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. 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)?
Solution :
For four passengers, the possiblities are:
(x-ing out the ones that are not possible--because a "normal" passenger would sit in their own seat if it is open)
1234
1243 x
1324 x
1342 x
1423 x
1432 x
2134
2143 x
2314 x
2341 x
2413 x
2431 x
3124
3142 x
3214
3241 x
3412 x
3421 x
4123
4132
4213
4231
4312 x
4321 x
Of the 8 valid possibilities, 1/2 of them have player 4 sitting in the his own seat.
An easier way to think about it:
There are only two (relevant) possible outcomes of a random seat choice:
1) Seat 1 (1/100 chance)
2) Seat 100 (1/100 chance)
3) Some other seat (98/100 chance)
Case 1 - Each other person can sit in their assigned seat, and the last passenger can sit his own. Let's call this a "win"
Case 2 - Each other person can sit in their assigned seat, but the last passenger's seat is taken, and he has to sit in seat 1. Let's call this a "loss"
Case 3 doesn't particularly interest us, because it simply defers the "decision" point to a passenger later, who will make another random choice.
At each point, the chances of randomly choosing seat 1 is the same as randomly choosing seat n, so the chances will always be 1/2, no matter how many seats/passengers we're talking about.
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