A line of 100 airline passengers is waiting to board the plain. 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. All 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)?
The solution is correct, but over-complicated.
Consider a plane with only two seats.
It is clear that the probability that passenger 2 gets his own seat in this case is 50%.
Either the mad passenger picks his own seat, and all is well, or he picks seat 2, and the second passenger is unlucky.
For a plane with any more than 2 seats remaining, there are really ony three cases to consider.
1.The mad passenger picks his own seat [1/N]
2.The mad passenger picks seat N [1/N]
3.The mad passenger picks any other seat [(N-2)/N}]
Whenever case 3 occurs, seating proceeds in an orderly manner, until the 'stolen' seat is reached.
At this point, the situation is identical, save only that N is smaller.
The process repeats until either case 1 or case 2 occurs, and the probabilities of these are always equal. One or other will occur eventually, if only when N=2.
Case 3 may recurr several times, but this cannot affect the outcome.
An equivalent problem is this.
Put one black ball, one white ball, and 98 red balls in a bag.
Pick a ball at random.
If the ball is red, remove a random number, 1 to N, of red balls from the bag, where N is the number of red balls remaining (initially 98).
Repeat until either the black ball or the white ball is picked.
What is the probability that the white ball is picked before the black ?
Simpler solution
