Is it tough, huh?

Probability Level 3

One hundred people line up to board an airplane. Each has a boarding pass with assigned seat. However, the first person to board has lost his boarding pass and takes a random seat. After that, each person takes the assigned seat if it is unoccupied, and one of unoccupied seats at random otherwise. What is the probability that the last person to board gets to sit in his assigned seat? (assume that there are only 100 seats in the plane.)

0.5 o.46 0.33 0.1 0.12 0.003 0.45 0.008

3 solutions

Cheah Chung Yin
Apr 20, 2015

The easiest way to think about this is noticing that the final seat can only be either the first person's seat or the last person's actual seat. If for example, the last seat happens to be the 10th person's seat, then by the time the 10th person got on board he would have occupied that seat instead of other seats already. This is a contradiction. Hence only the first and last person's seat could be left, thus 1/2 probability

Panshul Rastogi
Apr 16, 2015

Look at the situation when the k’th passenger enters. Neither of the previous passengers showed any preference for the k’th seat vs. the seat of the first passenger. This in particular is true when k = n . But the n’th passenger can only occupy his seat or the first passenger’ s seat. Therefore the probability is 1/2.

If there would have been 200 or 300 people, then the answer would have remained the same?

Archit Boobna - 6 years, 1 month ago

yes, the answer would have remained the same in any no. of people.

Panshul Rastogi - 6 years, 1 month ago

Ok Thanks.

Archit Boobna - 6 years, 1 month ago

@Archit Boobna – How did you do it?

Adarsh Kumar - 6 years, 1 month ago

@Adarsh Kumar – I tried to use induction.

I thought that the number "100" doesn't play a significant role in the solution so the answer shouldn't depend on the no. of people.

Let the probability be x if there are 100 people, then by induction, the probability should also be x for 101 people.

Then I assumed k people are already sitting and then found the probability for (k+1)th person as f(k).

Then I assumed k+1 people are already sitting and then found the probability for (k+2)nd person as g(k).

I put f(k)=g(k) and got P=1/2

Archit Boobna - 6 years, 1 month ago

@Archit Boobna – nice try...

Panshul Rastogi - 6 years, 1 month ago

@Archit Boobna – What motivated you to use this method?Have you seen something like this before?

Adarsh Kumar - 6 years, 1 month ago

@Adarsh Kumar – Not really, but I just thought that it should remain same for all numbers

Archit Boobna - 6 years, 1 month ago

@Archit Boobna – oh!k thanx!

Adarsh Kumar - 6 years, 1 month ago

Also.. one more method is that we could try the same case with lesser no. of people, say 3 and 4. Here as we would get the probability as 0.5 only. so, the probability would be same for 100 also i.e., 0.5 :)

Panshul Rastogi - 6 years, 1 month ago

Yes, that is also a way.

Archit Boobna - 6 years, 1 month ago

Ayush Agrawal
Apr 21, 2015

I saw it like this: either one of the 100 people occupy the last seat or they don't. In case any one of the 99 people do, the last person doesn't get to sit on it. Or if they don't, he does get to sit on it! There are only two options: he does or he doesn't (Yay Yoda!). Hence a 0.5 probability.

Is it tough, huh?

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