# Taking Seats on a Plane: Extension

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?

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.

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.

• You wish to show they are pairwise independent or mutually independent? These are different things. As for mutually independent... $0<\Pr(A_i)<1$ for all $i$ and yet $A_1\cap A_2\cap \dots \cap A_{n-1}\cap A_n^c$, the intersection of all of the $A_i$ except for the last which is the complement, is clearly the impossible event and has probability exactly zero. You can't have exactly one person in the wrong seat with all others in their correct seat, showing that $\Pr(A_1\cap A_2\cap \dots\cap A_{n-1}\cap A_n^c)=0\neq \Pr(A_1)\Pr(A_2)\cdots\Pr(A_{n-1})(1-\Pr(A_n))$ that they are dependent. Commented Aug 24, 2021 at 12:22
• Seems to me that $A_{2},A_{3},…,A_{n}$ are mutually independent, I forgot to mention this in the post Commented Aug 24, 2021 at 12:59
• @JMoravitz sorry forgot to tag Commented Aug 24, 2021 at 15:04
• "Seems to me that"... but my first comment shows they cannot be. Commented Aug 24, 2021 at 15:04
• I will emphasize again my very first question for clarification in the comments... Are you wishing to show that $\Pr(A_i\cap A_j)=\Pr(A_i)\cdot \Pr(A_j)$? That the events are pairwise independent taking only two of the events at a time? Or are you wishing to show that $\Pr(\bigcap\limits_{i\in \Delta} A_i)=\prod\limits_{i\in\Delta}\Pr(A_i)$ for any subset $\Delta$ of indices of any size including but not limited to $\Pr(A_1\cap A_2\cap A_3\cap \dots \cap A_n)$? I suspect there may be a language issue or misunderstanding/misuse of terminology at play here. Commented Aug 24, 2021 at 15:20

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)