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Consider there's a flight from some place to another. The flight has $100$ passengers. The passengers alight the plane one by one. The first passenger doesn't have his boarding pass so he goes and seats on a random seat. The next passenger then goes to his seat if its unoccupied otherwise takes a random seat. Find the probability that the last person seats in the seat assigned to him. $$\text{Approach 1}$$ The seat can either be occupied or unoccupied so the last person seats in his place is $\frac{1}{2}$. However I felt this logic had a flaw so I took another route. $$\text{Approach 2}$$ first consider the case that the first person seats in the correct place then the probability is $\frac{1}{100}$ the probability that last person seats in correct place is just $1$ as its mentioned in the question. Now consider the case where the first person doesn't seat in his or last person's seat is $\frac{1}{98}$. Let the last person get the correct seat so the number of ways of $98$ de arrangements is $98!\sum_0^98 \frac{(-1)^n}{n!}=p$ . The total no of ways of placing 98 passenger is $98!$ hence total probability $\frac{1}{100}+\frac{1.p}{98.98!}=0.01+\frac{0.3678}{98}\neq 0.5$ . Where is the flaw in the $2$and approach?

JMoravitz
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Archis Welankar
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    "The seat can either be occupied or unoccupied, so ... $\frac{1}{2}$" Compare that statement to "You either win the lottery or you lose the lottery." – JMoravitz Jun 20 '19 at 14:22
  • Why is the flaw expected to be on the $2^{nd}$ aproach rather than the $1^{st}$ There was a Vsauce2 video on YouTube about a very similar problem, but I failed to find it – David Jun 20 '19 at 14:26
  • @David presumably because the OP has already seen the answer but not the explanation and knows that the answer should wind up being $\frac{1}{2}$ but not why. – JMoravitz Jun 20 '19 at 14:27
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    Many passengers end up in the rights seats, so derangements aren't relevant. If passenger $1$ sits in seat $10$, for example, then passengers $2$ through $9$ occupy the correct seats. – saulspatz Jun 20 '19 at 14:29
  • @Jmoravitz I didn't know the answer I while I was writing the other question clears my doubt. – Archis Welankar Jun 20 '19 at 14:29
  • This question is indeed a duplicate and it was very cleverly solved back then: Every sit from 2 to 99 will be occupied by the time passenger 100 takes his turn. Also, at every step, sits 1 and 100 are equally likely to be taken – David Jun 20 '19 at 14:30

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