Explain to a 15 y.o. Byron Schmuland's answer that uses Summation and Product notations to solve the Crazy Lady Airplane Seat probability problem?

Question

Byron Schmuland's answer¹ is too abstruse for a 15 y.o. student who needs details! Please see my enumerated questions below.

Let's find the chance that any customer ends up in the wrong seat.

For $2\leq k\leq n$, customer $k$ will get bumped when he finds his seat occupied by someone with a smaller number, who was also bumped by someone with a smaller number, and so on back to customer $1$.

This process can be summarized by the diagram $$1\longrightarrow j_1\longrightarrow j_2\longrightarrow\cdots\longrightarrow j_m\longrightarrow k.$$
Here $j_1<j_2<\cdots <j_m$ is any (possibly empty) increasing sequence of integers strictly between $1$ and $k$. The probability of this sequence of events is $${1\over n}\times{1\over(n+1)-j_1}\times {1\over(n+1)-j_2}\times\cdots\times{1\over(n+1)-j_m}.$$

1. What exactly does $1/n$, the first term, signify?

2. Where do each of the ${1\over(n+1)-j_m}$ terms hail from and signify? How would you divine or forebode to formulate this term?

Thus, the probability that customer $k$ gets bumped is $$p(k)={1\over n}{\color{red}{\sum}}\prod_{\ell=1}^m {1\over(n+1)-j_\ell}$$ where the sum is over all sets of $j$ values $1<j_1<j_2<\cdots <j_m<k$. That is,

3. Where did the $\color{red}{\sum\limits_{1<j_1<j_2<\cdots <j_m<k}}$ stem from? It appears to come out of left field!

\begin{eqnarray*} p(k)&=&{1\over n}\color{limegreen}{\sum_{J\subseteq\{2,\dots,k-1\}}}\ \, \prod_{j\in J}{1\over (n+1)-j}\cr &=&{1\over n}\ \, \prod_{j=2}^{k-1} \left(1+{1\over (n+1)-j}\right) \cr &=&\color{sienna}{{1\over n}\ \,\prod_{j=2}^{k-1} {(n+2)-j\over (n+1)-j}}\cr &=&\color{sienna}{1\over n+2-k}. \end{eqnarray*}

4. How does $\color{red}{\sum\limits_{1<j_1<j_2<\cdots <j_m<k}} \equiv \color{limegreen}{\sum\limits_{J\subseteq\{2,\dots,k-1\}}}$?

5. How does $\color{sienna}{{1\over n}\ \,\prod\limits_{j=2}^{k-1} {(n+2)-j\over (n+1)-j}={1\over n+2-k}}$?

In the case $k=n$, we get $p(n)=1/2$ as in the other solutions. Maybe there is an intuitive explanation of the general formula; I couldn't think of one.

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Added reference: Finding your seat versus tossing a coin by Yared Nigussie, American Mathematical Monthly 121, June-July 2014, 545-546.

¹I currently see no answer on this page by someone called "Byron Schmuland", but this other question refers to the answerer as Byron Schmuland, and there's merely one deleted user. Then I deduced that user940 (the deleted user) was University of Alberta Prof. Byron Schmuland PhD Carleton University 1987.

Answer 1

So basically the original author meant to find the probability that:

• First passenger randomly choose passenger $j_{1}$’s seat. Since there are $n$ seats to choose, the probability is $\frac{1}{n}$. This is the answer to your first question.
• Passenger $j_{i}$ who got kicked from his original seat randomly choose passenger $j_{i+1}$’s seat. By the time passenger $j_{i}$ need to sit $j_{i}-1$ passengers have sit, so there are $n+1-j_{i}$ to choose. The probability of choosing a particular seat is then $\frac{1}{n+1-j_{i}}$. This is the answer to your second question.

Recall that $\frac{1}{n}\prod_{i=1}^{m}{\frac{1}{n+1-j_{i}}}$ is only the probability of a single possibilities i.e. the probability that the first passenger along with passenger $j_{i}\phantom{x}\forall\phantom{x}1\leq i\leq m$ sit in the wrong seat.So you need to add the probability of all possibilities. This is the anwer to your third question.

Since $1<j_{1}<…<j_{m}<k$ then $\{j_{1},…,j_{m}\}$ is a subset of $\{2,3,…,k-1\}$. This is the answer to your fourth question.

Hint for your fifth question: telescoping multiplication.