The Last Seat : Car Talk Puzzler

Yet another great car talk puzzler, The Last Seat:

RAY: You’re one of a hundred people standing in line to get onto an airplane that has 100 seats. There’s a seat for every person who’s in line, and each of you has a boarding pass for your assigned a seat. The first person to walk onto the plane drops his boarding pass and, instead of picking it up, decides, “I’m just going to sit anyplace.” He takes a seat at random.

Now, every other passenger will take either his assigned seat or, if that seat is taken, that passenger will take any seat at random.

TOM: I’ve been on that flight!

RAY: You are the last passenger to walk onto the plane. Obviously, there’s going to be one seat left, because everyone else is sitting in his correct seat, or not.

The question is: What are the chances that you get to sit in your assigned seat? I’m going make this multiple choice.

A: 1 out of 2.
B: 1 out of 10.
C: 2 out of 50.
D: 1 out of 100
E: Zero.

What strategies could you give students to help them reason their way through this puzzle?

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5 Responses to The Last Seat : Car Talk Puzzler

  1. Since it’s about 100 years ago I last looked at this sort of slightly annoying (!) problem I found myself in the position of your students. My suggested strategy is “Read the question, and now read between the lines, and now read the mind of he who thought this one up.”

  2. Michael says:

    I’d suggest that they take a look at this scenario with just two passengers. Then three. And four. And so on. Notice any patterns?

    • Dan says:

      Hmm.. I’m not a math person but I tried to solve this using Michael’s advice and ended up with:
      (N-2)(N-2)! out of (N-1)(N-1)!

      which yields ~-9.146912e+157 for N = 100. Assuming I didn’t do something wrong (which I probably did), I’m going to pick E), for [Effectively] Zero.

      Did you ever post a solution?

      • Dan says:

        I just noticed the solution in the link.. but I’m not sure I agree. My original answer assumed the first guy *never* sits in his own seat, but even if he does the posted solution seems incorrect.

        If you consider two people, me (M) and him (H), you get two possibilities:
        M H
        M H (in my seat)
        H M

        which is 50% chance of me in my seat. However, if you consider three people and add Other guy (O), then there are 6 possibilities:
        M H O
        M H O (in my seat)
        M O H (in my seat)
        H M O
        H O M
        O H M
        H M H

        only 2 of which end up with me in my seat (~33%). For 4 people, you get 6 / 24 times I’m in my seat (25%), etc. Which gives us 1/N or “1 out of N” where N is the number of people. So in this case, I would say D) 1 out of 100.

        • Julius Nadas says:

          MHO and HMO can be eliminated since the O will sit in his own seat if it is available.
          you only have to look at four possibilities:
          MOH, HOM, OHM and OMH and therefor I get to sit in my own seat 2/4 = 50% of the time.
          the answer is (A) 1/2

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