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:: Crazy guy on the plane::

A line of 100 airline passengers is waiting to board a plane. They each hold a ticket to one of the 100 seats on that flight. (For convenience, let's say that the nth passenger in line has a ticket for the seat number n.)

Unfortunately, the first person in line is crazy, and will ignore the seat number on their ticket, picking a random seat to occupy (equally likely). All of the other passengers are quite normal, and will go to their proper seat unless it is already occupied. If it is occupied, they will then find a free seat to sit in, at random.

What is the probability that the last (100th) person to board the plane will sit in their proper seat (#100)?

Answer: The jumping stops if someone sits on the crazy man's seat before the 100th persons, hence, the 100th will sit in his/her seat for sure then. Similarly, if anyone sits on the 100th passengers seat at any point, then he/she won't get to sit his/her seat for sure. If at any point someone picks a random seat other than #1 or #100, then the jumping is postponed till that seat number is reached. At any jumping point there is a equal probability that seats #1 or #100 will be chosen (other choices will just delay the problem till later down the line). By symmetry, the probability of the 100th passenger sitting in their seat is 50-50, since at all jump points there's an equal probability of choosing seats #1 or 100.

EXTRA:

e.g. let there be 4 seats in total.

case 1: crazy guy sits in place #2 -
Prob[2nd guy gets crazy seat]= 1/3
Prob[2nd guy doesn't sit in crazy guy seat nor last seat]*
Prob[3rd guy sits in crazy guy seat] = 1/3*1/2= 1/6
Sum= 3/6 = 0.5

case 2: crazy guy sits in place 3 -
Prob[3rd guy sits in crazy guy seat] = 1/2 =0.5

both cases are equally probable thus overall prob = 0.5.

e.g. 2: Let there be 5 seats in total.

One extra first case : crazy guy sits in place #2 -
Prob[2nd guy gets crazy seat]= 1/4
Prob[2nd guy sits in 3rd guy seat]*
Prob[3rd guy sits in crazy guy seat] = 1/4*1/3= 1/12
Prob[2nd and 3rd guys end up in each others seat]*
Prob[4th guy sits in crazy guy seat] = (1/4*1/3)*1/2= 1/12
Prob[2nd guy sits in 4th guy seat]*Prob[4th guy sits in crazy guy seat] = 1/4*1/2=1/8
Sum= 12/24 = 0.5

Having cosidered this extra case the problem then reduces to the case with total seats.

Of course you can reach a general formulae using combinatorics for the case with arbitrary number of seats - but why do that when the problem can be solved as simply as discussed above?

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