Most likely outside the scope of anything you will see on

I did this problem in two ways: the simple way and the complicated way. Each yielded a different solution !

The simple way
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The first problematic person will create a wave of mess and randomness, in which case the last person will definately not sit in his own seat as the incorrect seating point will touch someone eventually. So the probability that the last person will sit in his correct seat, as it shows in his ticket, is merely the probability that the first person will randomly choose his own seat = 1/100 = 1 %

So the answer is 1 % <--- more likely to be correct for me


The complicated way
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Assuming we're talking about 2 people only, then there is 50 % the first person, the one who doesn't read, will correctly select his seat, although at random. The second person will not effect any probabiliy number because he will select whatever seat that's left. This says that the last event does not count.

let's examine the case of 4 pessengers. The first person, who doesn't read, has a probability of selecting his correct seat of 0.25 . Now, the second person has a probability of 1/3 x 3/4 that he'll find his seat taken and thus will select another seat at random.

The 3rd person will have a probability of 1/2 x 1/3 x 3/4 of finding his seat being taken. the 4th and last pessenger will have a probability of 1/16 that will have his seat taken --> 15/16 probability that he'll sit in his own seat, as it shows in his ticket.

It looks like the probability that the last person will end up sitting in his seat is 1- (1/2^n) where n is the total number of pessengers/seats.

so for 100 pessengers, it is 1 - (1/2^100) ~ 100% !!!