Start by reframing the problem: the only seats that matter are seat 1 and seat 100, and the process ends when either is taken. Use symmetry to argue that at each critical step, seat 1 and seat 100 are equally likely to be chosen, leading to a probability of 1/2. Then generalize to n passengers to show the probability is always 1/2, and derive the closed-form expression P(n) = 1/2.
Pro tip: Emphasize the symmetry argument clearly and concisely; it's the key insight that impresses interviewers. Also, mention that the result is independent of n, which is a surprising and elegant outcome.
Restate the problem to ensure clarity: Passenger 1 picks a random seat, and each subsequent passenger takes their assigned seat if available, otherwise picks randomly from remaining seats. Focus on the final outcome for passenger 100.
Recognize that the only seats that affect the final outcome are seat 1 and seat 100. Once either is occupied, the chain of random choices stops, and the remaining passengers (including passenger 100) will either get their own seat or not.
At each step where a random choice is made, the set of available seats includes seat 1 and seat 100 (unless one is already taken). By symmetry, seat 1 and seat 100 are equally likely to be chosen. Thus, the probability that seat 100 is taken before seat 1 is 1/2.
For n passengers and n seats, the probability that passenger n gets seat n is 1/2. So the closed-form expression is P(n) = 1/2 for all n ≥ 2. For n=100, P(100) = 1/2.
Check n=2: Passenger 1 picks seat 1 or 2 with equal probability; if seat 1, passenger 2 gets seat 2; if seat 2, passenger 2 gets seat 1. So probability = 1/2. This confirms the pattern.
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This went better because the invariant argument scales directly.
First, clearly state the problem: n passengers board a plane with n seats, the first passenger picks a random seat, and each subsequent passenger takes their assigned seat if available, otherwise picks a random empty seat. The probability that the last passenger gets their own seat is 1/2. Then, provide a proof by induction or by identifying an invariant, such as focusing on the last passenger's seat and the first passenger's seat, and showing symmetry.
Pro tip: Emphasize the symmetry between the first passenger's seat and the last passenger's seat: the process ends when either is chosen, and by symmetry each is equally likely to be chosen first, giving probability 1/2. This elegant argument impresses interviewers.
Clearly define the scenario for n passengers and n seats, ensuring the interviewer knows you understand the setup.
Claim that the probability is 1/2 for all n >= 2, and briefly mention that it is independent of n.
Select either induction or an invariant-based argument. For induction, establish the base case n=2 and then assume true for n-1 to prove for n. For invariant, focus on the set of seats that remain available and the symmetry between the first and last seats.
If using induction: For n, consider the first passenger's choice. If they pick their own seat (prob 1/n), last gets their seat. If they pick the last seat (prob 1/n), last does not. If they pick seat k (2 <= k <= n-1), the problem reduces to a smaller instance with n-k+1 passengers, where the displaced passenger acts as the new 'first' passenger. By induction, probability is 1/2. Summing gives 1/2. If using invariant: Note that the process continues until either the first passenger's seat or the last passenger's seat is chosen. At each step, the probability of choosing either is equal, so by symmetry the last passenger's seat is chosen first with probability 1/2.
Summarize that the probability is 1/2, and optionally verify with small n (e.g., n=2, n=3) to build confidence.
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First, clarify the probability problem and its theoretical solution, then implement a vectorized Monte Carlo simulation in Python (using NumPy) with 1 million trials. Compute the estimate, standard error, and 95% confidence interval, and compare the estimate to the theoretical value to confirm it is within 0.01.
Pro tip: Use vectorized operations (e.g., NumPy) for speed and memory efficiency, and set a random seed for reproducibility. Also, explicitly state the theoretical probability and show the absolute difference to demonstrate rigor.
Restate the probability problem to ensure understanding, and derive or state the theoretical probability. For example, if it's the probability of at least one match in a birthday problem with n=100, the theoretical value is approximately 0.9999997.
Determine how to simulate one trial: generate random outcomes (e.g., birthdays) for n=100, and check if the event occurs. Plan to repeat this 1,000,000 times efficiently.
Write vectorized code to run all trials at once. For example, in Python: generate a 2D array of random integers, compute the event for each row, and count successes. Use a fixed seed for reproducibility.
Calculate the estimated probability (mean of successes), standard error (sqrt(p*(1-p)/N)), and 95% confidence interval (estimate ± 1.96*SE).
Compare the estimate to the theoretical probability. Compute the absolute difference and confirm it is less than 0.01. Discuss any discrepancies and the role of Monte Carlo error.
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Naive simulation is O(n) per trial in both time and space since you're maintaining and scanning a set of available seats.
Start by clearly defining the naive simulation: it likely iterates through all seats or all possible pairs, leading to O(n^2) time and O(n) space. Then describe the optimized approach: by only tracking the two boundary seats (seat 1 and seat n), we can compute the result in O(1) time and O(1) space, as the answer depends only on those two values. Conclude by discussing trade-offs and when each approach is appropriate.
Pro tip: Emphasize that the optimized approach works because the problem's constraints or objective function often depend only on the boundary seats, so tracking them suffices. This shows you can identify problem-specific structure to reduce complexity.
Describe what the naive simulation does: e.g., iterating over all seats or all pairs of seats to compute the result. State its time and space complexities, typically O(n^2) time and O(n) space.
Explain that by only tracking the two boundary seats (seat 1 and seat n), we can compute the result directly. This reduces time complexity to O(1) and space complexity to O(1).
Explain why tracking only the boundary seats is sufficient: the problem's objective or constraints often depend only on these seats, so other seats do not affect the outcome.
Discuss the trade-offs: the naive approach is simpler to implement but inefficient for large n; the optimized approach is more efficient but requires insight into the problem structure.
Mention when each approach might be preferred, such as using naive for small n or for validation, and optimized for large-scale or real-time systems.
AI-generated suggestions, not part of the candidate's original notes. May be inaccurate — verify before relying on them.