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Passenger Boarding Probability Problem

RICHMOND  —   April 27, 2018 — Establishing the probability of events involves a special type of mathematics often overlooked in many endeavors except games of chance like dice and cards.  The reason for that is that it depends on the assumption that each member of a set of possibilities has an equal probability of occurring.  In most real life scenarios there are other factors involved for various possibilities that can make them unequal.  In dice however it is safe enough to assume that the roll of a 2 has an equal probability as the roll of a 1,3,4,5 or 6.  In cards with a well shuffled deck it is safe to assume that the chance of the top card being a queen of clubs is the same as it being a king of spades.

Example problems in probability mathematics usually involve such games of chance where the necessary assumptions are safe.  However the math can also be useful with more real to life problems in identifying where some hidden factors might have an effect on outcomes.  It can provide a background against which they will show up.

Statistical analysis sometimes employs probability mathematics, but is less about predicting events than measuring large quantities of data with less effort and organizing it in ways that can make clear what factors (random or otherwise) might be involved.

For consideration here is a problem in purely probability mathematics.  The problem is stated in a way that eliminates real life complications.  I cannot take credit for writing the problem.  Finding a solution was difficult enough for me.

One hundred passengers buy assigned seats on a plane with one hundred seats.  They take their seats one passenger at a time such that a passenger cannot board the plane until the passenger before him, if there is one, is seated (real life?).  The first passenger to board fails to take his assigned seat and takes someone else's assigned seat instead.  All subsequent passengers take their assigned seat without error (real life?) unless theirs has been taken already.  If and only if theirs has been taken already they choose an empty seat randomly (real life?).  The question is what is the probability the last passenger gets his assigned seat?

Perhaps the real life solution is that the first passenger to find his seat taken would resolve the matter by consulting documentation.  That is not however the answer in this somewhat unreal problem.

As with most problems in probabilities the solution involves correctly identifying the dynamics.  Notice that the need for further calculations ceases if at any time all the remaining passengers' assigned seats are not taken.  In that case the last passenger is guaranteed to get his seat.  The need for further calculations also ceases if at any time a passenger other than the last one chooses the seat assigned the last passenger.  Call these concluding conditions "success" and "failure."

Throughout the boarding process there are two critical seats.  One of them is the empty, seat with wrong assignment, originally the seat assigned to the first passenger.  The other is the seat assigned to the last passenger.  If the empty, seat with wrong assignment is chosen than the success condition occurs because thereafter all the passengers will get their assigned seats.  If the last passenger's seat is chosen before him, then obviously the failure condition has occurred.

A third important seat to consider is the seat with wrong assignment.  Although it can change during the boarding process, there will be one and only one passenger of those remaining to board whose seat is taken until either the success or failure condition occurs.  If a passenger is the one whose seat is taken, and he chooses neither the success seat nor the failure seat, he must take a seat belonging to someone else and there will again be only one passenger remaining whose seat is taken, albeit a different one.

Having thus put the dynamics of the problem in a manageable way the probability of success can be determined at each boarding in the process.

In order to choose the seat guaranteeing success or failure, a passenger must be eligible.  He must be the one remaining passenger whose assigned seat is taken.  For the second passenger to board that is a 1/99 probability.

Since only one of 99 seats guarantees success, the probability of success is 1/99 times 1/99 or 1/9801.

Since only one of the 99 seats guarantees failure, the probability of failure is also 1/9801

There is a third possibility that neither success nor failure occurs at this boarding. It is 9801/9801 - 1/9801 - 1/9801 = 9799/9801.

At each boarding the probability of success is 1/n^2 where 'n' is the number of passengers (and seats as well) left.  The probability of failure is also 1/n^2. There is however a special condition at n=2.

The probability of neither success nor failure at the nth boarding is the sum of the probabilities of neither success nor failure at each previous boarding and the nth one.

Interesting as that might be, it is irrelevant to the question asked.  When neither success nor failure occurs the selection of either success or failure is merely postponed until the next boarding.  At all times the probability of success is equal to the probability of failure.  That remains true even after postponement is no longer possible.  Such a condition means that the probability of success is one half, the answer to the question.

This math is correct through 98 passengers.  The probability of eligibility of the 99th passenger is not 1/2 (1/n where n=2) it is 1.  If neither success nor failure has previously occurred he must be eligible.  The calculations necessarily conclude because the last passenger has no choice.