Tuesday, November 29, 2016

11/28/16: The Happenin'


  1. 11/28/16
    1. Well, today has been a very eventful return to campus.
    2. Spend the morning finishing StatMech hw, run to class.
    3. Hang out for a bit in SPS lounge, work with Jacob.
    4. Head home, make myself lunch (+ 2 extra sandwiches) and read E&M.
    5. Spend a productive Leroy-isolated 20 minutes crunching emails and researching classes.
    6. Work on quantum hw while simultaneously planning several lunches with friends.
    7. Go to gym with Xin, play some badminton, work out a bit.
    8. Run home, shower, off to Putnam session! Work some very interesting problems, most of which end up simpler than they appear. Here’s a fun one that with sufficient thought, any one of y’all could solve (Thomas’s genius solution requires no computation). There are 100 people boarding a plane. They all have assigned seats (out of the 100 total on the plane). But the first guy forgets his boarding pass. So he sits in a random seat (with high probability, displacing someone else).
    9. Each subsequent boarder then either takes his seat, if it is unoccupied, or moves to a random open seat, if their seat is taken.
    10. The question is: what is the probability that the last boarder will get the seat she was originally assigned? (Comment if you’re stuck and want the answer).
    11. Good problems (probability, lots more fun for me than dry series summations and geometry), good discussion--though the session leader withholds pizza from us for an extra 20 minutes :)
    12. Afterwards, chat a bit with Peter, Thomas, and Kelsey who’s majoring in physics.
    13. Then it’s off to swing! To my surprise, the place is packed. Sophie is back from Europe; she’s taking the coordinator role, and she invited all of the old Monday night regulars. I get lost in the dancing pretty fast; work on some more following with Alex, invent a few new moves…
    14. Return to 2104, work a bit more on QM homework (I killed the last exam so I don’t have to write this one up too carefully) write this log and hit the hay.
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4 comments:

  1. joan0716@sbcglobal.netNovember 30, 2016 at 10:05 AM

    Off hand, it is 1 in 2 chances. I need the answer and a brief explanation. It was so fun to see you and family, albeit short.

    ReplyDelete
  2. Spencer PetersDecember 2, 2016 at 1:34 AM

    Wow, impressive! You are correct. Do you have an argument to support/prove your conclusion?

    ReplyDelete
  3. joan0716@sbcglobal.netDecember 2, 2016 at 6:00 PM

    It was easy when I did it, but I had doubts. One seat is taken by the person without a seat assignment, but the other 98 passengers have assigned seats so there is only one passenger whose seat is taken and one seat left not taken. Now explain the puzzle to me so I understand.

    ReplyDelete
    Replies
    1. Spencer PetersDecember 6, 2016 at 12:24 AM

      The problem with this argument is that, suppose the first guy takes the second guy's seat. Then the second guy takes a random seat, and chaos breaks loose. It could be that he sits in the third seat, and the third guy sits in the fourth seat, and so on, and so nobody (certainly not 98 people) get their assigned seats.
      The clearest rigorous argument for the 1/2 result, I think, is like this. Think of it like a game: if the 100th guy gets his seat, we win; else we lose.
      Also, change the rules and make the people more aggressive: so if you find your seat is taken, you don't move; you kick the offending fellow out to a random seat. This doesn't change which seats are occupied at any time--the thing which actually matters: we're trying to figure out whether the 100th seat will be occupied when the 100th guy boards. But it makes it so that there's only one guy in the wrong seat at a given time--namely, the original confused guy.
      So if the original confused guy gets kicked out and lands in his own seat, at that point everyone who has boarded has their assigned seat. So everyone else can board without confusion, and the 100th guy gets his seat--we win!
      On the other hand, if the original confused guy gets kicked out and lands in the 100th guy's seat, he won't be kicked out again until the 100th guy comes to board and finds his seat taken. So we lose.
      And in fact, this is the only way we can win or lose, because by the time we get to the 100th guy, either his seat, or the original guy's seat, will be taken by someone who got kicked there.
      But, (the punchline), the win condition at each step (kick into first seat) is exactly as probable as the loss condition (kick into last seat)
      So the total probability of winning must be the same as the total probability of losing--the situation is symmetric.
      QED! :D

      Delete

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