Plane Probability Problem

Originally posted by geepamoogle
What are the odds Passenger #X gets displaced by a previous passenger?

I get P(n,N) = 1/(N+2-n), where P(n,N) is the probability that passenger #n gets displaced in a plane with N places.

Originally posted by eldragonfly
first approach: It seems that the intuitive solution is that person-100 can only be bumped by either person-1 or person-99, in either case he must take seat-1.

Nope.

Person 1 randomly picks seat 3, person 3 randomly picks seat 100.

Person 100 is bumped by person 3.

Originally posted by TheMaster37
Nope.

Person 1 randomly picks seat 3, person 3 randomly picks seat 100.

Person 100 is bumped by person 3.

good point, but the same result, person-100 ends up in seat#1.

Originally posted by eldragonfly
good point, but the same result, [b]person-100 ends up in seat#1.[/b]

THAT is indeed true 🙂

He always ends up in seat 1 or 100, like you said.

Originally posted by Palynka
I get P(n,N) = 1/(N+2-n), where P(n,N) is the probability that passenger #n gets displaced in a plane with N places.

Anyone confirms this?

Originally posted by Palynka
Anyone confirms this?

I was giving people some time to respond, but I plugged some numbers in and it works in my book.

You can reason this like the last seat, only you have seats beyond the specific person to add in the mix. One of those gets stolen, and all passengers sitting down before that person, including the specific passenger in question, get to sit in their own seats.