$\begingroup$ thank you for your explanation. for point 1, after drawing it out i finally understand it. but for point 2, could you please elaborate a bit more? scenario A: the first person in the line bumped into seat #1, customer #1 then bumped into seat #5, this is $1\longrightarrow j_1=5\longrightarrow j_2\longrightarrow\cdots\longrightarrow j_m\longrightarrow k$; Scenario B: the first person in the line bumped into seat #5, customer #5 then bumped on, this is $j_1=5\longrightarrow j_2\longrightarrow\cdots\longrightarrow j_m\longrightarrow k$ -- these are 2 different scenarios right? $\endgroup$
Commented Oct 1, 2013 at 15:05 $\begingroup$ +1 for the reference! here's a link to it $\endgroup$ Commented Aug 19, 2014 at 2:09 $\begingroup$This analysis is correct, but not complete enough to convince me. For example, why is the fate of the last person settled as soon as the first person's seat chosen? Why will any other seat but the first person's or the last person's be taken by the time the last person boards?
I had to fill in the holes for myself this way.
The last person's fate is decided as soon as anybody chooses the first person's seat (nobody is now in a wrong seat, so everybody else gets their assigned seat, including the last person) or the last person's seat (the last person now won't get their correct seat). Any other choice at any stage doesn't change the probabilities at all.
Rephrasing. at each stage, either the matter gets settled and there is a 50/50 chance it gets settled each way for the last person's seat, or the agony is just postponed. The matter can thus be settled at any stage, and the probabilities at that stage are the only ones that matter -- and they are 50/50 no matter what stage. Thus, the overall probability is 50/50.
answered Oct 1, 2010 at 17:03 David Lewis David Lewis 2,180 1 1 gold badge 15 15 silver badges 23 23 bronze badges$\begingroup$ Your answer is correct. I just would like to add that people can still be in the wrong seats. However, all unoccupied seats will have their people from that point on. $\endgroup$
Commented Feb 26, 2018 at 7:18$\begingroup$ For example, 1 can seat in 2. 2 can seat in 3. And 3 can sit in 1. So all 3 are in the wrong seats. But seat 4-100 will all be unoccupied $\endgroup$
Commented Feb 26, 2018 at 7:18$\begingroup$ -1. "at each stage, either the matter gets settled and there is a 50/50 chance it gets settled each way for the last person's seat." That is just plain wrong. It is not $1/2$ at every stage except the last. $\endgroup$
Commented Jun 10, 2019 at 6:43$\begingroup$ @Hans I think the OP means this: "at each stage, either the matter gets settled and (conditioned on it getting settled) there is a 50/50 chance it gets settled each way for the last person's seat." The punctuation of the sentence supports this interpretation. I do think the formulation of the first paragraph, "nobody is now in the wrong seat" is confusing, at best. Better to say that nobody still waiting is now displaced from their seat. $\endgroup$
Commented Jul 27, 2021 at 15:51 $\begingroup$Let $P(n)$ denote the probability of the last passenger getting his seat if we begin with $n$ passengers.
Consider the simple case for just $2$ seats:
$P(2) = \frac12$ (first boarder picks his own seat with 1/2 probability)
For $n$ seats: (i) With $\frac1n$ probability, the passenger picks the seat of the first passenger, the n'th seat from the end (in which case the last passenger would definitely get his seat). (ii) With 1/n probability, the current passenger picks the seat of the last passenger, first seat from the end (and now, the last passenger can definitely not get his own seat). (iii) Otherwise, the passenger picks some other seat (say #i from the end) among the n-2 remaining seats (with probability 1/n), continuing the dilemma. The problem now reduces to the initial problem with i seats.
Therefore, $$ P(n) = \frac1n \times 1 + \frac1n \times 0 + \frac1n\sum_^ P(i) $$ or $$ nP(n) = 1 + \sum_^ P(i).$$ So $$nP(n)-(n-1)P(n-1)=P(n-1)\Longleftrightarrow P(n)=P(n-1),$$ and $P(n)=P(2) = \frac12, \,\forall n \ge 2$ .
