Timeline for Taking Seats on a Plane
Current License: CC BY-SA 4.0
18 events
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| Sep 7, 2022 at 23:31 | comment | added | Will Orrick | A way to fix the argument while keeping your definition of $p_n$ is to write $p_n=\frac{1}{n}+\frac{1}{n}\sum_{j=2}^{n-1}p_j$. Clearly $p_2=\frac{1}{2}$. Taking the induction hypothesis to be that $p_j=\frac{1}{2}$ for $2\le j<n$, compute $p_n=\frac{1}{n}+(n-2)\frac{1}{n}\frac{1}{2}=\frac{1}{2}$. You don't need to use the $q_j$ in the argument. | |
| Aug 23, 2021 at 7:45 | comment | added | user851668 | 3. And pls expatiate why the probability that the nth person DOESN'T get the nth seat = $q_n=\underbrace{1/n}_{\text{1-st person sits in n-th seat}}+\underbrace{(n-2)/n}_{\text{1-st person does not sit in the 1-st or n-th seat}} \times q_{n-1}$? To me, this appears to come out of the blue! | |
| Aug 23, 2021 at 7:44 | comment | added | user851668 | 2. More details pls! Pls expatiate why the probability that the nth person gets the nth seat = $p_n=\underbrace{1/n}_{\text{1-st person sits in 1-st seat}}+\underbrace{(n-2)/n}_{\text{1-st person does not sit in the 1-st or n-th seat}} \times p_{n-1}$? To me, this appears to come out of the blue! | |
| Aug 23, 2021 at 7:42 | comment | added | user851668 | 1. Your equations are ambiguous. What exactly are $p_{n - 1}, q_{n - 1}$ multiplying? Pls edit your answer to clarify. | |
| Aug 23, 2021 at 7:36 | comment | added | user851668 | If you need more detail and the intermediate steps typed out, see this other answer on this same page and in this thread. | |
| Jun 11, 2019 at 18:29 | comment | added | Hans | @ely: Exactly as you did, he showed no derivation of the $\frac{n-2}np_{n-1}$. If you do not think this comes from a summation, you need to justify that term. The justification is not there. It does not constitute a proof. | |
| Jun 11, 2019 at 18:03 | comment | added | ely | @Hans this is incorrect. Hans' comment stating that the first equality comes from an assumtion leading to the summation is an incorrect statement. | |
| Jun 11, 2019 at 0:46 | comment | added | Hans | @Shashank has already derived long before the constancy $\frac12$ of $p_n$ for all $n$. | |
| Jun 10, 2019 at 20:12 | history | edited | user103828 | CC BY-SA 4.0 |
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| Jun 10, 2019 at 20:02 | comment | added | user103828 | ok, I agree (and added an edit to this effect). I don't have permission to delete your comments | |
| Jun 10, 2019 at 17:18 | comment | added | Hans | Do you agree with my comment regarding the flaw of you solution above? Why did you or someone else delete my first comment? | |
| Jun 9, 2019 at 18:12 | comment | added | Hans | The term in the first equation comes from assuming $\underbrace{\frac{n-2}{n}}_{\text{1-st person does not sit at the 1-st or n-th seat}} \times p_{n-1}=\sum_{i=2}^{n-1}\underbrace{\frac1n}_\text{1-st person sits at the i-th seat} p_i$ where $p_i$'s are all the same $\forall 2\le i\le n-1$. | |
| Jun 9, 2019 at 11:11 | history | edited | user103828 | CC BY-SA 4.0 |
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| Jun 9, 2019 at 11:10 | comment | added | user103828 | @Hans Could you provide a bit of extra explanation as I don't understand where I (and some of the other answers) make this assumption? | |
| Apr 27, 2018 at 12:45 | history | edited | user103828 | CC BY-SA 3.0 |
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| Mar 6, 2018 at 7:43 | history | edited | user103828 | CC BY-SA 3.0 |
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| Mar 6, 2018 at 7:36 | history | edited | user103828 | CC BY-SA 3.0 |
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| Aug 28, 2017 at 16:17 | history | answered | user103828 | CC BY-SA 3.0 |