Well,

I have read this puzzle a couple of times but never did I put my mind to it. Once deepanjan and I sat together to solve it. We both came up with two different solution, both correct. My laziness made me procrastinate posting the solution here. Anyways, now that I am on it, let me directly go to the puzzle.

**Problem : **There is an airplane and 100 passengers are given boarding pass for the same. The first passenger lost his boarding pass and he did not remember the seat number assigned to him. So, when this first passenger boards the plane he chooses one seat uniformly randomly. Rest of the passengers, board the plane in the sequence of 2 to 100. If a passenger finds his seat vacant, he will sit there, if not, he chooses a vacant seat uniformly randomly. The question is, what is the probability that the last person ( 100th in this case ) will find his seat vacant when he boards the plane?

Number 100 has nothing to do with the puzzle and we can as well state that there are passengers.

**Solution 1 : **Well, I have come up with this solution. I started initially with 2,3,4,5 passengers and found the probability manually. So I conjectured that the probability is 1/2 in all the cases. Now, the task remains is to prove that it is indeed the case for any . I remembered my favorite professor sundar vishwanathan and a very powerful mathematical tool that he taught us to use in a very elegant way. And mathematical induction indeed helped me proving my conjecture.

Base case : n=2.

Clearly, the first person sits in the 2nd person’s seat with probability 1/2. So the last person ( 2nd person in this case ) finds his seat vacant with the probability 1/2.

Hypothesis : For any probability that the last person finds his seat vacant is 1/2.

Induction Step : Let there be n+1 passengers.

Let = probability that n+1 th person finds his seat vacant when there are total n+1 seats.

Let = probability that at least one person of the first n persons ( say k th person ) found his seat vacant hence he sat in his own seat.

It is clear that is 1/2. Because, when none of the first n person find their seat vacant, it must be the case that 1st person seats in 2nd seat, 2nd person seats in 3rd seat and so on. Because they arrive in sequence, if 1st person seats in 3rd seat, then 2nd person would find his seat vacant. Or if 2nd person seats in 1st seat, then 3rd person would find his seat vacant. So after n-1 person take their seats, only two seat ( 1st and n+1 th ) remains vacant. Now the n th person chooses the 1st seat with the probability 1/2 which is precisely the probability that the last person ( n+1 th ) person would find his seat vacant given that none of the first n could find their seat vacant.

. The reason is that, if at least one person of the first n persons could find his seat, then we can view it as if k-th seat was already taken when the first person boarded the plane. We can view it like that because we are given that event B has happened and now we have to find the probability of all other events that are occurring. So original problem of n+1 person and n+1 seats reduces to n person and n seats for which we can use induction hypothesis.

Let .

So, .

So irrespective of how many persons and seats are there, the probability remains constant.

**Solution 2 : **Deepanjan came up with this solution. Given n person, his solution looks like this.

Let denote the fact that i-th person does not find his seat vacant.

because the first person chooses the 2nd seat with probability 1/n.

because, either first person choose the 3rd seat with the probability 1/2 or the 2nd person find his seat taken and chooses 3rd seat with probability 1/(n-1).

Similarly,

In general, it should be easy to see that for . Hence, for the last person when i=n we have . The probability that n-th person finds his seat vacant would be 1 – (1/2) = 1/2.

–Saurabh Joshi