9,967 4 4 gold badges 19 19 silver badges 53 53 bronze badges answered Aug 4, 2011 at 17:17 221 2 2 silver badges 2 2 bronze badges$\begingroup$ Please expatiate on your answer? Where did $nP(n)-(n-1)P(n-1)=P(n-1)$ in your last para. stem from? How did you prognosticate this equation? It appears to come out of the blue. $\endgroup$
– user851668 Commented Aug 23, 2021 at 7:28$\begingroup$ @Intellectuallydisabled plug in the RHS for $nP(n)$ and $(n-1)P(n-1)$ from the preceding equation. Then things cancel! $\endgroup$
Commented Sep 19, 2021 at 19:28 $\begingroup$I tried to synthesize the proof for myself from stuff I've read to get rid of all calculations (somehow I found the argument that "each person's choice is 50-50 between good and bad once we throw away the irrelevant stuff" convincing but hard to formalize).
Claim 1: when the last passenger boards, the remaining empty seat will either be his own or the first passenger's.
Proof: If the remaining empty seat belongs to passenger $n \neq 1, 100$, then passenger $n$ should have sat there.
Claim 2: if at any time a passenger other than the final passenger finds her seat occupied, then both the seat assigned to the first and to the final passenger will be free.
Proof: If not, then there is a nonzero probability that after this passenger makes a decision, both the first and last seats will be occupied. This contradicts Claim 1.
Claim 3: There is a bijection between the set of admissible seatings in which the final passenger gets his seat and the set where he doesn't.
Proof: Suppose for an admissible seating $S$ that passenger $n$ is the first to choose one of . By claim $2$, there is a unique admissible seating $T$ which agrees with $S$ except that passenger $n$ and the final passenger make the opposite decision ($T$ matches $S$ until passenger $n$ sits, then by Claim 2, $T$ must continue to match $S$ until the final passenger).
answered Dec 4, 2013 at 9:42 31.1k 3 3 gold badges 42 42 silver badges 74 74 bronze badges $\begingroup$ This is the best proof so far! $\endgroup$ Commented Oct 31, 2016 at 13:06$\begingroup$ 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 $\frac<1>$, whereas the seating where everyone in not in their own seat--this is unique--has probability $\frac<1>$. The additional step that is needed is to prove that any two admissible seatings related by the bijection are equally likely. (They are.) $\endgroup$
Commented Jul 27, 2021 at 13:52 $\begingroup$I don't really have the intuition for this, but I know the formal proof. This is equivalent to showing that the probability that in a permutation of $[n]$ chosen uniformly at random, two elements chosen uniformly at random are in the same cycle is $1/2$. By symmetry, it's enough to show that the probability that $1$ and $2$ are in the same cycle is $1/2$.
There are many ways to show this fact. For example: the probability that $1$ is in a cycle of length $k$ is $1/n$, for $1 \le k \le n$. This is true because the number of possible $k$-cycles containing $1$ is $ (k-1)! = (n-1)!/(n-k)!$, and the number of ways to complete a permutation once a $k$-cycle is chosen is $(n-k)!$. So there are $(n-1)!$ permutations of $[n]$ in which $1$ is in a $k$-cycle. Now the probability that $2$ is in the same cycle as $1$, given that $1$ is in a $k$-cycle, is $(k-1)/(n-1)$. So the probability that $2$ is in the same cycle as $1$ is $$ \sum_^n = \sum_^n (k-1) =
Alternatively, the Chinese restaurant process with $\alpha = 0, \theta = 1$ generates a uniform random permutation of $[n]$ at the $n$th step; $2$ is paired with $1$ at the second step with probability $1/2$. This is a bit more elegant but requires some understanding of the CRP.
answered Nov 19, 2010 at 19:40 Michael Lugo Michael Lugo 23.6k 3 3 gold badges 49 49 silver badges 95 95 bronze badges$\begingroup$ Dear @Michael Lugo, as someone used to random permutatiom, I also did the same reasoning and was happy with it. Later I realized I was unable to justify this modelisation applies to this problem. In the arirplane setting, symmetry between the passengers is broken (passengers $2$ and $n$ do not have the same probability to sit at their place). Also the cycle containing $1$ here is relatively small, with expectation of order $\log(n)$, whereas the cycle containing $1$ has expectation of order $n$ in a random permutation. Hence the parallel between the two settings remains somewhat obscure to me. $\endgroup$
Commented Aug 11, 2019 at 14:01 $\begingroup$There are many ways to come up with this answer, but here’s one that makes sense to me. For ease of explanation we’ll say that I’m the first person to sit down and you’re the last. Also if you sit in your own seat then you “win”, otherwise you “lose”.
