Taking seats on a plane: probability that the last two persons take their proper seats 100 men are getting on a plane (containing 100 chairs) one by one. Each one has a seat number but the first one forgot his number. So he randomly chooses a chair and sits on it. Others do know their own number. Therefore if their seat is not occupied, they sit on it and otherwise they randomly choose a chair and sit. What is the probability that the last two persons sit on their own chair !?
Edit: As @ByronSchmuland mentioned, a similar but different problem is HERE
 A: Using the idea on the rephrased solution Byron linked in the comments. Assume the first guy keeps getting evicted from his seat. Then there will come a time when the first guy is sent to one of 3 positions. Of these only 1 leaves the 2 last guy's places available so the probability of this happening is $\frac{1}{3}$ and if he does get the correct one then the last two passengers will naturally take their seats correctly. So the probability is $\frac{1}{3}$
In fact you can generalize this to see the probability the last $n$ persons get in their places is $\frac{1}{n+1}$ when $n<m$ where $m$ is the total number of seats. For $n=m$, it is trivially $\frac{1}{n}=\frac{1}{m}$.
A: So if there are $n$ seats, the probability of the last $k$ passenger will take their own seats is $P(n,k)$. 
When the first passenger randomly takes his seat, there are three possible situations.
1) he takes his own seats with probability $1/n$
2) he takes the seats belongs to one of the last $k$ passengers with probability $k/n$
3) he takes the seats of the $m$th passenger with probability $1/n$ for each $m$.
In case 1, then every one takes their own seat naturally.
In case 2, then it is impossible that last $k$ passengers still take all their seats.
In case 3, everyone before $m$th passenger take their seats naturally. the $m$ the passenger have to randomly choose one seat from $n-m+1$ seats and the seat assigned to the first passenger can be viewed as his newly assigned seat. So the probability of the last $k$ passengers takes their own seats is $P(n-m+1,k)$.
In summary
$$P(n,k)=\frac{1}{n}+\frac{1}{n}\sum_{m=2}^{n-k} P(n-m+1,k)$$
It is easily transformed to
$$
Q(n,k)=\frac{1}{n}\sum_{m=2}^{n-k} Q(n-m+1,k), \quad \text{where } Q(n,k)\equiv P(n,k)-\frac{1}{k+1}
$$
For each $k$, it is easy to see that $P(n=k+1,k)=1/(k+1)$. So $Q(n,k)=0$ for any $n\ge k$.
So $$P(n\ge k,k)=\frac{1}{k+1}$$
So in your problem, $n=100, k=2$, the answer would be 1/3.
In this similar problem, $n=100, k=1$, the answer would be 1/2.
