The Plane Seat Problem

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I came across this riddle about aeroplanes and probability (two of my favourite topics!):

There are 100 passengers lined up (in a random order) to board a plane. The plane is fully booked, meaning there are exactly 100 seats available, assigned to the 100 passengers. The first passenger chooses a seat at random, with all seats equally likely. Then, each of the other passengers proceeds as follows: if their assigned seat is free, they will sit in it; otherwise, they will take a random available seat. What is the probability that the last passenger will sit in their assigned seat?


Here's my analysis (using the terminology P1-P100 for the passengers, S1-S100 for the assigned seats, and DP1-DPn for the “displaced” passengers):


For passenger P1: 

P1 (randomly) makes one of three choices:

  1. Sits in his own seat (S1), in which case everyone sits in their assigned seat. 
  2. Sits in the last passenger’s seat (S100), in which case everyone else sits in their assigned seat, except P100 (who has to sit in S1).
  3. Sits in the seat of another passenger (S2-S99), which means all passengers from P2 to DP1 [excluding DP1] sits in their assigned seats. 

For passenger DP1: 

At this stage, all the passengers from P2 up to DP1 are sitting in their assigned seats, and therefore the number of empty seats will be (100 – DP1 + S1).

Now, DP1 (randomly) makes one of these three choices:

  1. Sits in S1, in which case all remaining passengers sit in their assigned seat. 
  2. Sits in S100, in which case also everyone else sits in their assigned seat, except P100 (who has to sit in S1).
  3. Sits in the seat of passenger (DP1 + 1)-P99, which means all remaining passengers after DP1 up to DP2 sits in their assigned seats, and DP2 will also have to make one of the same three choices.  

 

For DP2-DPn:

The same options as for passenger DP1 apply, with a smaller set of empty seats to choose from (If everyone gets displaced, for the last displaced passenger (DP98), only options 1 and 2 are available).

 

Notice the pattern – The selecting passenger decides the fate of the last passenger in only two cases: one where the last passenger gets his seat, and the other where the last passenger does not (which is 1. and 2. above). Both of these have equal probabilities, and therefore there is (somewhat surprisingly!) a 50-50 chance that the last passenger gets his seat. 

 

 

 

 

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