Dodgy Lift 1

Probability Level 5

You are on the 2nd/top floor of a building and you wish to ride the elevator to the 0th/ground floor. Unfortunately the elevator is dodgy. You enter the lift and the doors behind you close. The lift then proceeds to:

Move up 1 floor.
Move down 1 floor.
Shake violently and then Open the doors.

Each movement lasts 10 seconds and at each floor each possible movement occurs with equal probability. (e.g. the first operation is either the opening of the lift, or moving down 1 floor, each with probability 0.5).

Let $P$ be the probability that the first time the doors open you are on the ground floor.

The doors end up opening first on the ground floor.

Let $E$ be the expected amount of time taken in minutes.

If $P = \frac{a}{b}$ for coprime $a,b$ , and $E = \frac{c}{d}$ for coprime $c,d$ .

Find $a + b + c + d$ .

The answer is 14.

2 solutions

Alex Burgess
Jun 2, 2019

Let $P_n$ be the probability the door opens first on the ground floor | given that you are currently on the $n^{th}$ floor.

$\begin{matrix} P_2 &=&\frac{1}{2} P_1&+&(\frac{1}{2} 0)\\ P_1 &=&\frac{1}{3} P_2 &+&\frac{1}{3} P_0&+&(\frac{1}{3} 0) \\ P_0 &=&\frac{1}{2} P_1&+&\frac{1}{2} 1\\ \end{matrix}$

Hence, $P_1 = \frac{1}{6} P_1 + \frac{1}{6} P_1 + \frac{1}{6} = \frac{1}{3} P_1 + \frac{1}{6}$ .

So, $\frac{2}{3} P_1 = \frac{1}{6}$ ,

$P_1 = \frac{1}{4}$ ,

$P = P_2 = \frac{1}{8}$ .

The probability of the doors opening after $3$ lift-movements requires the lift to go from floor 2 to 1 and 1 to 0 and then opening, and is equal to: $\frac{1}{2} \times \frac{1}{3} \times\frac{1}{2} = \frac{1}{12}$ . The probability of the doors opening after $3 + 2n$ lift-movements requires the lift to go a combination of floor 1 to 2 and 2 to 1, or floor 0 to 1 and 1 to 0, $n$ times in some order, and the movements previous. This is equal to $\frac{1}{12} \times (\frac{1}{3} \frac{1}{2} + \frac{1}{3} \frac{1}{2} )^n = \frac{1}{12} \times (\frac{1}{3})^n$ . Hence the expected amount of movements is $1 + ($ expected number of movements over $1) = 1 + \frac{\frac{1}{12} \times 2\sum_{n=0}^{\infty} n (\frac{1}{3})^{n-1}}{\frac{1}{8}} = 1 + \frac{\frac{1}{6} \times (\frac{3}{2})^2}{\frac{1}{8}} =4$ .

$E = \frac{40}{60} = \frac{2}{3}$ .

$a + b + c + d = 1 + 8 + 2 + 3 = 14$ .

This question is also a bit ambiguous. It's not clear if the "if possible" statement means that there is a 33% chance of doing nothing in the case that it is not possible or that the other two cases have a 50% chance of occurring.

Malcolm Rich - 2 years ago

Thanks for the comment. Would saying something like: Does the subset of possible moves from {move up 1, move down 1, open door} with equal probability?

Alex Burgess - 2 years ago

The example adds clarity on this point.

Malcolm Rich - 2 years ago

Roni Edwin
Jun 6, 2019

Let $A_n, B_n,$ and $C_n$ be the probability that it takes $20n-10, 20n,$ and $20n+10$ seconds to reach the ground floor and have the elevator open there from the ground, 1st and 2nd floors respectively. Then

$A_n = \frac{1}{2}B_{n-1}$

$B_n = \frac{1}{3}(A_n+C_{n-1})$

$C_n = \frac{1}{2}B_n$

Solving recursively leads to $C_n=\frac{1}{3}C_{n-1}$ . $C_1= \frac{1}{12}$ so $C_n=\frac{1}{12}\left(\frac{1}{3}\right)^{n-1}$ . Note that $P=\sum_{n=1}^{\infty}\frac{1}{12}\left(\frac{1}{3}\right)^{n-1}=\frac{1}{8}$ The doors end up opening on the first floor so this becomes $C_n= \frac{\frac{1}{12}\left(\frac{1}{3}\right)^{n-1}}{\frac{1}{8}} = \frac{2}{3^n}$ . The expected value $E$ is $\sum_{n=1}^{\infty}\left(10+20n\right)\left(\frac{2}{3^n}\right)$ , which evaluates to 40. This is $\frac{2}{3}$ in minutes. $1+8+2+3$ is 14, which gives us the desired answer.

Dodgy Lift 1

The answer is 14.

2 solutions

1 pending report

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