## Puzzle #149: The Funny Plane Seating

A very catchy puzzle contributed by my friend and entrepreneur, Pallav Pandey – thanks Pallav!

Imagine there are a 100 people in line to board a plane that seats 100. The first person in line realizes he lost his boarding pass so when he boards he decides to take a random seat instead. Every person that boards the plane after him will either take their “proper” seat, or if that seat is taken, a random seat instead.

Question: What is the probability that the last person that boards will end up in his/her proper seat?

As always, please send your answers as comments within the blog (preferred), or send an e-mail to alokgoyal_2001@yahoo.com. Please do share the puzzle with others if you like, and please also send puzzles that you have come across that you think I can share in this blog.

Happy seating!

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### 7 Responses to Puzzle #149: The Funny Plane Seating

1. pch967 says:

Sits on his designated seat in 1st attempt – P(1) – 1/100
Sits on his designated seat in 2nd attempt – P(2) – 98/100×1/99
and so on….
we get P(n)=(100-n)/(100*99)

Probability of last person boarded to sit on his designated seat = P(1)+P(2)+P(3)+ ….. +P(98)
which comes out to be = 4949/9900 = 0.499898…

2. Prakhar says:

I am guessing that the first person picks up a random seat with a probability of 99/100 that it is not the last person’s seat. The second person’s probability of not having the last person’s seat is 98/99. Likewise for the rest. The resulting probability will be 99/100*98/99*97/98……. So the answer should be 1/100

3. Prakhar says:

My son says it is 1/10000.

4. Prakhar says:

Disregard my previous responses. I need to rethink as I read the puzzle again ๐

5. Prakhar says:

Finally , I think I got it right. But in a rather crude manner. Took a case of 2 , then 3, and then 4. Got 50% favorable cases. I am sure there is an elegant explanation for it which we will see soon from some of the smarties ๐

6. Abhinav Jain says:

the probability that the last person that boards will end up in his/her proper seat is 0.5

7. 1/2

Very famous puzzle. Only one of the two seats can be remaining when the last guy comes – either the seat allocated to the first person or the last person (anything else cannot be there can be easily proved by contradiction)

Now, since for all the people taking the seat after the first person and before the last person, there is no difference between the two seats.

So probability that the last person will get his correct seat is 0.5