Friday Puzzler -- Airplane seats

Re: Friday Puzzler -- Airplane seats

Joseph Mulherin

10%

Re: Friday Puzzler -- Airplane seats

Larry Barrett

No answer yet, but an approach. Start with a simpler problem and look for a pattern.

If there are just 2 passengers (P) and 2 seats (S), P1 will sit in S2. P2 will have to sit in S1. So the probability that P2 sits in S2 is 0.

If there are 3 passengers and 3 seats, P1 has option of sitting in S2 with probability = 1/2 or S3 with probability = 1/2. If P1 sits in S3, then P2 will sit in S2; P3 must sit in S1. But if P1 sits in S2, then P2 has option of S1 or S3, with probability = 1/2 for each. If P2 sits in S3, then P3 must sit in S1. If P2 sits in S1, then P3 will sit in S3. So with 3 passengers and 3 seats, the probability that P3 sits in S3 is 1/2 x 1/2 = 1/4.

Repeat this with 4 passengers and 4 seats. It gets a little complicated, I think the probability that P4 sits in S4 = 1/3.

And with 5 passengers and 5 seats, I think the probability that P5 sits in S5 = 3/8.

I haven't gone further, but note that the probability is increasing, from 0 to 1/4 to 1/3 to 3/8. The rate of increase is decreasing (1/4 - 0 = 1/4; 1/3 - 1/4 = 1/12; 3/8 - 1/3 = 1/24). Next step is to try this with 6 passengers and 6 seats and see if a pattern emerges.

Correction to Friday Puzzler -- Airplane seats

alexy

My apologies, guys.

Here is how the problem should have read, with the change highlighted:

100 people have assigned seats for a plane with a 100-passenger capacity.

The first passenger to get on the plane ***ignores his boarding pass and sits in a seat at random.*** The rest of the passengers board one at a time and sit in their assigned seat if available; if not, they sit in a random unoccupied seat.

What is the probability that the last person on the plane gets to sit in their assigned seat?

Hint: Neither the answer, nor [at least one] path to the answer involves factorials or powers (other than the first power if you want to be picky ;-) )

Sorry, no :-(

Good approach

alexy

And applying that approach to the corrected statement of the problem (sorry about that) will probably yield the answer, and maybe the "why"

Once you have that, it is a small uninteresting step to get the answer to the original mis-stated puzzle.

Re: Correction to Friday Puzzler -- Airplane seats

Larry Barrett

Suppose just 2 passengers and 2 seats. The last passenger to board will sit in the correct seat if:

- P1 boards first (probability 1/2) and randomly sits in S1 (probability 1/2)

- P2 boards first (probability 1/2) and randomly sits in S2 (probability 1/2).

Thus, the probability that the last to board sits in the correct seat is 1/2x1/2 + 1/2x1/2 = 1/4 + 1/4 = 1/2.

Suppose that are 3 passengers and 3 seats. The last passenger to board will sit in the correct seat if:

- P1 boards first (p 1/3) and randomly sits in S1 (p 1/3). Then P2 and P3 will sit in the correct seats regardless of boarding order.

- P1 boards first (p 1/3), randomly sits in S2 (p 1/3), P2 randomly boards second (p 1/2) and randomly sits in S1 (p 1/2). Then P3 will sit in S3.

- P1 boards first (p 1/3), randomly sits in S3 (p 1/3), P3 randomly boards second (p 1/2) and randomly sits in S1. Then P2 will board and sit in S2.

Similar arguments apply if P2 randomly boards first and if P3 randomly boards first.

So the probability that the last passenger to board will sit in the correct seat is:

3 x [1/3(1/3 + 1/3(1/2x1/2) + 1/3(1/2x1/2))] = 1/3 + 1/3(1/4) + 1/3(1/4) = 1/3 + 2/12 = 1/2.

If I did it correctly, the result for 4 passengers and 4 seats is also 1/2. So by extension, I think the same result applies for 100 passengers.

If the first passenger to board sits in the correct seat, then the last to board will also sit in the correct seat. The probability that the first passenger randomly sits in the correct seat is 1/100 for this problem. Thus, if the first to board randomly sits in an incorrect seat, the probability that the last to board sits in the correct seat is 1/2 - 1/100 = 49/100.

Correct

alexy

Good INduction to determine the correct answer, "1/2". (If it's true for n=2,3,and 4, it must be true for all n! g,d,& r )

What is very hard to see here is that the only seats that matter are the one assigned to passenger #1 and the one assigned to passenger #100. If at any point one of the passengers (including passenger #1 or any other passenger choosing at random since his seat was previously taken) sits in seat assigned to #1, then all passengers boarding subsequently, including #100, will sit in their own seat. If any such passenger chooses to sit in #100's seat, then obviously he won't get to sit there. With no reason to assume a bias in the selection of seats, #1 and #100 are equally likely choices for anyone choosing at random.