Let’s say that there are only two seats, yours and mine. If I sit in my own seat, you win. If I sit in your seat, you lose. So you have a $50\%$ chance of winning.
Now let’s go back to $100$ seats. The previous paragraph still holds true: you have a $50\%$ chance of winning if we only consider your seat and mine. Now if I sit anywhere else, I’m just postponing the decision. Let’s say I sit in the seat of the person who’s 13th in line. Persons $2$ through $12$ will sit in their own seats, then when person $13$ comes in he can either sit in my original seat (and you win) or yours (and you lose). Or of course he could sit anywhere else and postpone the decision again.
If this keeps going, then eventually there are only two seats left and person $99$ is forced to choose either your seat or mine, again with 50% chance. There are only two seats that matter throughout the game: yours and mine. Any sitting in other seats is just postponing the decision of which of the two interesting seats gets sat in first. Note also that you’ll only ever end up in your seat or mine, no one else’s.
It’s a bit like flipping a coin, except that you can postpone flipping, but not indefinitely. What’s the chance of coming out heads? Well $50\%$ , the postponement doesn’t change that.
Here’s a mathematical way to see it. Define $f(n)$ to be the chance that the last person in an airplane of n seats will get his own seat. It can be defined recursively like this: $$f(n)=\frac1n\cdot1+\fracn\cdot f(n−1) + \frac1n\cdot 0$$
The first term is the chance that the first person will sit in his own seat $\frac1n$ multiplied by the chance, then, that the last person will sit in his own $1$ . The last term is the chance that the first person will sit in the last person’s seat $\frac1n$ multiplied by the chance, then, that the last person will get his own seat $0$ . The middle term counts every other seat. There are $n−2$ other seats, and there’s a $\frac1n$ chance of each, and they all simplify to the $f(n−1)$ case. Also $f(2)=0.5$ .
If you plug in $0.5$ for the $f(n−1)$ term, you find that $f(n)=0.5$ , so it’s true for any $n>1$ .
9,967 4 4 gold badges 19 19 silver badges 53 53 bronze badges answered Oct 8, 2018 at 18:52 329 2 2 silver badges 8 8 bronze badges$\begingroup$ Doesn't this answer merely duplicate math.stackexchange.com/a/55628 on this same page and in this same thread? $\endgroup$
– user851668 Commented Aug 23, 2021 at 7:33 $\begingroup$This problem appears in Blitzstein's Introduction to Probability (2019 2 ed) Ch 1, Exercise 61, p 42. Here's the answer from p 10 of the Selected Solutions PDF.
Solution: The seat for the last passenger is either seat 1 or seat 100; for example, seat 42 can’t be available to the last passenger since the 42nd passenger in line would have sat there if possible. Seat 1 and seat 100 are equally likely to be available to the last passenger, since the previous 99 passengers view these two seats symmetrically. So the probability that the last passenger gets seat 100 is 1/2.
This solution doesn't satisfy me, because I couldn't get the symmetry argument and I didn't understand why "seat 42 can't be available". Here's how I convinced myself.
Let's call the first passenger $P_1$ . There are 100 possible seats $S_i\:(i=1,2. 100)$ she can randomly select. First let's cover the two "edge" cases, seat 1 and seat 100. Suppose she selects $S_1$ by chance. For notation purposes we'll say $S_1=P_1$ using equality operator as seat selection. If this happens there is no change in seating order, $P_2=S_2$ , $P_3=S_3$ , and eventually $P_=S_$ . In the other "edge" case, $P_1=S_$ . In this case it should be clear that $P_2=S_2$ , $P_3=S_3$ , and so on. When $P_$ gets on the plane she realizes that her seat is taken and the only available option is $S_1$ so $P_1=S_1$ . Notice that the probability of $P_=S_1$ and $P_=S_$ are the same for these two cases. Further notice I am not saying the probability of each occurence is 1/2. Clearly the probability of each of these events is 1/100 or 1/n, where n is in general the number of seats. It will turn out that n does not matter due to symmetry, which will be the fundamental insight into this problem. For now let's continue with more concrete cases.
