There are 100 people in line to board a plane with 100 seats. The first person has lost their boarding pass, so they take a random seat. Everyone that follows takes their assigned seat if it's available, but otherwise takes a random unoccupied seat. What is the probability the last passenger ends up in their assigned seat?

Event:

S=success for last passenger to get his assigned seat.
1=case 1
2=case 2
Note that cases 1 and 2 are complementary.

Case 1: P(1)=1/100
If the first person chose (randomly) his assigned seat
Then all the rest of the passengers will also get their assigned seats.
=> P(S|1)=1

Case 2: P(2)=99/100
If the first passenger did NOT choose the right seat.
Then
case 2a: the seat occupied by first passenger is never taken by the next 98 passengers, so the last passenger will be left with the first passenger's assigned seat => P(S)=0
case 2b: if the assigned owner of the seat taken by the passenger boards the plane, then he will have to occupy another seat. His own assigned seat will always be occupied by another passenger, including the last one, so again P(S)=0.
Conclusion: P(S)=0 for either case 2a or case 2b, -> P(S|2)=0 for case 2.

Therefore, by the law of total probability,
P(S)=P(S|1)*P(1)+P(S|2)*P(2)
=1/100*1 + 99/100*0
=1/100
(i.e. if the first passenger happened to have chosen his own seat)

Fancy answer. Too bad its wrong. The correct answer turns out to be 1/2.

To determine the probability that the last passenger ends up in their assigned seat, we can consider the scenario step by step.

Step 1: The first passenger takes a random seat.
Since the first passenger has lost their boarding pass, they choose a random seat from the 100 available. At this point, there is a 1/100 probability that they would end up in their assigned seat, as there is only one seat assigned to them out of the 100.

Step 2: The second passenger boards.
If the second passenger's assigned seat (the first passenger's seat) is free, they will take it. The probability of this happening is 1/100, as mentioned in Step 1.

Step 3: The rest of the passengers board.
From the third passenger onward, there are two possibilities for each passenger:

(i) Their assigned seat is free: In this case, they would take their assigned seat. The probability of this happening is 1/100 at each step, considering the previous passengers have followed this pattern.

(ii) Their assigned seat is occupied: In this case, they would choose randomly from the remaining unoccupied seats. The probability of successfully taking a random unoccupied seat is (100 - n + 1) / 100, where n represents the number of seats occupied so far.

Let's calculate the probability of the last passenger ending up in their assigned seat.

The probability of the last passenger ending up in their assigned seat can be written as:
P(last passenger ends up in their assigned seat) = P(assigned seat is available in each step)

Using this approach, we can develop the following equation:
P(last passenger ends up in their assigned seat) = (1/100) * (1/100) * (1/100) * ... * (1/100)

Since there are 100 passengers, we need to multiply (1/100) by itself 99 times:

P(last passenger ends up in their assigned seat) = (1/100)^99

Evaluating the expression (1/100)^99 gives us an extremely small number:

P(last passenger ends up in their assigned seat) ≈ 0.366032
(or approximately 0.366%, rounded to three decimal places)

Therefore, the probability that the last passenger ends up in their assigned seat is approximately 0.366%.