Some Amazing Mathematical Thought

[Alcoholic Actuary]

Ingenious, but simple:

You are in line to board an airplane. The seats are numbered 1-100 and passengers line up with ticket numbers 1-100 such that person 1 takes seat 1, and person 2 takes seat 2 etc... and unfortunately you are passenger 100. The following situation ensues - passenger 1 takes a random seat (like Southwest Airlines) instead of seat 1. The rest of the passengers boarding follow the same algorithm: if their seat number is available they take it and if someone is sitting in their seat, they take a different remaining seat at random. What is the probability that you (at seat 100) gets your assigned seat?

If the passengers line up in RANDOM order, and we replace "passenger #1" and "passenger #100" with "passenger first to board" and "passenger last to board", the puzzle has the exact same solution as the stated puzzle.

However the strict (1,2,3,4,...) ordering Alcoholic Actuary imposes makes the puzzle somewhat easier to visualize.

Which seat(s) are certainly already taken when passenger #3 boards? When passenger #4 boards? When passenger #100 boards?

Actually, a good way to solve the problem is to start with a 2 seat airplane (you are seat 2). Then work on a 3 seat plane and maybe 4.

Then try to extend the results to an 'N' numbered seat plane to prove your results (this will save you the effort of attempting to write code for a 100 seat plane).

Spoiler: Passenger Last to Board

passenger last to board has the same probability on a 100 person aircraft and a 1000 person aircraft - as Swami noted.

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Can you write out the sequence of probabilities?

Sure - start with a 2 seat plane and you are passenger 2. Passenger 1 gets on and takes a seat at random, so there is a 50% chance he sits in his seat, and a 50% chance he sits in yours. If he sits in his seat the probability you get yours is 100% and if he sits in your seat, the probability you get yours is 0%. I would write it this way P(2) = 0.5 x (1.0) + 0.5 x (0.0) = 0.5.

Now try a 3 seat plane: (1) Person 1 takes their correct seat and you automatically get yours. (2) person 1 takes your seat and you won't get yours. (3) person 1 takes seat 2, turning the problem into the 2 seat plane that we already solved.
P(3) = (1/3) x (1.0) + (1/3) x (0.0) + (1/3) x (P(2)) = (2/6) + (1/6) = 0.5.
P(4) = (1/4) x (1.0) + (1/4) x (0.0) + (2/4) x (P(3)) = (2/8) + (2/8) = 0.5
etc..


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