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$100$ women board a plane with $100$ seats. Each of them has a seat assigned in advance. For some reason the first woman who gets in takes a seat at random. Then the second passenger takes he allocated seat if it is not occupied (by the first), and picks a seat at random if her own seat is occupied. The third passenger takes her own if not occupied by one of the first two, and a random seat if it is. And so on. What is the probability that the last passenger will sit in her own seat?

I am bad at probability questions, help please.

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    $\begingroup$ Does this answer your question? Taking Seats on a Plane $\endgroup$
    – user53259
    Commented Aug 20, 2021 at 5:35

2 Answers 2

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I'm not sure about the proper mathematics behind it, but I can give an intuitive answer to this:

This problem is equivalent to the first passenger choosing a random seat, and when the passenger meant to be sitting in that seat arrives, she replaces the first passenger to retake her seat and the first passenger then chooses another random seat.

At any time, she will choose her own seat and the last passenger's seat always with the same probability (ignoring other choices that will be "cancelled" later). Whenever she chooses her own or the last passenger's seat, it's final. In other cases, the choice will be repeated. Therefore, at the end, the probabilities she will be in her own seat and the last passenger's seat are equal. Therefore the answer is $\frac{1}{2}$.

(help from @PeterFranek)

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  • $\begingroup$ I don't understand the "equal probability" at the end. How did you come to that? $\endgroup$
    – Peter Franek
    Commented Nov 20, 2014 at 0:20
  • $\begingroup$ At the end, when there are only two seats left for the first passenger to choose from, she chooses one to sit in and since she does this randomly, there is a probability of $\frac{1}{2}$ of her ending up in either seat $\endgroup$
    – Piri
    Commented Nov 20, 2014 at 0:24
  • $\begingroup$ How do you know that she is not sitting all the time (or at least, long time already) in her place, when the second-last passenger enters? $\endgroup$
    – Peter Franek
    Commented Nov 20, 2014 at 0:25
  • $\begingroup$ Ok, I think I got it now. At any time when she has to choose something, she will choose seat Nr. 1 and seat Nr. 100 always with the same probability (ignoring other choices that will be "canceled" later). Whenever she will choose seat Nr. 1 or 100, it's final. In other cases, the choice will be repeated. Therefore, at the end, the probabilities she will be on seat 1 or 100, are equal. If this works, it is a really nice answer. $\endgroup$
    – Peter Franek
    Commented Nov 20, 2014 at 0:42
  • $\begingroup$ Yes, sorry I didn't explain it more thoroughly. I'll edit my answer now, thank you. $\endgroup$
    – Piri
    Commented Nov 20, 2014 at 1:03
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Replacing 100 by $n$, we get a recursive equation $P(n)=\frac{1}{n}+\frac{1}{n}(P(2)+\ldots, P(n-1))$.

This is because the first one chooses the position of the $k$'th with probability $1/n$ (for any fixed $k$). If he has chosen his own seat, then the last one will sit in his own seat as well: that's the first $\frac{1}{n}$. Otherwise, people with indices $2,3,\ldots, k-1$ will sit in their seats and the $k$th will either choose seat Nr. 1---then all remaining passengers will sit on their places---or some other from the $n-k+1$ remaining seats. This is completely equivalent to the original problem for $n:=n-k+1$. The only exception is $k=n$ in which case the probability is $0$: therefore, $P(n-n+1)=P(1)$ is not included in the sum.

You can prove by induction that $P(1)=1$ and $P(2)=P(3)=\ldots =P(100)=1/2$.

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