Airplane Boarding Puzzle Answer: 50% [Solved]

Answer: 50%. The airplane boarding puzzle answer is 50%, so the last passenger gets their own seat exactly half the time, no matter how many passengers board.

One hundred passengers board a plane with one hundred seats, and the first passenger has lost his boarding pass. He picks a seat at random, and every passenger after him takes their own seat if it is free, or picks a random free seat if it is taken. The airplane boarding puzzle answer stays at 50% for any size of plane, and the reason is a short chain of logic.

The Puzzle

Picture the scene in full. A plane has 100 seats, numbered 1 to 100, and 100 passengers board one at a time in order. Passenger 1 has lost his boarding pass, so he chooses a seat at random, with each of the 100 seats equally likely. Passenger 2 boards next. If seat 2 is free, he sits in it. If seat 2 is already taken, he chooses a random free seat. Every passenger after him follows the same rule: take your own seat if it is free, otherwise pick a random free seat.

You are passenger 100, the last to board. What is the probability that you find your own seat, seat 100, still free?

The question is known as the lost boarding pass puzzle. Peter Winkler posed it in his book Mathematical Puzzles: A Connoisseur’s Collection, and it has since become a staple of quant and tech interviews, where it rewards finding the structure of a problem instead of grinding through cases.

The Airplane Boarding Puzzle Answer With 2 Passengers

Shrink the plane. With only 2 passengers and 2 seats, the whole puzzle fits in your head.

Passenger 1 picks seat 1 with probability 1 in 2, and passenger 2 then sits in seat 2. Passenger 1 picks seat 2 with probability 1 in 2, and passenger 2 is left with seat 1. So the last passenger gets their own seat half the time.

That is the answer for 2 passengers: 1 in 2. The surprise is that the answer does not change when the plane grows.

The Airplane Boarding Puzzle Answer With 3 Passengers

Now try 3 passengers and 3 seats. Passenger 1 has three choices, each with probability 1 in 3.

Passenger 1 picks	What happens next	Passenger 3 gets seat 3?
Seat 1	Passengers 2 and 3 sit in their own seats	Yes
Seat 2	Passenger 2 picks randomly between seats 1 and 3	Yes, half the time
Seat 3	Passenger 2 sits in seat 2, passenger 3 is displaced	No

The first row contributes 1 in 3, the second row adds 1 in 3 times 1 in 2, which is 1 in 6, and the third row adds nothing. The total is 1 in 3 plus 1 in 6, which is 1 in 2.

So with 3 passengers the answer is still 50%. Two sizes in, a pattern is forming. The next step is to see why it holds for any size.

Why the Last Seat Is Always Seat 1 or Seat 100

Watch what happens after passenger 1 sits down. Suppose he picked seat 7. Passengers 2 through 6 find their own seats free and sit in them. Passenger 7 finds seat 7 taken, so he picks a random free seat. Seat 1 ends the chain kindly: every passenger after him finds their own seat free, and passenger 100 gets seat 100. Seat 100 ends it the other way: passenger 100 is already displaced before boarding. A middle seat, say seat 40, keeps the chain going: passengers 8 through 39 sit in their own seats, and passenger 40 picks at random.

The chain only ever ends in two ways: somebody picks seat 1, or somebody picks seat 100. Until one of those two seats is picked, the chain keeps going, and passenger 100 has not lost yet.

Now the key point. At every random pick in the chain, seats 1 and 100 are both still free, and both are equally likely to be the one picked. So the first time the chain lands on either of them, it is a fair coin flip: seat 1 with probability 1 in 2, seat 100 with probability 1 in 2.

Seat 1 means passenger 100 gets their seat. Seat 100 means they do not. That is why the airplane boarding puzzle answer is 50% for any number of passengers, whether the plane holds 3 seats or 300.

Put it another way: take any boarding where passenger 100 gets their seat, and swap seats 1 and 100 in every random choice. You get an equally likely boarding where passenger 100 loses, so winning and losing boardings pair up exactly, and the split is even.

The Trap Most Candidates Fall Into

The most common wrong answer is 99 in 100. The reasoning sounds convincing: passenger 1 takes my seat with probability 1 in 100, so I keep mine the other 99 times. But that reasoning treats passenger 1 as the only threat, and it misses the chain. Passenger 1 might take seat 7, and passenger 7 might take seat 100. Your seat can be taken by a passenger who was displaced, not only by passenger 1.

The second wrong answer is 1 in 100, from the opposite instinct: with 100 seats, surely the last passenger is nearly always out of luck. That instinct ignores the other half of the chain. Passenger 1 might take seat 1 himself, and then nobody is displaced at all.

The third trap is to start computing: adding up the probability over all 100 seats passenger 1 might pick, then over every seat the next displaced passenger might pick, until the arithmetic swallows the interview. The question rewards the candidate who stops and asks what the last free seat can be.

What the Interviewer Is Testing in the Airplane Boarding Puzzle

This question tests three skills at once. First, it tests small-case thinking. The candidate who tries 2 passengers, then 3, spots the pattern in a minute. The candidate who attacks 100 passengers head on gets stuck.

Second, it tests whether you can find the invariant: the last free seat is always seat 1 or seat 100. Good probability questions have one clean observation that collapses the whole problem.

Third, it tests whether you can explain a symmetry argument out loud. The strongest answer walks through the chain, names the two ways it ends, and says plainly that both are equally likely because neither seat is favoured at any random pick.

A Follow-Up Worth Knowing

Interviewers who like this puzzle often follow up: what is the chance that passenger 37 gets their own seat? For any passenger after the first, the chance of sitting in the right seat is the number of seats still ahead of them, divided by one more than that number.

Passenger 2 sits in the right seat 99 times in 100, because the only way to lose it is for passenger 1 to pick seat 2. Next in line, passenger 3 sits in the right seat 98 times in 99. Near the back, passenger 99 sits in the right seat 2 times in 3. And passenger 100 sits in the right seat 1 time in 2, which is the answer to the original puzzle.

One more number falls out of the same pattern. Add up the chances for all 100 passengers, and the expected number sitting in their own seats is about 95. A single lost boarding pass barely dents the plane: almost everyone still ends up in the right seat.

Quick Answers to Common Questions

Is the airplane boarding puzzle answer always 50%?

Yes. For any number of passengers from 2 upward, the last passenger gets their own seat with probability 1 in 2. The chain argument does not depend on the size of the plane.

What if the first passenger is not random?

Then the puzzle changes completely. If passenger 1 takes their own seat, everyone sits in their own seat. The randomness of the first passenger is what starts the chain.

Does the answer change if displaced passengers pick the nearest free seat instead of a random one?

Yes. With a fixed rule like “nearest free seat”, the outcome is determined by the first pick alone, and there is no 50% answer. The random choice is load-bearing: it is what makes seats 1 and 100 equally likely at the end.

Try Another Puzzle

If you liked this one, try the hardest logic puzzle, where three gods answer in a language you do not know, and one of them answers at random. Then test yourself on the Monty Hall problem, where switching doors doubles your chance of winning the car, and on the 100 prisoners boxes puzzle, where a simple chain rule wins about 31 percent of the time.