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I recently came across a quite interesting problem: "One hundred passengers are lined up to board a full flight. The first passenger lost his boarding pass and decides to choose a seat randomly. Each subsequent passenger (responsible enough to not lose their boarding pass) will sit in his or her assigned seat if it is free and, otherwise, randomly choose a seat from those remaining. What is the probability that the last passenger will get to sit in his assigned seat?"

I figured out that the probability would be 0.5. If you would like to know why I would link you to another topic on this site where this problem was discussed. Taking Seats on a Plane

However, I kept thinking about what the probability would be if there were multiple people who lost their boarding pass. So for example, the first 5 people out of the 100 passengers lost their boarding pass so they take a random seat. Every person that boards the plane after him will either take their "proper" seat, or if that seat is taken, a random seat instead. Then what is the probability that the last passenger will get to sit in his assigned seat? I have trouble figuring it out and I hope that some of you could help me.

Thanks in advance!

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Suppose that the first $n$ passengers have lost their boarding passes. Consider the $n+1$ seats consisting of the $n$ seats assigned to the first $n$ passengers, along with the last person's seat. The order in which these $n+1$ seats get filled is entirely random, as nobody will take any of these seats based on what their boarding pass says. The last passenger will get to sit in her correct seat if and only if that seat is the last of the $n+1$ seats to get filled, so the probability that the last passenger gets her correct seat is $$ \frac{1}{n+1}. $$

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  • $\begingroup$ When you say "as nobody will take any of these seats based on what their boarding pass says," doesn't the last person take the last seat based on what their boarding pass says? $\endgroup$
    – roulette01
    Commented Mar 31, 2021 at 18:24
  • $\begingroup$ Or is this saying that the order in which n out of these n+1 seats are filled by the first n passengers are random? $\endgroup$
    – roulette01
    Commented Mar 31, 2021 at 18:29
  • $\begingroup$ I think another way (or maybe it is what you're saying and I'm just misunderstanding) that we can look at this problem is consider the $n+1$ seats and how many ways we can distribute the first $n$ passengers among them. There are $\binom{n+1}{n}$ ways. Exactly 1 of these ways leaves the last seat empty. So $1/(n+1)$ probability. $\endgroup$
    – roulette01
    Commented Mar 31, 2021 at 18:33

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