The first passenger chooses his seat at random so the probability of choosing Seat 1 is 1/n and choosing the wrong seat is (n - 1)/n.
For a two seat aircraft the probability of passenger 2 getting the correct seat is 0.5.
For a three seater the probability of the 3rd passenger getting Seat 3 is
probility of passenger 1 choosing Seat 1 + (probability of passenger 1 choosing wrong seat x probability of passenger 3 getting Seat 3)
To find the probability of pasenger 3 getting Seat 3 enumerate the number of possible seating arrangements with Seat 1 not occupied by Passenger 1. These are
2 1 3
3 1 2
2 3 1
3 2 1
so probability of passenger 3 getting Seat 3 = 1/4. If Passenger 1 chooses Seat 1 all other passengers get their correct seats.
So the probability of Passenger 3 getting Seat 3 = 1/3 + (2/3) x (1/4) = 0.5.
For four passengers the probability of passenger 4 getting Seat 4 = 1/4 + (3/4) x (4/18) = 5/12 from enumeration as shown for three seats.
Generalising to n seats we have
the probabilty of Passenger 1 choosing Seat 1 = 1/n
the probability of passenger 1 not choosing Seat 1 = (n-1)/n.
To calculate the probability of the nth passenger getting the nth seat we need to calculate the number of completely or partially correct seating arrangements 1. This is
(n -1) x (n - 1)!
from this we need to subtract the number of these seating arrangements in which Passenger 1 is in the nth seat is
(n - 1)!.
So the number of seating arrangements where the nth passenger can be in the nth seat passenger 1 not in Seat 1 is
(n -1) x (n - 1)! - (n - 1)! = (n - 2) x (n - 1)!
The number of ways that the nth passenger can be in the nth seat is
(n - 2) x (n - 1)! = (n - 2) x (n - 2)!
(n - 1)
so the probability of the nth passenger getting the nth seat is
(n - 2) x (n - 2)! = (n - 2)
(n - 1) x (n - 1)! (n - 1) x (n - 1)
Therefore the probability of the nth passenger getting the nth seat is
1 + (n - 1) x (n - 2) = 1 + (n - 2)
n n x (n - 1) x (n -1) n n x (n - 1)
Using this formula for 2 , 3 and 4 gives
p(2) = 0.5
p(3) = 0.5
p(4) = 5/12 as found above.
For the 100th passenger p(100) = 197/9900
This is not an very elegant but then neither is air travel. I always seem to find Passenger 1 sitting in my seat!
Best Wishes
John
----- Original Message -----
From: "Suhal Bux" <[log in to unmask]>
To: <[log in to unmask]>
Sent: Wednesday, January 30, 2008 4:23 PM
Subject: Airplane riddle
Hello
I know this isn't strictly the right sort of post but it's interesting
and is doing my head in...
100 passengers are queuing to get on a plane.
Each has a ticket with a seat number and they are standing in order.
The first man in the queue is crazy and will sit anywhere at random.
The rest of the passengers will sit in their own seat, unless it is not
available, in which case they will sit in another seat at random.
What is the probability that the 100th passenger sits in his own seat?
I'd appreciate proofs or good reasoning.
The best most correct answer gets a special prize (a McDonald's
voucher!).
Regards
Suhal Bux
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