The Puzzle: 100 people are boarding a plane with 100 seats.
- The first passenger is drunk and picks a seat completely at random.
- Every subsequent passenger takes their assigned seat if it’s open.
- If their seat is taken, they pick a random seat from the remaining ones. What is the probability that the 100th passenger sits in their own assigned seat?
1. The Strategy: Simplification
Let’s try for N=2 passengers.
- Drunk picks seat 1 (Correct): Last person gets seat 2 (Success).
- Drunk picks seat 2 (Wrong): Last person gets seat 1 (Fail).
- Probability = 1/2.
Let’s try for N=3 passengers.
- Drunk picks 1: Success.
- Drunk picks 3: Fail.
- Drunk picks 2: Now passenger 2 is “the new drunk” for the remaining 2 seats. We already know that the probability for 2 people is 1/2.
- Probability = 1/2.
2. The General Rule
No matter how many seats there are, the decision always “boils down” to whether Seat 1 or Seat N is picked first.
- If any passenger picks Seat 1, the cycle of “taking someone else’s seat” ends, and the last passenger gets their correct seat.
- If any passenger picks Seat N, the last passenger is guaranteed to fail.
- All other seats just “kick the can down the road” to the next person.
Result: The probability is always 1/2.
Interview-Focused Questions
Q: Why isn’t it something complex like (1/N)?
A: Because for the last passenger, there are only two “special” seats: their own and the first drunk’s. Every other seat just displaces someone who then has the same choice. It’s essentially a symmetrical choice that resolves when one of those two special seats is finally sat in.
Q: What if the first 10 passengers were drunk?
A: That would significantly complicate the math, but the principle of Symmetry across the remaining “unoccupied” seats would still be the starting point for the derivation.
Key Takeaway
This puzzle tests your ability to simplify a problem using small test cases to find a pattern (N=2,N=3) and then prove it logically.
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