Let's assume that $P_1=S_$ . Again, note that there is a 1/n chance of this seat selection, same as all the other seats. When $P_2$ boards the plane, she is able to take her seat $P_2=S_2$ . Similarly, $P_3=S_3$ and so on until $P_=S_$ . Ok, now $P_$ boards the plane and sees $S_1$ and $S_$ are the only two available seats. Her own seat was taken by $P_1$ . She now must randomly choose. Clearly, given the previous seating order there is now a 50/50 chance that either $P_=S_$ or $P_=S_$ . If $P_=S_$ then $P_$ gets to sit in her assigned seat ( $P_=S_$ ), otherwise she must sit in seat 1, so $P_=S_$ . Note that the final 50/50 choice between $S_1$ and $S_$ is the "game" that other answers have referred to here. It also provides the recursive base case that many answers here discuss. To see this, let's play another "game" and see what happens.
Let's now assume that $P_1=S_$ . Similar to above, $S_1$ is left open, $P_2=S_2$ , $P_3=S_3$ and so on until $P_=S_$ . Now, $P_$ boards the plane and see that $S_1$ , $S_$ , and $S_$ are available. Well, let's go through these 3 possible choices. Assume $P_=S_1$ . Now, $P_=S_$ and $P_=S_$ . The "game" ends with Passenger 100 getting their assigned seat. Continuing, imagine $P_=S_$ . In that scenario, $P_=S_$ and $P_=S_1$ . In this case the game ended with Passenger 100 sitting in Seat 1. Notice that the probability of each outcome was exactly the same. Ok, one last possibility. Let's imagine that $P_=S_$ . Ahhhh, now we've "kicked" $P_$ out of her seat, and when she boards the plane she must choose between $S_1$ and $S_$ . Hey wait a minute. Didn't we already cover that case? Yes! And so here we are. Hopefully, by now you can recognize the recursion in this problem, the base case, the symmetry, and the "game" that every single simulation will play out the same way with Passenger 100 having equal probability of sitting in Seat 1 or Seat 100, with absolutely zero probability of sitting anywhere else. Therefore P=1/2!
answered Feb 18, 2020 at 18:06 Evan Zamir Evan Zamir 221 2 2 silver badges 5 5 bronze badges$\begingroup$ if every single passenger ends up sitting incorrectly such that once S98 is seated only seat 5 and 54 are empty, then what happens? it is not necessary that 1 or 100 will have am equal probability for P100 $\endgroup$
Commented Jan 12, 2022 at 4:26$\begingroup$ @Arslán In the vast majority of cases most passengers will sit correctly. For every passenger (up to certain point) to sit incorrectly an extremely unlikely sequence of events has to occur, namely P1 has to sit in P2's seat, P2 has to sit in P3's seat, P3 has to sit in P4's seat, and so on. If, for example, P1 sits in P3's seat instead of P2's seat, then P2 will sit correctly; if P1 sits in P73's seat, P2 through P72 will sit correctly (and other, later passengers will likely sit correctly as well). $\endgroup$
Commented Oct 16, 2022 at 1:32$\begingroup$ It's not possible that P99 and P100 find only P5 and P54's seats available: once P1's seat gets taken, everyone after that sits correctly. $\endgroup$
Commented Oct 16, 2022 at 1:32 $\begingroup$I thought I'd add another solution that uses recursion. Let $p_n$ be the probability that the n-th person gets the n-th seat (his own seat) and $q_n=1-p_n$ the probability that he does not. Then,
Edit: I should add that initially I just thought it was as simple as plugging one equation into the other but indeed an inductive argument is needed like @ely's answer.
