# Boarding the aeroplane

100 people are in line, boarding an aeroplane with 100 seats, one at a time.
They are in no particular order.
The first person has lost his boarding pass, so he sits in a random seat.
The second person does the following:

Goes to his seat (the one it says to go to on the boarding pass).
If unoccupied, sit in it.
If occupied, find a random seat to sit in.

Everyone else behind him does the same.
What is the probability that the last person sits in his correct seat?

## 10 comments for “Boarding the aeroplane”

1. January 23, 2017 at 18:58

There will be a probability of 0.5 (ie 50% chance) that the 100th boarder gets his/her originally allocated seat. There will also be a 0.5 probability (also 50% chance) that the 100th boarder will get the seat originally allocated to the first boarder. There is no chance that the 100th boarder will get any other seat.

This is because any future wrongful seat choices will stop as soon as any boarder sits in either of these two seats – apart from that affecting the last boarder, if it was his/her seat and not that of boarder 1 that was chosen first out of these two relevant seats. The seat originally allocated to boarder 1 will remain unoccupied until the end if any earlier boarder takes the seat originally allocated to boarder 100.

And, on the single (non-final) seating turn where one of these two seats is sat upon, it is a 50/50 choice between the only two seats that matter: those originally allocated to boarder 1 and to boarder 100.

Best regards

• January 23, 2017 at 19:49

Nigel, I’m relying on Chuckles here, it really is his post and I do not have the answer. Presumably you have now given it.

2. January 23, 2017 at 19:50

Ah, it’s come through now:

“To quote Lawrence Kesteloot, the publisher thereof –

Solution: There are many ways to come up with this answer, but here’s one that makes sense to me. For ease of explanation we’ll say that I’m the first person to sit down and you’re the last. Also if you sit in your own seat then you “win”, otherwise you “lose”.

Let’s say that there are only two seats, yours and mine. If I sit in my own seat, you win. If I sit in your seat, you lose. So you have a 50% chance of winning.

Now let’s go back to 100 seats. The previous paragraph still holds true: you have a 50% chance of winning if we only consider your seat and mine. Now if I sit anywhere else, I’m just postponing the decision. Let’s say I sit in the seat of the person who’s 13th in line. Persons 2 through 12 will sit in their own seats, then when person 13 comes in he can either sit in my original seat (and you win) or yours (and you lose). Or of course he could sit anywhere else and postpone the decision again.

If this keeps going, then eventually there are only two seats left and person 99 is forced to choose either your seat or mine, again with 50% chance. There are only two seats that matter throughout the game: yours and mine. Any sitting in other seats is just postponing the decision of which of the two interesting seats gets sat in first. Note also that you’ll only ever end up in your seat or mine, no one else’s.

It’s a bit like flipping a coin, except that you can postpone flipping, but not indefinitely. What’s the chance of coming out heads? Well 50%, the postponement doesn’t change that.

Here’s a mathematical way to see it. Define f(n)f(n) to be the chance that the last person in an airplane of nn seats will get his own seat. It can be defined recursively like this:
f(n)=1n1+n−2nf(n−1)+1n0f(n)=1n1+n−2nf(n−1)+1n0

The first term is the chance that the first person will sit in his own seat (1n1n) multiplied by the chance, then, that the last person will sit in his own (11). The last term is the chance that the first person will sit in the last person’s seat (1n1n) multiplied by the chance, then, that the last person will get his own seat (00). The middle term counts every other seat. There are n−2n−2 other seats, and there’s a 1n1n chance of each, and they all simplify to the f(n−1)f(n−1) case. Also:
f(2)=0.5f(2)=0.5

If you plug in 0.50.5 for the f(n−1)f(n−1) term, you find that f(n)=0.5f(n)=0.5, so it’s true for any n>1n>1. “

3. January 23, 2017 at 19:50

https://www.teamten.com/lawrence/puzzles/airplane_seating.html

And yes, Nigel is correct in his answer.

4. January 23, 2017 at 20:27

It is, I know, rather dodgy to say this: but I was surprised and pleased with myself on this puzzle. More importantly, I realised something that is either a truth or a very likely hypothesis.

Being somewhat aged, I know I no longer have the intellectual ability of my youth and early adult years. However, I strongly suspect that peaks of thought still occur: and my hypothesis is that the peaks are not lower; it is that they are less frequent. [Though being less sustained is also likely, which affects their utility.]

Anyway, this should give us oldies some optimism on future achievements and contributions. They are not impossible, just (much) less likely than before.

Best regards

• January 23, 2017 at 20:49

Being somewhat aged, I know I no longer have the intellectual ability of my youth and early adult years.

Could have fooled us.

5. FrankC
January 23, 2017 at 21:34

How did the first person manage to board without a boarding pass?

• January 23, 2017 at 21:39

Stop bringing reality into it.

6. Lord T
January 24, 2017 at 12:12

It’s 100%. The pilot and the flight attendants are the last people in their seats.

• January 24, 2017 at 12:22

Cat among the pigeons, yes? 🙂