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A line of 100 airline passengers is waiting to board a plane. They each hold a ticket to one of the 100 seats on that flight. For convenience, let's say that the nth passenger in line has a ticket for seat number 'n'. Being drunk, the first person in line picks a random seat (equally likely for each seat). All of the other passengers are sober, and will go to their assigned seats unless it is already occupied; If it is occupied, they will then find a free seat to sit in, at random. What is the probability that the last (100th) person to board the plane will sit in their own seat (#100)?

I got this question from here. I understand how we got an answer of $0.5$. There's a follow-up question which I've stated as follows:

What's the probability that the second-last person sits on their seat?

And the answer to this is $2/3$. Why is it $2/3$ and not $1/3$?

Shs Tht
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    Standard version of the question with answer $\frac12$ at https://math.stackexchange.com/questions/5595/taking-seats-on-a-plane – Henry Dec 15 '24 at 02:29

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At each random selection only 3 things of significance can happen:

  1. The drunk person or a person with their seat taken randomly picks seat 1.

  2. The drunk person or a person with their seat taken randomly picks seat 99.

  3. The drunk person or a person with their seat taken randomly picks seat 100.

One of these will happen first, so we may condition on one of these happening, and get that they each have an equal chance of happening.

Scenario 1. and 3. give the second to last person sitting in their own seat. Thus there is a $\frac23$ probability of that person sitting in their own seat.

tkf
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  • +1 though I think it might be clearer if if you said "The drunk person or a person with their seat taken ..." with the addition of "a" helping my understanding of what you are saying. – Henry Dec 15 '24 at 02:32
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    An extension of this is that the expected number of people in their correct seat is $99 +\frac1{100} - H_{99} \approx 93.83$ – Henry Dec 15 '24 at 02:39