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

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 $\frac{1}{n}=\frac{1}{m}$.
