The Classic Probability Puzzle: Boarding a Bus with Random Seat Assignments

Have you heard of the intriguing probability puzzle involving passengers boarding a bus? This puzzle has challenged the minds of mathematicians and enthusiasts for decades, revealing fascinating insights into probabilities and logical deductions.

Initial Conditions and Key Observations

The puzzle involves a bus with 50 seats and 50 passengers, each assigned a specific seat. However, passengers board the bus in a peculiar manner: the first passenger randomly chooses a seat, and each subsequent passenger either takes their assigned seat if available, or a random seat if their assigned seat is taken.

Key Observations:

If the first passenger, whom we call Passenger 1, takes their assigned seat, everyone else will sit in their assigned seat, resulting in no passengers sitting in the wrong seat. If Passenger 1 takes any seat other than their own, the scenario becomes complex. The chain of randomness begins, and the seating arrangement can lead to passengers getting wrong seats, with the critical passengers being Passenger 1 and Passenger 50. The fate of Passenger 50 depends on whether Passenger 1 takes the first seat or the last seat. If Passenger 1 takes the last seat, Passenger 50 will be stuck in the wrong seat.

Expected Outcomes and Simplification

The final determination of which passengers end up in the wrong seats is heavily influenced by Passenger 1's initial choice. By simplifying the problem, we can focus on the two critical seats: Passenger 1's seat (Seat 1) and Passenger 50's seat (Seat 50).

Passenger 1 has a 1/50 chance of sitting in their assigned seat. If this happens, no one ends up in the wrong seat. Conversely, if Passenger 1 ends up in Seat 50, Passenger 50 is guaranteed to end up in the wrong seat. For any other seat, the situation creates a random selection, leading to a 50% chance that Passenger 50 will end up in the wrong seat.

Calculation and Conclusion

To calculate the expected number of wrong seats, we can use a probabilistic approach:

The probability that Passenger 1 sits in Seat 1 is 1/50, resulting in 0 wrong seats. The probability that Passenger 1 sits in Seat 50 is 1/50, resulting in 1 wrong seat. For any other seat (Seats 2 to 49), there is a 50% chance that Passenger 50 will end up in the wrong seat.

We can express this calculation as:

$$ EX frac{1}{50} cdot 0 frac{1}{50} cdot 1 frac{48}{50} cdot frac{1}{2} 0 frac{1}{50} frac{24}{50} frac{25}{50} frac{1}{2}$$

This means that, on average, half of the passengers end up in the wrong seats. This elegant solution demonstrates the power of mathematical reasoning and probability in solving complex puzzles.

Conclusion

The classic probability puzzle involving passengers boarding a bus with random seat assignments is a fascinating problem that reveals the unexpected simplicity in complex scenarios. By understanding the key observations and simplifying the problem, we can derive powerful conclusions about probabilities. This puzzle serves as a delightful reminder that sometimes, the answers to complex problems can be surprisingly straightforward.