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Here is the problem:
One hundred people line up to board an airplane. Each has a boarding pass with assigned seat. However, the first person to board has lost his boarding pass and takes a random seat. After that, each person takes the assigned seat if it is unoccupied, and one of unoccupied seats at random otherwise. What is the probability that the last person to board gets to sit in his assigned seat?

I came up with a solution but I could not find it anywhere on the internet, so I would appreciate if someone could check the legitimacy of my work:

Let's say $p(x_{n})$ is the probability that the $n^{th}$ person to board the plane takes the seat of the last person to board the plane.

Obviously the first person to board has a 1 in 100 chance of randomly picking the last person's seat, so $p(x_{1})$ = $\frac{1}{100}$.

The second person to board the plane will sit in their assigned seat unless the first person takes their seat. There is a 1 in 100 chance that the second person's seat got taken. Assuming that the second person's seat got taken, there is a 1 in 99 chance that the second person takes the last person's seat, so $p(x_{2})$ = $\frac{p(x_{1})}{99}$.

Following this logical progression, $p(x_{3})$ = $\frac{p(x_{1})+p(x_{2})}{98}$, $p(x_{3})$ = $\frac{p(x_{1})+p(x_{2})+p(x_{3})}{97}$, etc.

We can define a new variable $p({y_n})$ = $\sum_{i = 1}^{n}p({x_n})$.

It doesn't take too much work to find that $p({y_n})$ = $p({y_{n-1}})(1+\frac{1}{101-n})$

Solving this recurrence relation, we get $p({y_n})$ = $\frac{1}{101-n}$

Plugging in $n=99$, we see that the probability that one of the first 99 takes the last person's seat $p({y_{99}})$ = $\frac{1}{2}$. Therefore the probability that the last person to board sits in their assigned seat is $\frac{1}{2}$.

I am aware that this is the correct solution. However, I'm not sure that my work is logically sound. Any validation (or rejection) of my work would be appreciated.

  • I have this feeling that this has been asked before on this website, but I don't know when or where....It has been dealt with because I remember having seen this problem before. – imranfat Oct 03 '17 at 22:04
  • @imranfat I have seen this problem on stackexchange as well, but I don't know anyone who has verified this as a solution. –  Oct 03 '17 at 22:08
  • While not a verification of your particular solution, the answer is indeed 1/2 as shown in https://math.stackexchange.com/questions/5595/taking-seats-on-a-plane – ConMan Oct 03 '17 at 22:50
  • https://medium.com/i-math/solving-an-advanced-probability-problem-with-virtually-no-math-5750707885f1 – David G. Stork Oct 04 '17 at 00:28
  • I think the maths checks out. Personally I think this problem is most easily seen as calculating the probability of "permutations of $[n]$ with only one possible non-unit cycle which must contain $1$" not having $n$ in the same cycle as $1$ e.g the permutations like: $(1,3,4\ldots ,9)(n)$. There is a bijection between these permutations and those having $n$ and $1$ in the same cycle: add on the $n$ to the cycle $(1,3,4\ldots ,9,n)$. The probability that $n$ is not in this cycle with $1$ is therefore $1/2$. Here $1$ is the passenger with the lost boarding pass and $n$ is the last passenger. – N. Shales Oct 04 '17 at 01:54

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