If p is the probability that the 100th passenger sits in his assigned seat, then what’s $p^2+(1−p)^2$

Question

I was not able to solve this question on my test.

An aeroplane has $100$ seats (numbered $1$ to $100$) and $100$ passengers waiting to board each having a ticket with a number from $1$ to $100$. No number is on $2$ tickets or on $2$ seats. The rules of boarding are as below:

(i) Passengers board in the order of the number on their respective ticket.
(ii) The first passenger to board can sit on any seat.
(iii) A passenger with ticket number $i(i\ne 1)$, boards the plane and sits on seat number '$i$' if it is empty. However, if that seat is occupied he can sit on any empty seat.

If $p$ is the probability that the $100^{th}$ passenger sits in his assigned seat, then $p^2+ (1 – p)^2$ is equal to

So I contacted my teacher and he said to let no. of persons in plane be $n$ and put $n=2,3,4…$ and so on, use induction, and I will see that it does not depend on $n$ and is constant $=0.5$. This is the 1st problem I have encountered of this type. Isn't there any other good method to solve it?

Answer 1

Induction is easiest. But you can do a bit of logic.

If passenger $1$ takes the seat of another passenger it is a seat assigned to someone with a number higher then his, say $n_1$. Then all the passengers between $1$ and $n_1$ will take there seats. Passenger $n_1$ will enter and find her seat taken. She has two options: Take passenger $1$s seat, or take the seat of a person higher than hers.

If she takes passenger $1$s seat..... then it's over. All the rest of the passengers will will have find their seats available. If the takes a seat of someone higher. Say $n_2$ then all the passengers between $n_1$ and $n_2$ will take their seat and passenger $n_2$ will have the same choice and so on.

Now every time a displaced passenger $n_k$ enters and finds their seat taken there will be to following seats to choose from: Seat assigned to passenger $1$. .... Seats assigned to passengers: $n_k+1$ through $99$ and seat assigned to passenger $100$. Whichever seat they choose is equally likely.

Now eventually someone will choose either seat Pass. $1$ and end this. Or choose seat Pass. $100$ and displace passenger $100$. Given those two options each are equally likely. It's a matter as to whether a displaced person will choose p$1$ first or p$100$ first. And neither is weighted more or less than the other.

So the probability is $\frac 12$.