Meeting at the Coffee Shop

Probability Level 4

Two individuals Al and Bob go to a coffee shop every one or two days. After going to the coffee shop, Al will go back to the shop in one day with probability $p$ ; he goes back in two days with probability $1-p$ . After going to the coffee shop, Bob will go back in one day with probability $q$ ; he goes back in two days with probability $1-q$ .

Let $p=\frac{1}{3}$ , and $q=\frac{2}{5}$ .

Suppose Al and Bob both came to the coffee shop today. What is the expected number of days until Al and Bob both come to the coffee shop on the same day again?

Source: link text

The answer is 2.66667.

1 solution

Christopher Criscitiello
Dec 15, 2017

Hint: When $n>2$ , there are only two ways for Al and Bob to meet on day $n$ and not on any day before. Consider the cases when n is even and n is odd. The desired expectation comes out to

$\sum _{n=1}^{\infty } 2 n \left(p^2 (1-p)^{n-1} (1-q)^n+q^2 (1-p)^n (1-q)^{n-1}\right)+\sum _{n=0}^{\infty } (2 n+1) \left(2 p q (1-p)^n (1-q)^n\right)+2 (1-p) (1-q)-p q=(p-2) (q-2)$

The case where Al chooses to come to the coffee shop $k$ days later with probability $p_k$ , $k>0$ is much more difficult (define $q_k$ similarly for Bob). I am curious if anybody has any ideas. (For clarity: $\sum_{k>0} p_k = \sum_{k>0} q_k = 1$ .) The second easiest case to consider seems to be when Al and Bob visit the shop in one and three days (rather than one and two).

I am curious about how I made an error; if we think of $\begin{aligned} p_n&=p\times p_{n-1}+(1-p)\times p_{n-2},\;\; p_0=1,\;\; p_1=\dfrac 13\\ q_n&=q\times q_{n-1}+(1-q)\times q_{n-2},\;\; q_0=1,\;\; q_1=\dfrac 25, \end{aligned}$ then we can solve $\begin{aligned} p_n&=\dfrac 15\left(\left(\dfrac{-2}{3}\right)^n2^{n+1}+3\right)\\ q_n&=\dfrac 18\left(\left(\dfrac{-1}{5}\right)^n3^{n+1}+5\right). \end{aligned}$ Then the answer should be $\dfrac 25\times \dfrac 13 +\lim_{N\rightarrow \infty }\sum_{n=2}^N\left(n\times \left(\prod_{k=1}^{n-1}(1-p_kq_k)\right)\times p_nq_n\right)\approx 2.873.$

This interpretation does however ignore the condition that Al and Bob must go to the coffee shop every one or two days.

Miles Koumouris - 3 years, 5 months ago

Here's the mistake I think you made: I am asking for when they meet given that they do not meet any other time before. I think pn, qn do not take this into account.

Or maybe you are interpreting the question wrong. Here is an example of Al's behavior. Day 0 he is at the shop. On that day he decides to go to the coffee shop 2 days later (he could have decided one day too). So he shows up at the shop on day 2. On day 2 he also decides to go to the coffee shop one day later (again he could have decided 2 days later). So he shows back up at the coffee shop on day 3. This continues. Again the decisions are made randomly with probability p.

I am not positive my answer is right but does this help? If not let me know! Thanks for your interest in this problem

Christopher Criscitiello - 3 years, 5 months ago

Thank you, I misinterpreted the problem; I assumed that the probabilities 'add'; so if Al went to the shop one day, then the next day he would go to the shop with probability $p$ , and the next day he would go to the shop with probability $(1-p)+p^2$ . In other words, I misread the problem and overlooked the condition that they must go to the shop every one or two days. This made the formulae for $p_n$ and $q_n$ quite complicated - maybe this could be a follow-up problem?

Miles Koumouris - 3 years, 5 months ago

Meeting at the Coffee Shop

The answer is 2.66667.

1 solution

0 pending reports