Timeline for Taking Seats on a Plane
Current License: CC BY-SA 4.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 6, 2023 at 13:09 | comment | added | SBK | Then there is another claim that $\mathbf{P}\bigl(X_j = c(1)\bigr) = \mathbf{P}\bigl(X_j = c(100)\bigr)\ \ \forall\ j$ and so forth... This is what I mean by how these solutions aren't rigorous in the same way we expect arguments in analysis or number theory to be rigorous. We excuse ourselves from having to actually turn it into math and I'm not sure why that is | |
| Jul 6, 2023 at 13:08 | comment | added | SBK | Then something about how $X_{100}$ is independent of certain other things... | |
| Jul 6, 2023 at 13:07 | comment | added | SBK | I just had a question closed that claimed these sorts of proof were not rigorous and so I'll illustrate what I mean. Suppose that the 'correct' seat for the $j^{th}$ person is $c(j)$. There is a set of mathematical claims here that are not written down. Something like: You can construct a set of random variables $\{X_j\}_{j=1}^{100}$, all taking values in $\{1,2,...,100\}$ and such that $\mathbf{P}\bigl(X_j=k\bigr)$ is the probability that the $j^{th}$ person sits in seat $k$. | |
| Aug 21, 2021 at 20:54 | comment | added | Alex | @Joel I think Aryabhata counts this case also. Correct me if I am wrong | |
| Jul 28, 2021 at 4:04 | history | edited | Greg Martin | CC BY-SA 4.0 |
added 6 characters in body
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| Aug 21, 2020 at 1:57 | comment | added | Joel | Surely it’s 51%. There’s a 1% chance the first person in line actually chooses his own seat, in which case so does everyone thereafter. | |
| Mar 30, 2017 at 20:55 | comment | added | Aryabhata | @StellaBiderman: The number of pre-taken seats is actually immaterial for this answer. | |
| Mar 30, 2017 at 14:33 | comment | added | Stella Biderman | This answer is good, but I think fails to make a fundamental insight: at any given time there is only one person who is yet to board and whose seat has been taken. This is because when someone boards either they sit in their seat or they take someone else's seat, but remove themselves from the queue. In both cases the net number of pre-taken seats doesn't change. | |
| Mar 11, 2017 at 10:59 | comment | added | Geralt of Rivia | @Mathgeek by "first seat" or "last seat" he meant the right seat for the first and the right seat for the last person. So it connects to "Every person that boards the plane after ... will either take their "proper" seat, or if that seat is taken, a random seat instead." | |
| Jan 25, 2017 at 22:11 | comment | added | Aryabhata | @Mathgeek: Yes, it will be 1. | |
| Jan 24, 2017 at 2:45 | comment | added | user383014 | @Aryabhata What if question is that, "What is the probability that last guy sits on either his seat or first person's seat?" Will it be $1$? | |
| Jan 23, 2017 at 23:26 | comment | added | Aryabhata | @Mathgeek: Suppose the last guy gets seat X, which is neither the first seat, nor the last seat. What seat did person numbered X take? | |
| Jan 20, 2017 at 19:50 | comment | added | user383014 | Could you please elaborate how the fate of the last person is determined the moment either the first or the last seat is selected? and how any other seat will necessarily be taken by the time the last guy gets to 'choose'.? | |
| Sep 29, 2010 at 8:24 | vote | accept | crasic | ||
| Sep 29, 2010 at 8:23 | comment | added | crasic | This is a good intuitive way to think about it. A formal proof is too heavy for a over-a-cup-of-coffee discussion, this is just right. I'll give you the credit since nobody seems to want to take a shot at the physical application | |
| Sep 27, 2010 at 20:30 | history | answered | Aryabhata | CC BY-SA 2.5 |