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Crazy Woman in Front!
There is an island with 120 residents living on it, all friends, family... of which, one of them is crazy.
The island has 1 airport, and 1 plane, capable of seating 120 passengers. One day, the island leader decides to be extremely kind, and throws a party on a nearby island, all tickets purchased by him.
The people are obviously very excited, and come the day of the flight, people start to queue up to enter the plane. The craze girl, being the most excited, is the first in line, and the island leader having organized the event, ends up at the end of the queue.
Each person has their respective tickets, and each ticket has a unique seat number, however the crazy person couldn't care less which seat she gets. Each person that enters goes to their seat, in the event that their seat is occupied however, they then select the nearest seat to it and sit there.
Now, what is the probability that the island leader, being the last in the queue, actually gets to sit in his designated seat?
Comments
Probability is simply the probability that crazy girl sat in the right seat. Anywhere else and the last person on the plane will not get the right seat.....
So 1:120 :-)
The passengers are told their designated (randomly allocated) seats as they board the plane but the crazy woman can choose at random to sit in any seat and passengers whose designated seats are then occupied must sit in the nearest empty seat.
If any passenger sits in the crazy woman’s designated seat then all subsequent boarding passengers, including the leader, must sit in their own seat.
If any passenger other than the leader sits in the leader’s designated seat then all subsequent boarding passengers, except the leader, must sit in their own seats. In this case the leader must sit in the only unoccupied seat and that will be the crazy woman’s designated seat.
The passenger who selects one or the other of the above options determines the outcome, i.e. whether the leader actually gets to sit in his designated seat.
For the nth passenger boarding the plane (where 1 >= n > 120 and the passenger is either the crazy woman or a passenger whose designated seat is occupied), the probability that the nth passenger sits in the crazy woman’s designated seat = the probability that the nth passenger sits in the leader’s seat = 1/(121-n).
Therefore the probability that the nth passenger determines the outcome = 2/(121-n) and the probability that selecting the outcome is deferred to a subsequently boarding passenger = (119 – n)/(121 – n).
The probability that the nth passenger sits in the crazy woman’s seat divided by the probability that the nth passenger determines the outcome = 1/2.
Since there is only one available seat when the leader boards, one of the earlier boarding passengers must determine the outcome and therefore the probability that the leader sits in his designated seat = 1/2.