The two proofs below moot "bijection", but I don't see a formula. 1. Does the bijection have an explicit formula?
- Can this bijection be pictorialized? My 16 y.o. kid doesn't understand the abstract, Daedalian phrasing below.
hunter's answer dated Dec. 4 2013
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 {first passenger's seat, last passenger's seat}. 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).
azjps comments on A combinatorics problem that I was never able to solve.
Here's a bijective proof: let all $n$ individuals find a seat according to the rule. For each sequence in which the $n$th person sits in the seat assigned to the $k$th person, consider the corresponding sequence in which the $n$th and $k$th individuals are swapped (so the last person is in the right seat). Both sequences have equal probability and it is not difficult to verify that this correspondence is a bijection, so the answer is 1/2.
I don't know whether this counts as picturing the bijection, but let's write out all the admissible seatings for a four-passenger plane. We will represent each admissible seating by the cycle of displaced passengers, numbered according to their position in line, with passenger $1$ counting as displaced whether they end up sitting in their own seat or not. The cycle $(1,a,b,c,d)$, for example, will mean $1$ sits in $a$'s seat, $a$ sits in $b$'s seat, $b$ sits in $c$'s seat, $c$ sits in $d$'s seat, and $d$ sits in $1$'s seat. It is necessary that $1<a<b<c<d$. $$ \begin{array}{r|r||r|r||r} \text{cycle} & \text{seating} & \text{cycle} & \text{seating} & \text{probability}\\ (1) & 1234 & (1,4) & 4231 & \frac{1}{4}\cdot1\cdot1\cdot1=\frac{1}{4}\\ (1,2) & 2134 & (1,2,4) & 4132 & \frac{1}{4}\cdot\frac{1}{3}\cdot1\cdot1=\frac{1}{12}\\ (1,3) & 3214 & (1,3,4) & 4213 & \frac{1}{4}\cdot1\cdot\frac{1}{2}\cdot1=\frac{1}{8}\\ (1,2,3) & 3124 & (1,2,3,4) & 4123 & \frac{1}{4}\cdot\frac{1}{3}\cdot\frac{1}{2}\cdot1=\frac{1}{24}\\ \hline & & & & \frac{1}{2} \end{array} $$ In the first row of this table, the cycle $(1)$ represents the situation where passenger $1$ sits in her own seat, while the cycle $(1,4)$ represents the situation where passenger $1$ sits in passenger $4$'s seat and passenger $4$ sits in passenger $1$'s seat. Each of these situations occurs with probability $\frac{1}{4}$. Similarly in row $2$, the cycle $(1,2)$ represents the situation where $1$ sits in $2$'s seat and $2$ sits in $1$'s seat, while the cycle $(1,2,4)$ represents the situation where $1$ sits in $2$'s seat, $2$ sits in $4$'s seat, and $4$ sits in $1$'s seat. Each of these situations occurs with probability $\frac{1}{12}$. The remaining two rows can be understood similarly. Observe that the total probability of the situations represented by the first column and the total probability of the situations represented by the third column both equal $\frac{1}{2}$. In column $1$, passenger $4$ sits in his own seat; in column $3$, passenger $4$ sits in passenger $1$'s seat.
The admissible seatings for a five-passenger plane are below. $$ \begin{array}{r|r||r|r||r} \text{cycle} & \text{seating} & \text{cycle} & \text{seating} & \text{probability}\\ (1) & 12345 & (1,5) & 52341 & \frac{1}{5}\cdot1\cdot1\cdot1\cdot1=\frac{1}{5}\\ (1,2) & 21345 & (1,2,5) & 51342 & \frac{1}{5}\cdot\frac{1}{4}\cdot1\cdot1\cdot1=\frac{1}{20}\\ (1,3) & 32145 & (1,3,5) & 52143 & \frac{1}{5}\cdot1\cdot\frac{1}{3}\cdot1\cdot1=\frac{1}{15}\\ (1,4) & 42315 & (1,4,5) & 52314 & \frac{1}{5}\cdot1\cdot1\cdot\frac{1}{2}\cdot1=\frac{1}{10}\\ (1,2,3) & 31245 & (1,2,3,5) & 51243 & \frac{1}{5}\cdot\frac{1}{4}\cdot\frac{1}{3}\cdot1\cdot1=\frac{1}{60}\\ (1,2,4) & 41325 & (1,2,4,5) & 51324 & \frac{1}{5}\cdot\frac{1}{4}\cdot1\cdot\frac{1}{2}\cdot1=\frac{1}{40}\\ (1,3,4) & 42135 & (1,3,4,5) & 52134 & \frac{1}{5}\cdot1\cdot\frac{1}{3}\cdot\frac{1}{2}\cdot1=\frac{1}{30}\\ (1,2,3,4) & 41235 & (1,2,3,4,5) & 51234 & \frac{1}{5}\cdot\frac{1}{4}\cdot\frac{1}{3}\cdot\frac{1}{2}\cdot1=\frac{1}{120}\\ \hline & & & & \frac{1}{2} \end{array} $$