Taking Seats on a Plane

Question

This is a neat little problem that I was discussing today with my lab group out at lunch. Not particularly difficult but interesting implications nonetheless

Imagine there are a 100 people in line to board a plane that seats 100. The first person in line realizes he lost his boarding pass so when he boards he decides to take a random seat instead. Every person that boards the plane after him will either take their "proper" seat, or if that seat is taken, a random seat instead.

Question: What is the probability that the last person that boards will end up in his/her proper seat.

Moreover, and this is the part I'm still pondering about. Can you think of a physical system that would follow this combinatorial statistics? Maybe a spin wave function in a crystal etc...

Answer 1

This is a classic puzzle!

The answer is that the probability that the last person ends in up in his proper seat is exactly $\frac{1}{2}$

The reasoning goes as follows:

First observe that the fate of the last person is determined the moment either the first or the last seat is selected! This is because the last person will either get the first seat or the last seat. Any other seat will necessarily be taken by the time the last guy gets to 'choose'.

Since at each choice step, the first or last is equally probable to be taken, the last person will get either the first or last with equal probability: $\frac{1}{2}$.

Sorry, no clue about a physical system.

Answer 2

Here is a rephrasing which simplifies the intuition of this nice puzzle.

Suppose whenever someone finds their seat taken, they politely evict the squatter and take their seat. In this case, the first passenger keeps getting evicted (and choosing a new random seat) until, by the time everyone else has boarded, he has been forced by a process of elimination into his correct seat.

This process is the same as the original process except for the identities of the people in the seats, so the probability of the last boarder finding their seat occupied is the same.

When the last boarder boards, the first boarder is either in his own seat or in the last boarder's seat, which have both looked exactly the same (i.e. empty) to the first boarder up to now, so there is no way the poor first boarder could be more likely to choose one than the other.

Answer 3

This analysis is correct, but not complete enough to convince me. For example, why is the fate of the last person settled as soon as the first person's seat chosen? Why will any other seat but the first person's or the last person's be taken by the time the last person boards?

I had to fill in the holes for myself this way...

The last person's fate is decided as soon as anybody chooses the first person's seat (nobody is now in a wrong seat, so everybody else gets their assigned seat, including the last person) or the last person's seat (the last person now won't get their correct seat). Any other choice at any stage doesn't change the probabilities at all.

Rephrasing... at each stage, either the matter gets settled and there is a 50/50 chance it gets settled each way for the last person's seat, or the agony is just postponed. The matter can thus be settled at any stage, and the probabilities at that stage are the only ones that matter -- and they are 50/50 no matter what stage. Thus, the overall probability is 50/50.

Answer 4

I don't really have the intuition for this, but I know the formal proof. This is equivalent to showing that the probability that in a permutation of $[n]$ chosen uniformly at random, two elements chosen uniformly at random are in the same cycle is $1/2$. By symmetry, it's enough to show that the probability that $1$ and $2$ are in the same cycle is $1/2$.

There are many ways to show this fact. For example: the probability that $1$ is in a cycle of length $k$ is $1/n$, for $1 \le k \le n$. This is true because the number of possible $k$-cycles containing $1$ is ${n-1 \choose k-1} (k-1)! = (n-1)!/(n-k)!$, and the number of ways to complete a permutation once a $k$-cycle is chosen is $(n-k)!$. So there are $(n-1)!$ permutations of $[n]$ in which $1$ is in a $k$-cycle. Now the probability that $2$ is in the same cycle as $1$, given that $1$ is in a $k$-cycle, is $(k-1)/(n-1)$. So the probability that $2$ is in the same cycle as $1$ is $$ \sum_{k=1}^n {k-1 \over n-1} {1 \over n} = {1 \over n(n-1)} \sum_{k=1}^n (k-1) = {1 \over n(n-1)} {n(n-1)\over 2} = 1/2. $$

Alternatively, the Chinese restaurant process with $\alpha = 0, \theta = 1$ generates a uniform random permutation of $[n]$ at the $n$th step; $2$ is paired with $1$ at the second step with probability $1/2$. This is a bit more elegant but requires some understanding of the CRP.

Answer 5

Let's find the chance that any customer ends up in the wrong seat.

For $2\leq k\leq n$, customer $k$ will get bumped when he finds his seat occupied by someone with a smaller number, who was also bumped by someone with a smaller number, and so on back to customer $1$.

This process can be summarized by the diagram $$1\longrightarrow j_1\longrightarrow j_2\longrightarrow\cdots\longrightarrow j_m\longrightarrow k.$$
Here $j_1<j_2<\cdots <j_m$ is any (possibly empty) increasing sequence of integers strictly between $1$ and $k$. The probability of this sequence of events is $${1\over n}\times{1\over(n+1)-j_1}\times {1\over(n+1)-j_2}\times\cdots\times{1\over(n+1)-j_m}.$$

Thus, the probability that customer $k$ gets bumped is $$p(k)={1\over n}\sum\prod_{\ell=1}^m {1\over(n+1)-j_\ell}$$ where the sum is over all sets of $j$ values $1<j_1<j_2<\cdots <j_m<k$. That is, \begin{eqnarray*} p(k)&=&{1\over n}\sum_{J\subseteq\{2,\dots,k-1\}}\,\prod_{j\in J}{1\over (n+1)-j}\cr &=&{1\over n}\,\prod_{j=2}^{k-1} \left(1+{1\over (n+1)-j}\right)\cr &=&{1\over n}\,\prod_{j=2}^{k-1} {(n+2)-j\over (n+1)-j}\cr &=&{1\over n+2-k}. \end{eqnarray*}

In the case $k=n$, we get $p(n)=1/2$ as in the other solutions. Maybe there is an intuitive explanation of the general formula; I couldn't think of one.

Answer 6

Let P(n) denote the probability of the last passenger getting his seat if we begin with n passengers.

Consider the simple case for just 2 seats:

P(2) = 1/2 {first boarder picks his own seat with 1/2 probability}

For n seats: (i) With 1/n probability, the passenger picks the seat of the first passenger, the n'th seat from the end (in which case the last passenger would definitely get his seat). (ii) With 1/n probability, the current passenger picks the seat of the last passenger, first seat from the end (and now, the last passenger can definitely not get his own seat). (iii) Otherwise, the passenger picks some other seat (say #i from the end) among the n-2 remaining seats (with probability 1/n), continuing the dilemma. The problem now reduces to the initial problem with i seats.

Therefore, P(n) = (1/n * (1)) + (1/n * (0)) + Sum {(1/n) * P(i), 2 < i < n};

            = 1/n + 1/n {Sum P(i), 2 < i < n }    --(1)

Similarly,((n-1)/n)P(n-1) = 1/n + 1/n Sum { P(i) , 2 < i < n-1 }

                      = [ 1/n + 1/n Sum { P(i) , 2 < i < n } ] - 1/n P(n-1) 

   => (n-1)/n P(n-1) = P(n) -1/n P(n-1)
   => P(n) = P(n-1)
   => P(m) = P(n-1) , for all integers m > n-1

Since P(2) = 1/2, Therefore P(m) = 1/2 for all integers m > 2