his tent) = 1/4, then the P the next one gets it right is 1/3 because he can go to one less place, etc.


*P (3 go their own tent, and 1 to a foreign one) = 0 because impossible

*P (2 go their own tent, and 2 to a foreign one) = (4! / (2!*2!)) / 4! = 6/24 =  0.25
because we need to select two men who go to their right place, two that do not, and (in the denominator:) they can altogether be placed 4! different ways.  


*P (1 go their own tent, and 3 to a foreign one) = (4! / 1!*3!) / 4! = 4/24 = 0.16667


because the same logic as no.3.


*P (all men are mistaken) = 3/4 * 2/3 * 1/2 * 1 = 6/24 = O.25


because the P that the first man goes to a wrong place is 3/4, the probability that the second one goes to a wrong place is one less: 2/3, etc.
Question:
But there is something wrong with my attempt as the probabilities do not add up to 100%. Where do I go wrong? 


*

*0.04167

*0.00000

*0.25000

*0.16667

*0.25000

0.70834
 A: The number of ways for $k$ men to get to the correct tent is
$$
\overbrace{\ \ \ \binom{4}{k}\ \ \ }^{\substack{\text{ways to choose}\\\text{the correct tents}}}\overbrace{\ \ \ \mathcal{D}_{4-k}\ \ \ \vphantom{\binom{4}{k}}}^{\substack{\text{ways to choose}\\\text{the wrong tents}}}
$$
where $\mathcal{D}_k$ is the number of derangements of $k$ items. That is,
$$
\begin{array}{c|c}
k&\binom{4}{k}\mathcal{D}_{4-k}&P(k)\\\hline
0&9&\frac38\\
1&8&\frac13\\
2&6&\frac14\\
3&0&0\\
4&1&\frac1{24}
\end{array}
$$
A: Robjohn gives the right answer, here are the flaws in your reasoning :


*

*

*There are $4$ ways of choosing which guy goes to the right tent, but then there are $2$ ways (not only $1$) for the three other guys to all be in a wrong tent.


*

*When the first guy has chosen a wrong tent with probability $3/4$, the probability for the second guy to be wrong too is not always $2/3$. It is $1$ in one case out of three: when his own tent is already occupied by the first guy. 






*(after comment): 


In your calculation you assume that the first drunk man (Albert) has proba $3/4$ of choosing a wrong tent: this is correct. Then you believe the second guy (Bernard) has a probability $2/3$ of choosing a wrong tent. This is incorrect.
If Albert has gone to Charles' or Douglas' tent, indeed, Bernard has a choice between 2 wrong tents and his own. However, if Albert has gone to Bernard's tent, then Bernard can only go to wrong tents (Albert's, Charles' or Douglas').  
Overall, knowing that Albert has picked another tent than his own, Bernard's probability of choosing a wrong one isn't $2/3$ but $2/3*2/3+1*1/3=7/9$. (But this is not a great way of solving the exercise)
A: It is possible that this is a trick question.  Since it explicitly said that the women did not get drunk, we can expect that they will wake up when someone tries to enter their tents, and if it's the wrong man they will make him leave.  Therefore, all the men will end up in their own tent with probability 1.