Edit: my answer assumes that $p_i=p_j$ - basically I haven't taken into account that probabilities might be different depending on which seat the 1-st person sits in (as stated by @hans). The answer might be salvageable if one instead defines $p_n$ as the probability that everyone on a plane with $n$ -seats all sit in their own seat but not sure as this might make some other implicit assumption.
answered Aug 28, 2017 at 16:17 user103828 user103828 2,428 23 23 silver badges 52 52 bronze badges$\begingroup$ Do you agree with my comment regarding the flaw of you solution above? Why did you or someone else delete my first comment? $\endgroup$
Commented Jun 10, 2019 at 17:18$\begingroup$ ok, I agree (and added an edit to this effect). I don't have permission to delete your comments $\endgroup$
Commented Jun 10, 2019 at 20:02$\begingroup$ @Shashank has already derived long before the constancy $\frac12$ of $p_n$ for all $n$. $\endgroup$
Commented Jun 11, 2019 at 0:46$\begingroup$ @ely: Exactly as you did, he showed no derivation of the $\frac
$\begingroup$ A way to fix the argument while keeping your definition of $p_n$ is to write $p_n=\frac<1>+\frac<1>\sum_^p_j$. Clearly $p_2=\frac<1>$. Taking the induction hypothesis to be that $p_j=\frac<1>$ for $2\le j+(n-2)\frac<1>\frac<1>=\frac<1>$. You don't need to use the $q_j$ in the argument. $\endgroup$
Commented Sep 7, 2022 at 23:31 $\begingroup$It is possible to count the different configurations of interest to calculate the probability directly, by formalizing some of the ideas already presented.
In an allowed configuration, denote a displacement of one or more passengers with the diagram $i\rightarrow j$ whenever passenger $i$ displaces passenger $j$ from their assigned seat ( $i < j$ ).
Suppose C is an allowed configuration with a displacement D of passengers, say, $. i\rightarrow j. $ Clearly i has a predecessor (which can be added to the diagram) or i is passenger 1, since the problem states that only passenger 1 is free to displace passengers without being displaced themselves. Since each predecessor must represent a passenger who boarded earlier, of which there are a finite number, using this argument at most i times, shows that D must begin with passenger 1. By a similar argument, D must have a last passenger which chooses passenger 1’s seat to end the displacement.
If E is a displacement in C, by the same argument it must start with passenger 1, followed by the choices already indicated in D, so that E is the same as D. Clearly, two allowed configurations are the same if and only if their displacements are the same. Additionally, note that a displacement of the form, $\textstyle\ 1\rightarrow i \rightarrow … \rightarrow j$ where is a subset of in increasing order, always specifies a valid configuration. Hence,
There is a bijection between the set of allowable configurations and diagrams of the form $1\rightarrow i \rightarrow … \rightarrow j$ where is a subset of in increasing order.
Now count configurations by counting the diagrams. Write 1 as the diagram for the null displacement in the configuration where all passengers sit in their assigned seats. Note that this is same as counting all subsets of a two particular sets, so the probability that the final passenger is sitting in their assigned seat is:
$\textstyle \frac 1\rightarrow i \rightarrow … \rightarrow j\textrm < where is a subset of in increasing order>> 1\rightarrow k \rightarrow … \rightarrow l\textrm < where is a subset of in increasing order>> = \frac>> = \frac\\$
EDIT: This is not new, but for completeness sake . After the 4th paragraph second sentence, it's clear that each allowable configuration can be identified by its unique displacement (chain). (Take the allowable configuration--seating chart--that has each passenger in their own seat to be the null displacement, so that every allowable configuration has exactly one displacement).
Then, for each displacement that ends in 100, 100 sits in seat 1 (end of induction/paragraph 3). For each displacement not ending in 100, passenger 100 sits in their own seat. Since every allowable configuration contains one displacement, then passenger must sit in seat 1 or 100.
As other's have noted (e.g. Will, Hunter), we can pair up each displacement $D$ with another equally likely: $\textstyle\ 1\rightarrow i \rightarrow … \rightarrow j$ pairs up with $\textstyle\ 1\rightarrow i \rightarrow … \rightarrow j \rightarrow 100$ since at the point j chooses seat 1, they could have chosen seat 100 with equal probability. The result immediately follows then since in any allowable configuration it's equally likely that passenger 100 is in their own seat or seat 1.
