The Airplane Seat Problem: Why the Answer Is 1/2 (Three Proofs)

Q: What is the answer to the airplane seat problem?

The probability that the last passenger gets their own seat is exactly 1/2, regardless of the number of passengers (for any n of at least 2). The cleanest proof is a symmetry argument: every randomly-choosing passenger is equally likely to take seat 1 as the last passenger's seat, and whichever of those two seats is taken first decides the outcome.

Q: Why is the airplane seat problem answer 1/2 and not 1/100?

Because the outcome only depends on two seats: seat 1 and the last passenger's seat. Every passenger forced to choose randomly treats those two seats symmetrically, so seat 1 is claimed before the last seat with probability exactly 1/2. The answer 1/100 would only apply if a single passenger picked one seat uniformly at random, ignoring the cascade of displacements.

Q: Which firms ask the airplane seat problem in interviews?

It is a staple brainteaser across quant trading and research interviews, commonly reported at Jane Street, options market makers, and prop shops in phone screens and first rounds. Because it is well known, interviewers increasingly ask the follow-ups instead, such as the probability that passenger k gets their own seat, which equals (n-k+1)/(n-k+2).

Q: What is the probability that passenger k gets their own seat?

For any passenger k from 2 to n, the probability of getting their own seat is (n-k+1)/(n-k+2). This comes from a symmetry argument over the seats {1, k, k+1, ..., n}: passenger k is displaced only if seat k is claimed by a random chooser before seat 1. Setting k = n recovers the classic answer of 1/2 for the last passenger.

Symmetry, recursion, and Monte Carlo — and the follow-up questions that separate a memorized answer from a real one.

Updated July 3, 2026 · QuantVault

The problem

100 passengers board a plane with 100 seats, one at a time, in order. Passenger 1 has lost their boarding pass and sits in a uniformly random seat. Every passenger after that sits in their own assigned seat if it's free, and otherwise picks a random free seat. What is the probability that passenger 100 ends up in their own seat?

The answer is exactly $1/2$ — and it's $1/2$ for any number of passengers $n \ge 2$, which is the first hint that something structural is going on. This is one of the most-asked brainteasers in quant interviews, showing up regularly in first rounds at Jane Street and options market makers. Interviewers don't ask it because the answer is hard to find online; they ask it because your path to the answer reveals whether you hunt for structure or grind through cases. The strongest candidates can produce more than one proof and handle the follow-ups cold.

Proof 1: The symmetry argument (the one to lead with)

Here is the two-minute version. Track only two seats: seat 1 (the confused passenger's assigned seat) and seat 100 (the last passenger's seat).

Every passenger who is forced to choose randomly — passenger 1, or anyone who finds their seat taken — chooses uniformly among the free seats. At the moment of any such random choice, seat 1 and seat 100 are either both still free or both already resolved. If both are free, the random chooser is equally likely to take seat 1 as seat 100. The game ends, for our purposes, the instant one of those two seats gets taken by a random chooser:

If seat 1 is taken first, every subsequent passenger finds their own seat free, and passenger 100 gets seat 100.
If seat 100 is taken first, passenger 100 is displaced.

Since every random choice treats the two seats symmetrically, seat 1 is taken first with probability $1/2$. Done. This is the same flavor of clean conditioning that cracks the Monty Hall problem: strip the problem down to the only random events that matter.

Proof 2: Recursion (the one that survives follow-ups)

Let $p_n$ be the probability the last passenger gets their own seat with $n$ passengers. Condition on passenger 1's choice:

$$p_n = \underbrace{\frac{1}{n}\cdot 1}_{\text{picks seat }1} + \underbrace{\frac{1}{n}\cdot 0}_{\text{picks seat }n} + \frac{1}{n}\sum_{k=2}^{n-1} p_{\,n-k+1}$$

The key insight is the third term: if passenger 1 takes seat $k$, passengers $2,\dots,k-1$ all sit correctly, and passenger $k$ becomes the new "confused" passenger in a sub-problem with $n-k+1$ seats. The structure is self-similar.

