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There is an auditorium with $n$ distinguishable seats. There are $n$ people, each assigned to a unique seat such that all seats are filled and no $2$ people are assigned the same seat.

People start entering the auditorium one by one. Person $1$ enters the room, but forgets their assigned seat and decides to go to a random seat (any of $n$ seats equally likely to be chosen). Person $2$ enters the room and goes to their assigned seat if it is free and to a random seat (any free seat equally likely to be chosen) if not. Every person after follows person $2$'s procedure thereafter.

What is the probability that the last person goes to their assigned seat?

Context: teacher brought it up in a class as a problem.

I think the desired probability is $1/2$ because we can consider simple cases like $n=2$ or $n=3$. If we consider a string of length $3$, with the assigned seating being $123$, person one forgetting their seat results in $(1\vert\cdot\vert\cdot)$ or $(\cdot\vert1\vert\cdot)$ or $(\cdot\vert\cdot\vert1)$. In the first and third case person $2$ sits in their own seat because it's free and the last person goes to their seat in the first case only. In the second case it is equally likely whether the last person ends up in their seat or not, so the total probability is $1/2$.

I'd rather take partial solutions or hints because I want to solve it myself.

pjq42
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    This is just the airplane seating problem, eg at https://math.stackexchange.com/questions/5595/taking-seats-on-a-plane/4208862#4208862 – true blue anil Aug 17 '22 at 08:10

1 Answers1

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Consider the problem from the perspective of mathematical induction. WLOG, we assume that the assigned seats are $123\ldots (n-1)n$.

Let $P_n$ denote the answer (probability) w.r.t $n$.

When $n=2$, as the OP points out, the probability is clearly $1/2$. Now consider the case when $n = N + 1$ for some $N$. If Person $1$ chooses his seat (#1) correctly (with prob. $1/n$), then the conditional probability becomes $1$. Otherwise, if Person $1$ chooses the seat $k$, then the problem reduces to $n = k$ (Person $k$ comes and randomly chooses a seat). A special case is that if Person $1$ chooses the seat $n = N+1$, the probability is $0$. Therefore, the total probability is $$P_{N+1} = (1 + \sum_{k = 2}^{N} P_k) / (N+1).$$

Vezen BU
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