His Own

What is the probability of the last passenger finding his own seat free?
Imagine you have a plane with 100 seats and the first passenger to enter seats in another persons seat.
The following passengers seat in their own seat if they are free. If they find it occupied they take any seat.
What is the probability of the last passanger finding his own seat free?
49/99
Proof:
1) Probability that first passenger (P1) sits in last passengers (P100) seat (S100) is 1/99
2) probability that P1 sits in S99 , and P99 sits in S100 is (1/99) * (1/2)
3.1) probability that P1 sits in S98, and P98 sits in S100 is (1/99) * (1/3)
3.2) probability that P1 -> S98, P98->S99, P99->S100 is (1/99) * (1/3) * (1/2)
4.1) P1->S97, P97->S100 = (1/99)*(1/4)
4.2) P1->S97, P97->S99, P99->S100 = (1/99)*(1/4)*(1/2)
4.3) P1->S97, P97->S98, P98->S100 = (1/99)*(1/4)*(1/3)
4.4) P1->S97, P97->S98, P98->S99, P99->S100 = (1/99)*(1/4)*(1/3)*(1/2)
5.1) P1->S96, P96->S100 = (1/99)*(1/5)
5.2) P1->S96, P96->S99, P99->S100 = (1/99)*(1/5)*(1/2)
5.3) P1->S96, P96->S98, P98->S100 = (1/99)*(1/5)*(1/3)
5.4) P1->S96, P96->S98, P98->S99, P99->S100 = (1/99)*(1/5)*(1/3)*(1/2)
5.5) P1->S96, P96->S97, P97->S100 = (1/99)*(1/5)*(1/4)
5.6) P1->S96, P96->S97, P97->S99, P99->S100 = (1/99)*(1/5)*(1/4)*(1/2)
5.7) P1->S96, P96->S97, P97->S98, P98->S100 = (1/99)*(1/5)*(1/4)*(1/3)
5.8) P1->S96, P96->S97, P97->S98, P98->S99, P99->S100 = (1/99)*(1/5)*(1/4)*(1/3)*(1/2)
1) (1/99)
2) (1/99)(1/2)
3) (1/99)(1/3)(1 + ½)
4) (1/99)(1/4)(1 + 1/3)(1 + ½)
5) (1/99)(1/5)(1 + ¼)(1 + 1/3)(1 + ½)
i) (1/99)(1/i)product(j=2 to (i-1),(1+1/j))
99) (1/99)(1/99)product(j=2 to 98,(1+1/j))
Grand Sum = (1/99)*(1 + ½ +sum(i=3 to 99,(1/i)*product(j=2 to (i-1),(1+1/j))))
=(1/99)*(1 + ½ + sum(i=3 to 99,(1/2)))
=(1/99)*(1 + 98/2)
= 50/99
Probability that last passenger gets bumped = 50/99
Hence Probability that last passenger does NOT get bumped = 1 – 50/99 = 49/99
A bit less than 50%
Would be 1/2 like others suggest if first passenger was not constrained to sit in the wrong seat as given in the problem statement.
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