Three people get into an empty elevator at the first floor of a building that has 10 floors. Each presses the button for their desired floor (unless one of the others has already pressed the button). Assume that they are equally likely to want to go to floors 2 through 10 (independently of each other). What is the probability that the buttons for 3 consecutive floors are pressed? (question from 'introduction to probability' by Bliztstein)
I thought the answer might be $$\frac{7 \cdot3!}{9^3}$$… but then I thought maybe the total number of ways is the bose-einstein value $n+k-1 \choose k$ (how many ways there are to choose k times from a set of n objects with replacement, if order doesn't matter).. so then the answer could also be $$\frac{7}{165}$$ Which is the correct answer, and why is the logic for the other answer wrong?
Best Answer
I think there is only one reasonable interpretation of the problem. It leads to a probability model that produces your first answer.
Let our three passengers be A, B, and C. If A, B, C wish respectively to go to floors $p$, $q$, and $r$, record that fact as an ordered triple $(p,q,r)$. By the equally likely and independent part of the problem statement, all ordered triples $(p,q,r)$, where $p$, $q$, and $r$ range from $2$ to $10$, are equally likely. There are $9^3$ such triples. Whether a passenger, say C, happens not to press the elevator button to Floor $6$ because it has already been pressed is irrelevant.
We now count the favourables. There are $7$ ways to choose a collection of $3$ consecutive numbers in the interval from $2$ to $10$. For each of these collections, there are $3!$ ways in which A, B, and C might wish to get off at the floors in this collection, for a total of $7\cdot 3!$. For the probability, divide.