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There are 100 passengers with assigned seats. The first two people that boards chooses a seat at random uniformly. The remaining passengers will sit in their assigned seat if it is still open, but if it's not open, they will choose an open seat at random. What is the probability that the last person sits in his assigned seat?

The one person variant is posted in Taking Seats on a Plane where the solution was found to be $0.5$.

I am struggling with adapting that approach to the current problem.

My intuition is that in general, the $i$-th person, if their seat was taken by someone who boarded earlier, has exactly $100 - i + 1$ choices. So if there was a person with exactly 2 choices, it must be (a) the 99th person and (b) the 99th person's seat must be taken. (3) Both of the first and second person's seats must not both be seated in already because if they were the 99th seat would be open. (i) If both were not seated, then the 2 choices for the 99th person must be one of these two seats, and the seat assigned to the 100th person must have been taken. (ii) If exactly one of the first two person's assigned seats were not taken, then the 99th person has the other of the 2 seats as a choice and the 100th seat as a choice.

So I think maybe the key to the problem is figuring out the probability of (i) and (ii) occuring?

roulette01
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    Use the same logic as in the original problem that drunkards will be "booted out of their seat and must find a new seat" if there is ever any dispute. We find that the two drunkards and the last person will necessarily be seated in their three seats in some order with each order in those seats equally likely to occur. It follows that the probability our last person sits in their correct seat with probability $\frac{1}{3}$. – JMoravitz May 11 '21 at 19:14
  • Extending to $n$ drunkards, the probability would be $\frac{1}{n+1}$ – JMoravitz May 11 '21 at 19:15
  • @JMoravitz Can you expand a bit on the symmetry of the problem that resulted in each order being equally likely? That's not readily obvious to me – roulette01 May 11 '21 at 19:17
  • The first drunkard could be in any of the first, second, or last seats (after being ejected from their seat potentially many times) and since the drunkard can not tell those seats apart would end in each seat with an equal chance. That is the whole point of "The drunkard chooses a seat uniformly at random" – JMoravitz May 11 '21 at 19:20
  • The punchline is that it is better to think of it as the drunkard being booted out of their seat when a dispute occurs rather than the "drunken behavior passing on to the next passenger" in order to considerably simply things. – JMoravitz May 11 '21 at 19:21
  • We know that all the other non-timid passengers end in their own correct seat in this rewording of the problem, the only unknowns are where our drunkards finally end up and where our timid last person ends up, each outcome being equally likely as there is nothing distinguishing them in the eyes of the random processes involved (after all insistent passengers have booted the drunken passengers however many times were necessary), and since they constitute the entirety of the sample space we can calculate using just those three seats for describing the sample space, calculating as a ratio. – JMoravitz May 11 '21 at 19:26

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