answered May 26, 2016 at 6:42 127 2 2 silver badges 2 2 bronze badges$\begingroup$ 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>$, whereas $1\rightarrow2\rightarrow\ldots\rightarrow99$ and $1\rightarrow2\rightarrow\ldots\rightarrow99\rightarrow100$ both have probability $\frac<1>$. $\endgroup$
Commented Jul 28, 2021 at 4:19$\begingroup$ 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. $\endgroup$
Commented Apr 13, 2022 at 19:50$\begingroup$ 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? $\endgroup$
Commented Apr 14, 2022 at 1:33$\begingroup$ 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^$ elements in the sample space for an $N$ passenger plane, and (2) that I disagree with you about the probability distribution being uniform. $\endgroup$
Commented Apr 14, 2022 at 14:39$\begingroup$ 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. $\endgroup$
Commented Apr 14, 2022 at 18:46 $\begingroup$I know I am late to this question but I came here as a result of a link from this closed question: 100 seats Quant problem
I believe this is simpler than the solutions I see. Imagine that we simulate a coin toss with a wheel with three outcomes: red, white and blue.
We let red be heads and white be tails. If we get a blue, we spin again. That's exactly what is happening in this problem.
If the first person, chooses his own seat then everyone else, including the last person, will get his own seat. If the first person chooses the last seat then the last person doesn't get his seat. Otherwise, the first person chooses another seat and the game continues.
Everyone gets their own seat up to the person that was chosen in the first round. Let's say the first person chose $47$ , for example, Then seats $2$ through $46$ will have their rightful owners and person $47$ will have to choose. His choices are $1$ , $100$ or something else. In each case, if $1$ is chosen then the last person gets his seat, if $100$ is chosen, he doesn't get his seat and if something else is chosen then the game continues.
In this sense, we are essentially spinning the wheel until we get either a $1$ or $100$ so the probability that the last person gets his seat is $\frac$ .
answered Oct 5, 2022 at 16:39 John Douma John Douma 11.6k 2 2 gold badges 24 24 silver badges 25 25 bronze badges$\begingroup$ I really like this explanation, tyvm for sharing it. "His choices are 1, 100 or something else". The "or something else" is the part that I originally got stuck on, trying to compute the different ways it could proceed from there, but if you treat it like a continuing game. The only un-intuitive thing for me, "something else" is vastly more likely than 1 or 100 since seats are chosen at random it feels odd that eventually it comes down to 1 or 100. $\endgroup$
Commented Nov 15, 2022 at 4:47 $\begingroup$Call the passengers $P_1,\dots,P_n$ in the order in which they board, and let $S_i$ be the seat assigned to $P_i$.
Consider the first moment at which a passenger $P_i$ sits in either $S_1$ or $S_n$. The other one of those seats is empty, so the passengers haven't all boarded yet, so $i\ne n$.
If $i=1$, $P_i$ takes a random seat as per the OP.
Otherwise, $i\ne 1$ and $i\ne n$, so the reason why $P_i$ picked $S_1$ or $S_n$ is that $S_i$ was already occupied. So in this case, $P_i$ takes a random seat as per the OP.
Thus in both the above cases, this moment is when a passenger takes a random seat. They are equally likely to pick $S_1$ as $S_n$. Thus each of the two cases below occurs with probability $1/2$.
If $P_i$ picked $S_n$, then $P_n$ will not get their assigned seat $S_n$.
If $P_i$ picked $S_1$, then the cycle of displacement of Bruce's proof ends, and everyone who has not yet boarded gets their assigned seat, including $P_n$.
Thus the answer is $1/2$.
answered Mar 24, 2018 at 20:52 2,996 2 2 gold badges 12 12 silver badges 30 30 bronze badges$\begingroup$ Although several answers have been posted, I felt that those that were not over-complicated were unsatisfactory for one reason or another: Aryabhata's was too unclear; Matt posed a different problem and its solution, without proving that the OP had the same answer; David Lewis's is wrong (as J Chang pointed out); the "proofs" in hunter's are just further claims which lack proof; Bruce's fails to prove that the two possibilities are equally probable (on which point I agree with Michael Albanese). $\endgroup$
Commented Mar 24, 2018 at 20:52 $\begingroup$ Why don't you re-locate your comment into you answer? $\endgroup$ Commented Jul 27, 2021 at 0:15 $\begingroup$Here is a simple proof. For $k\ge 2$ , let $p_k$ be the probability that the $k$ -th person's seat is taken when he/she tries to sit.
