Timeline for Taking Seats on a Plane
Current License: CC BY-SA 3.0
11 events
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| Sep 7, 2022 at 17:22 | comment | added | Will Orrick | It's not clear at the outset that the conditional probability that passenger $n$ sits in their own seat given that passenger $1$ sits in passenger $j$'s seat, with $2\le j\le n-1$, is the same as $p_{n-1}$ since in the computation of $p_{n-1}$ any passenger might end up in the wrong seat whereas in the conditional probability computation passengers $2$ through $j-1$ certainly sit in their own seats. If you allow for this difference, your inductive argument still goes through, however. See my comment to user103828's post. | |
| Aug 23, 2021 at 13:43 | history | rollback | robjohn♦ |
Rollback to Revision 2
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| Aug 23, 2021 at 7:54 | review | Suggested edits | |||
| Aug 23, 2021 at 8:54 | |||||
| S Aug 23, 2021 at 7:52 | history | suggested | user851668 | CC BY-SA 4.0 |
Clarified referent of link.
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| Aug 23, 2021 at 7:38 | review | Suggested edits | |||
| S Aug 23, 2021 at 7:52 | |||||
| Jun 11, 2019 at 18:07 | comment | added | ely | Specifying the term $\frac{n - 2}{n}p_{n-1}$ is fully justified with no summation. It is simply the recursive statement of the problem details. There is no such thing as "deriving it" or "obtaining it." It is a direct mathematical statement of the problem. You do not need to make any assumption about what the term $\frac{n - 2}{n}p_{n-1}$ is equal to in order for the first equation to be valid. Whatever it is equal to, whether individual $p_{i}$ terms are equal or not, that equation is valid. Combined with the base case of the recursion, it then proves that they are equal. | |
| Jun 11, 2019 at 16:35 | comment | added | Hans | If you do not get the term $\frac{n - 2}{n}p_{n-1}$ from summation as I surmised, why don't you write out in detail how you obtained that term? There is no justification for it. The proof is still flawed when you do not justify that step. | |
| Jun 11, 2019 at 13:14 | comment | added | ely | @Hans no, I am not assuming $\frac{n-2}{n}p_{n-1} = \sum_{i=2}^{n-1}\frac{1}{n}p_{i}$. I don't know where you come up with that or why you think it is related to the equality. You do not need to assume any summands of anything are all equal. In fact, in my derivation, you prove they would be equal through the equations I have used (which do not themselves rely on it). You seem very fundamentally confused. It might be nicer if you open a new question where you link to the many answers of this thread and ask the community to help you sort it out, instead of adding spam comments everywhere. | |
| Jun 8, 2019 at 2:56 | comment | added | Hans | -1. This derivation presumes $p_i=p_j,\forall\, 2\le i<j\le n-1$. This is not right. | |
| Apr 17, 2018 at 21:05 | history | edited | ely | CC BY-SA 3.0 |
deleted 497 characters in body
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| Feb 11, 2018 at 20:18 | history | answered | ely | CC BY-SA 3.0 |