Timeline for Taking Seats on a Plane
Current License: CC BY-SA 3.0
3 events
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| Jul 27, 2021 at 13:52 | comment | added | Will Orrick | I also like this proof best, but it isn't complete. The bijection would be enough if all admissible seatings were equally likely, but they aren't. For example, the seating where everyone is in their own seat has probability $\frac{1}{100}$, whereas the seating where everyone in not in their own seat--this is unique--has probability $\frac{1}{100\,!}$. The additional step that is needed is to prove that any two admissible seatings related by the bijection are equally likely. (They are.) | |
| Oct 31, 2016 at 13:06 | comment | added | tuko | This is the best proof so far! | |
| Dec 4, 2013 at 9:42 | history | answered | hunter | CC BY-SA 3.0 |