Worked check. Base case: $p_2 = 1/2$ (passenger 1 picks seat 1 or seat 2, coin flip). Then for $n=3$:

$$p_3 = \tfrac{1}{3}(1) + \tfrac{1}{3}(0) + \tfrac{1}{3}\,p_2 = \tfrac{1}{3} + \tfrac{1}{6} = \tfrac{1}{2}.$$

By induction, if $p_m = 1/2$ for all $2 \le m < n$, the sum contributes $\frac{n-2}{n}\cdot\frac{1}{2}$, so $p_n = \frac{1}{n} + \frac{n-2}{2n} = \frac{1}{2}$. The recursion is more work than the symmetry proof, but it's the machinery you need when the interviewer perturbs the setup — the same skill tested throughout our probability question bank and in absorbing-state problems like gambler's ruin.

Proof 3: Simulation (the sanity check)

If you're asked this in a coding round, a ten-line Monte Carlo settles it:

import random
def trial(n=100):
    free = set(range(n))
    free.discard(random.randrange(n))      # passenger 1 sits randomly
    for k in range(1, n-1):
        if k in free: free.discard(k)      # own seat available
        else: free.discard(random.choice(list(free)))
    return (n-1) in free                    # is seat 100 still free?

print(sum(trial() for _ in range(10**5)) / 10**5)   # ~0.500

Runs converge to $0.500 \pm 0.003$ at $10^5$ trials. Mentioning the standard error ($\approx \sqrt{0.25/10^5} \approx 0.0016$) earns points: it shows you know what "converged" means.

Follow-ups interviewers actually ask

Getting $1/2$ is table stakes. These extensions are where offers are decided:

Follow-up	Answer	Why
P(passenger $k$ gets own seat), $k \ge 2$	$\frac{n-k+1}{n-k+2}$	Symmetry among seats $\{1, k, k+1, \dots, n\}$: passenger $k$ is displaced only if seat $k$ is taken before seat 1, one of $n-k+2$ equally likely "first picks."
Passenger 1 deliberately avoids seat 1	$\frac{n-2}{2(n-1)}$	Rerun the recursion with the first pick uniform on the other $n-1$ seats: $\frac{n-2}{n-1}\cdot\frac{1}{2}$. For $n=100$: $49/99$.
Expected number of displaced passengers	$H_n - 1 + \frac{n-1}{n}$	Sum the per-passenger displacement probabilities $\frac{1}{n-k+2}$; roughly $5.2$ people for $n = 100$.

The first row is the most common follow-up, and it's why the recursion proof matters: candidates who only memorized the symmetry one-liner often can't adapt it. Expected-value extensions like the third row lean on linearity of expectation — the single most reusable tool in this genre, covered in depth in our expected value question bank.

Traps to avoid

Answering $1/100$ or $99/100$. Both come from ignoring the cascading structure. If your gut says either, slow down.
Assuming passenger 1 avoids their own seat. The classic statement allows passenger 1 to sit in seat 1 by chance. The "deliberately avoids" version is a different problem with a different answer (see table).
Reciting the answer instantly. Interviewers know this problem is famous. If you've seen it, say so — then prove it two ways. That converts a red flag into a strength.
Grinding cases for $n=100$ directly. Solve $n=2$ and $n=3$ first. Small cases screaming "$1/2$" is what should send you hunting for the symmetry.

Ready to drill? Work through our full probability interview questions and brainteaser bank with worked solutions, then pressure-test your instincts in our trading games — the same structural thinking under a clock.

Frequently asked questions

What is the answer to the airplane seat problem?

The probability that the last passenger gets their own seat is exactly 1/2, regardless of the number of passengers (for any n of at least 2). The cleanest proof is a symmetry argument: every randomly-choosing passenger is equally likely to take seat 1 as the last passenger's seat, and whichever of those two seats is taken first decides the outcome.

Why is the airplane seat problem answer 1/2 and not 1/100?