We consider the disjoint events that person $j$ takes up this particular seat, for $1\le j \le n$ . In order for that to happen, $j$ 's seat has to be taken, and then $j$ is left with $n-j+1$ random choices of which $j$ must take $k$ 's seat. We can write: $$p_=<1\over n>+p_2<1\over n-1>+p_3<1\over n-2>\dots+p_<1\over n-k>.$$
answered Feb 1, 2021 at 9:39 66 4 4 bronze badges$\begingroup$ Just to clarify, each term (except the first) is "the probability that person j's seat is taken and he sits in seat k", $p_j \cdot (n - j + 1)^<-1>$ , which are a complete list of ways person k's seat could be taken and all of them are disjoint, so it is correct to add them. $\endgroup$
Commented Aug 1, 2021 at 18:04$\begingroup$ 1. In your equation for $p_k$, what does $1/n$ (the first term) signify? Pls edit your answer to clarify. $\endgroup$
– user851668 Commented Aug 23, 2021 at 7:57$\begingroup$ @DanielV 2. I know that "each term (except the first)" is "the probability that person j's seat is taken and he sits in seat k". But why's this $p_j \cdot (n - j + 1)^<-1>$? $\endgroup$
– user851668 Commented Aug 23, 2021 at 7:58$\begingroup$ @KSLLF $1/n$ is the probability that the first person sits in seat $k$. The probability that person $j$'s seat is taken is $p_j$, and there are are $(n - j + 1)$ seats then he will choose from, $1$ of them is person $k$'s seat. $\endgroup$
Commented Aug 23, 2021 at 11:49$\begingroup$ @DanielV Thanks! But how can I intuit $p_
The other answer that uses recursion to define $p_$ and $q_$ is very light on details and makes it sound as if you can directly solve of $p_ = q_$ from the two equations.
Note that the two equations
don't offer independent constraints on the variables since $p_ = 1 - q_$ . So subtract the equations $p_ - q_$ :
At this point we can state for the base case, $n=2$ , we know $p_ = q_ = \frac$ . Now if we assume $q_ = p_$ , clearly the right-hand side of our last equation becomes zero, and $q_ = p_$ .
Since $q_ + p_ = 1$ , then $\boxed
It's just important to note that after setting up the problem recursively, you still need to appeal to induction to derive the solution. It's not as simple as algebraic manipulation of the definition of $p_$ or $q_$ even though this is how the other solution makes it sound.
349k 37 37 gold badges 482 482 silver badges 878 878 bronze badges answered Feb 11, 2018 at 20:18 2,537 1 1 gold badge 23 23 silver badges 25 25 bronze badges$\begingroup$ -1. This derivation presumes $p_i=p_j,\forall\, 2\le i $\begingroup$ @Hans no, I am not assuming $\frac $\begingroup$ If you do not get the term $\frac $\begingroup$ Specifying the term $\frac $\begingroup$ 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_ A recursive solution: Once a passenger who has been displaced from their assigned seat sits in the seat assigned to passenger $1$ , no more passengers will be displaced. But if the displaced passenger takes some other seat, a passenger still standing in line will become displaced Let the plane have $N$ seats and let $r_n$ , $1\le n\le N-1$ , be the probability that with $n$ passengers standing in line none of the standing passengers is displaced from their assigned seat. We have $r_=\frac$ because that is the probability the first passenger happens to choose their own seat. If $n$ passengers are standing in line and somebody in line is displaced—it will always be exactly one passenger—the probability that the displaced passenger is at the head of the line is, by symmetry among the standing passengers, $\frac$ . And if the displaced passenger is at the head of the line, the probability that they take the seat assigned to passenger $1$ is also $\frac$ . It follows that $$ r_=r_n+\frac. $$ By induction, we find that $r_n=\frac$ and hence $r_1=\frac$ . Another recursive solution: This expands on Shashank's answer, emphasizing a key point that caused me some confusion at first. When a non-displaced passenger gets to the head of the line, they simply take their assigned seat, but when a displaced passenger gets to the head of the line, they choose uniformly at random from the set $$ \\>\cup\\>. $$ Although technically passenger $1$ is not displaced, they follow the same procedure as the displaced passengers do, so we