This is a rather interesting probability problem I saw, and which really has an absolutely deceiving answer. There is an airplane with 100 seats, and 100 flyers each have 1 seat number assigned to them. Additionally these passengers board the plane in order, so 1 goes first. But the first passenger unknowingly sits in any random seat, and from then on all the passengers sit in their correct position. That is, unless their seat is taken, in which case they also take a random seat (they do not want to disturb anyone seated: this is a flight to Canada). What is the probability that the final passenger sits in his/her correct seat (#100)?

Try out the problem yourself: give it at least a day of thought before scrolling below the picture to see the solution.

The solution to this problem is one that will almost guarantee your anger. All the deliberations, all the calculations, and all the scenarios, and yet the answer is so simple. One way to think about this problem is by using 3 passengers instead on 100 and trying to use basic induction from there onwards. Let’s think about it another way though.

Here is the question that drives the answer: Which seats can the last person sit in? The only possible seats for the last passenger to take are #1 and #100. Now let’s think about why. Every passenger from 1 to 99 has been seated, and they either sat in their own place, which means the number is occupied, or sat in a random place, which also means the number is occupied because someone else was sitting there. But the exception to this was passenger 1, who directly sat in a random place. So we know 2-99 are occupied for sure, and so #1 and #100 are the only possibilities for the last seat. Both of these seats have equal chances of being sat on as a random seat, and therefore there is an equal likelihood of either being the last seat. So we come to a very clean answer: the probability is 50%…

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While I figured out the answer without much trouble, I took a very complex route.

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