Outside a cinema, a queue of 100 film-goers are each given a ticket with their seat number on it. The seat numbers are allocated based on queue-position, so that, for example, the 21st person in the line has seat number 21, the 32nd has seat number 32, the 77th has seat no 77 etc etc.
Unfortunately, the first person in the queue loses their ticket and on entering the cinema decides to sit in a random seat. If everyone else tries to sit in their allocated seat, what is probability that the final (100th) person in the queue will sit in seat number 100?
Answer: The probability was exactly 50%, as neatly worked out (see solutions below) by: John Bowen, rubber industry consultant, Bromsgrove, Worcs, UK; Stephan Paischer, head of product management special products, Semperit AG Holding, Vienna, Austria; Amparo Botella, responsable de Compras y Calidad, Ismael Quesada SA, Elche, Alicante, Spain; Rohit Kalé, distribution strategy manager, AMN/V/B2C/DIS, Michelin North America Inc., USA: Michele Girardi, quality manager, Scame Mastaf Spa, Suisio, Italy; and everyone else who had a go.
The sample space of the last queue member, consisting of all possible free seats, has only two elements. Either the seat intended for the first member is free or the last member’s seat is available. The reason is, that if another seat is not occupied, it was also not occupied when the member with that seat number entered the cinema.
Hence it would be occupied by that member. Consequently, only the first and last members’ seats can be free. The queue would not be affected, in particular the behavior of the first 99 members would not be affected, if we switch the tickets of the first and last members.
So, the last member’s seat will be free with the same probability as in the original scenario. The last member’s seat will be free in either the original or the adapted scenario. As the probabilities are equal and add up to one. The probability is a half.
The probability for the last person to get the assigned seat is 50%.
To solve this, we need to build up the approach from very low numbers of users.
Case 1: 2 users
The first user may chose one of two available seats, the second user needs to take the remaining one – the chance that it’s his assigned one is 50%.
1 – 2
R – R
W(2) – W(1)
Case 2: 3 users
Here there are 4 possible scenarios – 2 of which the 3rd user gets his assigned seat, 2 of which he doesn’t:
1 – 2 – 3 (R = right seat, W = wrong seat)
R – R – R
W(2) – W(1) – R
W(2) – W(3) – W(1)
W(3) – R – W(1)
For each number of users, the same pattern shows: in 50% of the cases, the last user will get his assigned seat, in 50% of the case he won’t.
This applies also for a large number such as 100 users.
As 98 seats would have been taken by the correct seat holders leaving only the one who lost the ticket and the final person with 2 seats.
So there is a possibility of the person with without ticket seating on the correct seat or occupying the seat of the final person. Hence it is 50%
When 1 sits in A
- persons 2 to A-1 will sit correctly
- A will sit in B >A
- A+1 to B-1 will sit correctly
- B will sit in C>B
and so on.
At a certain point X will sit in 1 or 100. In the first case X+1 to 100 will sit correctly , in the second X+1 to 99 will too and 100 will sit in 1.
Since for every sequence A,B,C,...,X,1 there's a sequence A,B,C,...,X,100 and vice versa, there's a 50% probability that the final person in the queue will sit in the correct seat.
New teaser on Monday.