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Interesting probability problem – Crazy airplane guy

  • Writer: Rana Basheer
    Rana Basheer
  • May 19, 2011
  • 2 min read

This problem I found on the following website. I am just restating it below

A line of 100 airline passengers is waiting to board a plane. They each hold a ticket to one of the 100 seats on that flight. (For convenience, let’s say that the nth passenger in line has a ticket for the seat number n.)  Unfortunately, the first person in line is crazy, and will ignore the seat number on their ticket, picking a random seat to occupy. All of the other passengers are quite normal, and will go to their proper seat unless it is already occupied. If it is occupied, they will then find a free seat to sit in, at random.  What is the probability that the last (100th) person to board the plane will sit in their proper seat (#100)?

Their Answer

Now according to the solution provided by the website the answer is 0.5. Their reasoning is that the crazy situation in which people picking random chairs to sit ends when one of the non-crazy passenger who lost his seat picks the crazy first person’s seat. Hence the probability of the 100th person sitting on the right chair is the sum of probability of passenger 2 to 98 picking the first person’s chair

So that comes out according to them as

P(S_{100}=100)=\frac{1}{99}+\frac{1}{99\times 98}+\frac{1}{98\times 97}+\cdots+\frac{1}{2\times 3}=0.5

I feel the above reasoning is flawed. A passenger who lost his seat could pick the crazy first person’s seat but someone before him would have already picked the 100th person’s seat. I don’t think that scenario is taken into account in their reasoning.

My Answer

So in my view the probability can be solved in reverse way as

P(S_{100}=100) = 1 - P(S_{100}\ne 100)

In the above equation the notation

probability of

person sitting at

seat while 

probability of

person not sitting at

seat.

The probability that the

person will not sit in his chair is the sum of the probability of any other person before him sitting in that chair. Therefore, this probability is given by

P(S_{100}\ne 100)=P(S_1\ne 1\bigcap S_1=100)+P(S_2\ne 2\bigcap S_2=100)+P(S_3\ne 3\bigcap S_3=100)+\cdots+P(S_{98}\ne 98\bigcap S_{98}=100)+P(S_{99}\ne 99\bigcap S_{99}=100)

where

means the

person did not get to sit in chair 98 but then ended up sitting on chair 100.

Using the reltationship

we get 

and since the crazy dude is always going to sit in someone else seat,

and

. The reason for

 is because he has equal chance of picking one out of the 99 chairs ahead of him.

For the

person when his seat is occupied and he has to pick a random seat from the available pool of

 seats the chance of him picking any seat

are all equal and is given by 

P(S_i=j|S_i\ne i)=\frac{1}{101-i}
P(S_{100}\ne 100)=\frac{1}{99}+\frac{P(S_2\ne 2)}{99} +\frac{P(S_3\ne 3)}{98}+\cdots+\frac{P(S_{98}\ne 98)}{3}+\frac{P(S_{99}\ne 99)}{2}

Similarly

P(S_{99}\ne 99)=\frac{1}{99}+\frac{P(S_2\ne 2)}{99} +\frac{P(S_3\ne 3)}{98}+\cdots+\frac{P(S_{98}\ne 98)}{3}

Therefore,

P(S_{100}\ne 100) = P(S_{99}\ne 99)+\frac{P(S_{99}\ne 99)}{2}=P(S_{99}\ne 99)\frac{3}{2} =P(S_{98}\ne 98)\frac{3}{2}\frac{4}{3}=\frac{3}{2}\frac{4}{3}\cdots\frac{99}{98}\frac{100}{99}P(S_2\ne 2)=50P(S_2\ne 2)=\frac{50}{99}=0.5051

or

P(S_{100}=100)=1=P(S_{100}\ne 100)=0.4949

Expanding the problem to N seats

Now lets expand this problem to a scenario with N seats and the first person as previously is crazy and won’t sit in his chair and will select randomly a chair from 2 to N to sit. What is the probability that the last person gets to sit in his chair.

Using our previous reasoning the probability that

person won’t get to sit in his chair is given by

P(S_i\ne i)=P(S_1\ne 1)P(S_1=i|S_1\ne 1)+P(S_2\ne 2)P(S_2=i|S_2\ne 2)+P(S_3\ne 3)P(S_3=i|S_3\ne 3)+\cdots+P(S_{i-2}\ne i-2)P(S_{i-2}=i|S_{i-2}\ne i-2)+P(S_{i-1}\ne i-1)P(S_{i-1}=i|S_{i-1}\ne i-1)

Since the first guy is not going to sit in his chair P(S_1\ne 1)=1$ and then he is going to choose any of the N-1 seats in random with probability given by

. For the

person, the probability of him sitting on

seat when his seat is occupied is given by 

.

Therefore,

P(S_i\ne i)=\frac{1}{N-1}+\frac{P(S_2\ne 2)}{N-1} +\frac{P(S_3\ne 3)}{N-2}+\cdots+\frac{P(S_{i-2}\ne i-2)}{N+3-i}+\frac{P(S_{i-1}\ne i-1)}{N+2-i}

Similarly for the

person the probability that he won’t sit in his chair is

P(S_{i-1}\ne i-1)=\frac{1}{N-1}+\frac{P(S_2\ne 2)}{N-1} +\frac{P(S_3\ne 3)}{N-2}+\cdots+\frac{P(S_{i-2}\ne i-2)}{N+3-i}

That means

P(S_i\ne i)=P(S_{i-1}\ne i-1)+\frac{P(S_{i-1}\ne i-1)}{N+2-i}=P(S_{i-1}\ne i-1)\frac{N+3-i}{N+2-i}= \frac{N}{N+2-i}P(S_2\ne 2)=\frac{N}{(N-1)(N+2-i)}

For the last guy the chance of him not sitting in his chair is when

or alternatively the probability that he will sit in the chair is

P(S_N=N)=\frac{N-2}{2N-2}

In the limiting case

\lim_{N\rightarrow\infty}P(S_N=N)=\lim_{N\rightarrow\infty}P(S_N\ne N)=0.5

Verification

I always like to verify my answer with a montecarlo simulation. The result for

seats/passengers with the first guy being crazy and the probability of the last guy getting his seat is shown below. The montecarlo simulation used 100,000 runs to determine the probability.


Monte carlo simulation of crazy guy problem

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