Timeline for Taking Seats on a Plane
Current License: CC BY-SA 4.0
17 events
| when toggle format | what | by | license | comment | |
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| Apr 21, 2022 at 3:59 | history | edited | rsp | CC BY-SA 4.0 |
improved proof
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| Apr 21, 2022 at 0:36 | history | edited | rsp | CC BY-SA 4.0 |
added formatting
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| Apr 21, 2022 at 0:24 | history | edited | rsp | CC BY-SA 4.0 |
corrected proof
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| Apr 20, 2022 at 23:23 | history | edited | rsp | CC BY-SA 4.0 |
Bad editing
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| Apr 14, 2022 at 19:27 | comment | added | Will Orrick | I've deleted the reply since it's no longer relevant. | |
| Apr 14, 2022 at 18:46 | comment | added | rsp | Sorry, it was my intention to retract my most recent comment. I didn't see that you had replied. I acknowledge your point that not every allowable configuration is equally likely. | |
| Apr 14, 2022 at 14:39 | comment | added | Will Orrick | By the way, I agree with everything in your comment except for the first word, "No", and the second-to-last sentence. See my answer to a related post, where the probabilities of every seating arrangement are worked out for a four-passenger plane and a five-passenger plane. You will see (1) that I agree with you that there are $2^{N-1}$ elements in the sample space for an $N$ passenger plane, and (2) that I disagree with you about the probability distribution being uniform. | |
| Apr 14, 2022 at 1:33 | comment | added | Will Orrick | Are you saying that the probability that passenger 1 sits in their own seat (and hence everybody else does too) is $1/2^{99}$? That doesn't make sense to me since there are only 100 seats passenger 1 can sit in, and they're all equally likely. If that's not what you're saying, can you elaborate? | |
| Apr 13, 2022 at 19:50 | comment | added | rsp | No. Each diagram you've listed is a complete seating chart. While I do say the last passenger displaced has to sit in seat 1, perhaps I should have made this explicit in the diagram def as well by indicating passenger j →1 always. So 1 represents everyone in their assigned seat. 1→2→��→99 says passengers 1,…,98 are sitting in seats 2,...,99 resp. with the 99th passenger in seat 1 and the 100th in their own seat. Each is just a single distinct point in a sample space, one no more or less likely than another. The size of the sample space is not 100!, but 2^99 as there are rules for boarding. | |
| Jul 28, 2021 at 4:19 | comment | added | Will Orrick | The diagrams are not all equally probable, so you can't compute the probability as a ratio of diagram counts. What saves the computation is that for each diagram that does not end in $100$, there is an equally probable diagram that does end in $100$. For example, $1$ and $1\rightarrow100$ both have probability $\frac{1}{100}$, whereas $1\rightarrow2\rightarrow\ldots\rightarrow99$ and $1\rightarrow2\rightarrow\ldots\rightarrow99\rightarrow100$ both have probability $\frac{1}{100\,!}$. | |
| May 27, 2016 at 22:41 | history | edited | rsp | CC BY-SA 3.0 |
Corrected the proof. Clarified main idea.
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| May 26, 2016 at 17:52 | history | edited | rsp | CC BY-SA 3.0 |
fixed grammer
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| May 26, 2016 at 17:07 | history | edited | rsp | CC BY-SA 3.0 |
corrected grammer
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| May 26, 2016 at 6:59 | review | Late answers | |||
| May 26, 2016 at 7:01 | |||||
| May 26, 2016 at 6:51 | history | edited | rsp | CC BY-SA 3.0 |
corrected grammer
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| May 26, 2016 at 6:43 | review | First posts | |||
| May 26, 2016 at 6:45 | |||||
| May 26, 2016 at 6:42 | history | answered | rsp | CC BY-SA 3.0 |