andysif
Joined: Aug 8, 2011
• Posts: 430
March 19th, 2018 at 6:49:33 PM permalink
Quote:

This is one of my favorite problems but I ask it in the form of a 100-seat airplane.

LOL.

Someone translated it into a question in Chinese and I translated it back. Skipped the airplane part.
Last edited by: andysif on Mar 19, 2018
andysif
Joined: Aug 8, 2011
• Posts: 430
March 19th, 2018 at 6:54:26 PM permalink
Doc,

I was thinking along the same line.
It's hard to do it starting from 100, but if you do it starting from 3, and when you are up to 4 you use the result from 3 and build it up that way, it's more manageable.
andysif
Joined: Aug 8, 2011
• Posts: 430
March 19th, 2018 at 7:10:51 PM permalink
This is how I handle it

Assume there is 2 person instead of 100, then the chance of the last person siting in his own seat is 0.5
if there is 3 person, then p = 1/3*100%(1st guy picked his own seat) + 1/3*0% (1st guy picked the last seat and 1/3*0.5(pick any other seat, which reverts to the situation when there is 2 person involved, ie p(2))
when there is 4 person, p = 1/4*100% + 1*4*0% + 2/4 * (p(3))
And it always turn out to be 0.5 from 2 person all the way to 100 person
Nared
Joined: Jan 12, 2019
• Posts: 2
January 12th, 2019 at 8:27:51 AM permalink
I used to have this problem on my web site. It brings back happy memories.
A number of complicated or incorrect answers have been submitted but
gordonm888 has the best one. Here I state it in as few words as possible
to help with understanding.

A sits in the wrong seat, displacing B
B sits in the wrong seat displacing C
C sits in the wrong seat displacing D
This continues until a displaced person sits in A's seat,
displacing no one and everyone else sits in his own seat, or
a displaced person sits in the last person's seat,
displacing the last person.
Either event is equally likely to happen, so
The probability of the last person getting his own seat is 1/2.
charliepatrick
Joined: Jun 17, 2011