lump them in with that group. As a consequence, if passenger $k$ is displaced, then when that passenger gets to the head of the line, the situation is just like the original problem, but for a plane of $N-k+1$ passengers. Define $p_n$ to be the probability that, in the setup described, the last passenger to board an $n$ -seat airplane gets their assigned seat. Then for $n\ge2$ , $$ p_n=\frac\cdot1+\left[\frac\sum_^p_\right]+\frac\cdot 0. $$ Each term in this expression corresponds to a seat choice of passenger $1$ : if the choice is seat $1$ , passenger $n$ gets their assigned seat with probability $1$ ; if it's seat $k$ , for $2\le k\le n-1$ , then passengers $2$ through $k-1$ take their assigned seats and, when passenger $k$ comes to choose a seat, it is like the original problem but with $n-k+1$ seats; if the choice is seat $n$ , then passenger $n$ gets their assigned seat with probability $0$ . Multiplying by $n$ and reversing the order of summation gives $$ np_n=1+\sum_^p_k. $$ Subtracting this from the same recurrence with $n$ replaced by $n+1$ gives $$ (n+1)p_-np_n=p_n, $$ for $n\ge2$ . Hence $p_2=p_3=p_4=\ldots$ . One easily sees $p_2=\frac$ . The stumbling block for me was in seeing the similarity between passenger $1$ and the displaced passengers. Passenger $1$ chooses randomly among the available seats, which include their assigned seat, and I got hung up on the possibility that passenger $1$ might get their own seat, which can't happen for other displaced passengers. That, however, is not the correct parallel to draw. The right parallel is that both passenger $1$ and the displaced passengers can get passenger $1$ 's seat, which ends the cycle of displacement. Several other answers on this page, like (this, this, and this) —jump straight to an even simpler recurrence. The recurrence does hold and can be derived from this one, but haven't been able to understand how to get that recurrence directly. This issue was raised by Hans on this page in a number of his comments. If I stumble upon any insight that might help other readers with this point, I will add it here. Solution using conditional probability: we have seen that if seat $1$ is chosen before the last passenger is seated, then that passenger sits in their assigned seat; if seat $N$ is chosen before the last passenger is seated, then the last passenger is displaced and must take seat $1$ . Let $E_k$ be the event that the $k$ th passenger to be seated is the first to take one of seats $1$ and $N$ . Then $$ \Pr(\text)=\sum_^\Pr(\text\mid E_k)\Pr(E_k)=\frac\sum_^\Pr(E_k)=\frac. $$ Diagrammatically, this calculation is shown below. The probability that $N$ gets their assigned seat is the sum of the probabilities that each of passengers $1$ through $N-1$ takes seat $1$ . Solution using a bijection between cycles: (This completes a final, needed step in hunter's answer.) The seating process results in a permutation of the passengers, and this permutation always consists of a single cycle containing passenger $1$ followed by displaced passengers, an ascending order. So for $N=100$ the cycle $(1,46,53,75,88)$ represents the situation where passenger $1$ displaces passenger $46$ , $46$ displaces $53$ , $53$ displaces $75$ , $75$ displaces $88$ , and $88$ sits in passenger $1$ 's seat. Passengers $89$ and up, and passenger $100$ in particular, will sit in their assigned seats. The cycle $(1,46,53,75,88,100)$ represents a similar situation, except that passenger $88$ displaces passenger $100$ . Passengers $89$ through $99$ will sit in their assigned seats, and passenger $100$ will have only one choice of seat: passenger $1$ 's seat. The probability of occurrence of these two cycles is the same: $$ \frac\frac\frac\frac\frac. $$ In general, when displaced passenger $k$ chooses a seat, they choose uniformly at random from $101-k$ possible seats. Once passenger $88$ has chosen either seat $1$ or seat $100$ (each of which happens with probability $\frac$ ), the seating of the remaining passengers standing in line is determined. Now every cycle that does not contain $100$ is paired in this way with an equally probable cycle that does contain $100$ . Hence passenger $100$ sits in their assigned seat with probability $\frac$ .
