Taking Seats on a Plane: Extension

Question

This question is inspired by the infamous “Taking Seats on a Plane” problem: Taking Seats on a Plane

Summary of the Original Question

$n$ passengers board a plane that has $n$ seats. The first passenger forgot her boarding pass so she chose a random seat. If a passenger has his/her seat taken he/she also choose a random seat. What is the probability that the last passenger sit on his/her own seat?

Answer to the Original Question

The probability that:

• The first passenger sit in passenger $J_{1}$’s seat
• Passenger $J_{i}$ sit in passenger $J_{i+1}$’s seat
• Passenger $J_{m}$ sit in first passenger’s seat

is

$$ \frac{1}{n}\prod_{i=1}^{m}{\frac{1}{n+1-J_{i}}} $$

The first passenger had $n$ seats to choose so the probability of choosing passenger $J_{1}$’s seat is $\frac{1}{n}$. This is the first term in the above expression.

When passenger $J_{i}$ wanted to sit, $J_{i}-1$ seats had been taken and there were only $n+1-J_{i}$ seats left. The probability of this passenger to choose a particular seat is $\frac{1}{n+1-J_{i}}$. Hence the terms in above expression.

Since we want to calculate the probability that the last passenger sit in his/her own seat, we set $J=\{J_{1},…,J_{m}\}$ as a non - empty subset of $\{2,3,…,n-1\}$ then add the above expression for all possible subsets to obtain the desired probability

$$ \begin{align} p&=\sum_{J}{\frac{1}{n}\prod_{i=1}^{m}{\frac{1}{n+1-J_{i}}}}\\ \\ &=\frac{1}{n}\prod_{k=2}^{n-1}{\left(1+\frac{1}{n+1-k}\right)}\\ \\ &=\frac{1}{n}\prod_{k=2}^{n-1}{\left(\frac{n+2-k}{n+1-k}\right)}\\ \\ &=\frac{1}{2} \end{align} $$

Extension to the Original Question

I want to calculate $P(A_{i})$ - the probability that passenger $i$ sit on his/her own seat and I want to prove that the events $A_{i}$s are independent.

My Answer to the Extension

To calculate the probability that passenger $x_{1},…,x_{l}$ sit in their own seats, we set $J$ as a non - empty subset of $\{2,…,n\}$ that does not contain any of the $x_{1},…,x_{l}$. So we have the following expression:

$$ \begin{align} P\left(A_{x_{1}}\cap…\cap A_{x_{l}}\right)&=\frac{1}{n}\left(\prod_{i=2}^{n}{\left(\frac{n+2-i}{n+1-i}\right)}\right)\cdot\left(\prod_{j=1}^{l}{\left(\frac{n+1-x_{j}}{n+2-x_{j}}\right)}\right)\\ \\ &=\prod_{j=1}^{l}{\left(\frac{n+1-x_{j}}{n+2-x_{j}}\right)} \end{align} $$

From here we have $P(A_{i})$

$$ P\left(A_{i}\right)=\frac{n+1-i}{n+2-i} $$

As expected from the original question

$$ P\left(A_{n}\right)=\frac{1}{2} $$

Since the following is true, I also proved that the events are independent

$$ P\left(A_{x_{1}}\cap…\cap A_{x_{l}}\right)=\prod_{j=1}^{l}{P\left(A_{x_{j}}\right)} $$

I want to know if my result is correct and also welcome if there’s any alternative or discussion.

Answer 1

Very neat answer indeed : I did not think that the computational approach with the sum over the subsets was tractable :-)

I just wanted to point out the following solution for the marginal probability distribution :

The key observation is that the event

"Passenger $k$ taking its seat"

is decided by the first (remember the sequential seating plan) passenger taking a seat in one of the seats labelled $\{1,k, k+1,....,N\}$ (containing $N-k+2$ labels).

Precisely, passenger $k$ will take its seat iff the seat taken by the first passenger (defined above) is distinct from $k$.

Now this position is uniform (this requires some thought), hence the probability $$\frac{N-k+1}{N-k+2}.$$

(EDIT : a similar answer for the case $k=N$ is the validated answer to the question Taking Seats on a Plane)

(EDIT 2 : going through all the answers of that post, you can find both your argument and mine, for general $k$ as well; only the independence I have not seen at first glance)