# Probability of pushing buttons in elevator

I am considering the following problem from Introduction to Probability by Blitzstein and Hwang (Exercise 23 of Chapter 1).

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. Assume that are equally likely to want to get to floors 2 through 10 (independently of each other). What is the probability that the buttons for 3 consecutive floors are pressed.

My thoughts

There are 3 people A, B and C in the elevator. If A pushes a, B pushes b and C pushes c (a triple (a,b,c) where b=a+1 and c=b+1), then the number of total triples that can be formed from a sample of $$9$$ elements is $$\binom93=84$$. I do not see why the sample space should be $$9\cdot9\cdot9$$ since we are only interested in the triples and more specifically we are interested in the ordered triples.

The favorable outcomes are (2,3,4), (3,4,5), (4,5,6), (5,6,7), (6,7,8), (7,8,9), (8,9,10), i.e. 7 in total.

The probability that the buttons for three consecutive floors are pressed is therefore $$\dfrac{7}{84}$$.

Is the above correct?

• As the question is asked, you also have to count $(3,2,4)$ etc. – Arnaud Mortier Dec 8 '19 at 19:55
• These are includes in the binomial coefficient (9 choose 3), since this does not distinguish between the different orders in which 3 items can be chosen from 9. Feel free to correct me. – saleem Dec 8 '19 at 19:58
• Oh I see I overlooked that. – Arnaud Mortier Dec 8 '19 at 20:00
• The button press combinations aren’t equiprobable. E.g., there’s only one way to obtain $(2,2,2)$, but three ways to obtain $(2,2,3)$. Counting only the combinations without taking order into account undercounts most of the events in the sample space. – amd Dec 8 '19 at 23:04

The denominator in this case is $$9^3=729$$. This is because it is given that each person's choice of floor is independent, so there is a possibility that more than one of them will want to get off at the same floor.

As noted in comments, the numerator is $$7\cdot 3!=42$$. You are correct to note that there are exactly seven ways that the first of the three floors could be chosen. But you must also consider that A choosing 2, B choosing 3, and C choosing 4 is a different event than A choosing 3, B choosing 4, and C choosing 2 even though the floors visited are the same in both cases.

That gives us a total probability of $$42\over729$$.

• For clarity: The sample-space contains $9^{3}$ possibilities because this is the space of all possible triples? And the numerator must be multiplied with $3!$ because all combination of the favorable triples must be included? – saleem Dec 8 '19 at 20:15
• @saleem The rest of each paragraph justifies the calculation in the first sentence. Sorry if that was confusing. – Matthew Daly Dec 8 '19 at 20:17
• For clarity: The sample-space contains $9^3$ possibilities because this is the space of all possible triples? And the numerator must be multiplied with 3! because all combination of the favorable triples must be included? – saleem Dec 8 '19 at 20:32

The sample space has to include events such as $$(2,2,3)$$ and even $$(5,5,5)$$ since it was nowhere said that the buttons pressed are different. In fact, the fact that they are independent implies that they can be equal in this case.

So the sample space has naturally $$9\cdot9\cdot9$$ elements, and it's more natural to count ordered triples rather than unordered, because not all triples have the same symmetry.

I think the others have offered good explanations of the sample space. I'd like to offer a different approach that might be a little more intuitive.

I like to think of these elevator problems in terms of throwing dice. In other words, you throw a 9-sided die 3 times. What is the probability that the throws form a consecutive trio?

Well, first consider the probability of $$(2,3,4)$$, $$P(2,3,4) = \left(\frac{1}{9}\right)^3,$$ where it's the product by independence.

But it didn't have to be in that order. So we multiply by a factor of $$3!$$.

Next, it didn't have to be those three. As you noted, there are 7 triplets we could choose from. Therefore, the probability is

$$3!\cdot 7\left(\frac{1}{9}\right)^3 = \frac{42}{729}.$$