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The problem is such: two drunks start at either end of an alleyway of length n. Apart from at the ends, they each move one step forwards or one step backwards randomly. At the ends of the alley they move towards the centre with probability $1$. What is the expected number of steps before they meet?

In the problem I'm looking to solve $n = 99$, though it could be any odd number.

We are looking for the most elegant solution to this problem. It has been solved via Monte-Carlo and Markov Chains in our group, but I want to believe there is a simpler solution not involving $2400 \times 2400$ matrices. I am trying to take inspiration from the question below, but the boundary condition complicates matters.

Exact probability of collision of two independent random walkers after N steps

Links to papers or other questions dealing with this particular problem would be gladly accepted.

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Max
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  • As a start perhaps replace the two drunks with the distance between them. – JP McCarthy Jan 17 '14 at 12:50
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    @JpMcCarthy The distance between the two walks is not Markov because the transitions when one of the walkers is at 0 or at n are not the same as when none are. – Did Jan 17 '14 at 16:37
  • For the analogous problem with a single drunk (how long until the drunk reaches the other end of the alley), you can mitigate somewhat the issue of the behaviour at the endpoint by reflecting the alleyway (let the drunk walk on ${ -n, \ldots, -1,0,1, \ldots,n}$ instead) and ask how long until they reach $\pm n$. Now you can let them move left/right at the origin with 50-50 probability as at other points and perhaps use Markov chain methods. – Mike F Jan 17 '14 at 21:52
  • My attempts have been around taking the distance between the stochastic process, and reflecting this. Ie, terminating conditions are X=0 and X=2N-2. However, the doesn't capture the meetings when one drunk has moved out of the alley and has his position reflected, while the other has not. – Max Jan 20 '14 at 10:47

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Here are some easy remarks on the question (but not a solution). First, if $n$ is odd the two particles never meet, for parity reasons, hence one can assume that $n$ is even.

Say that the particle starting from $0$ hits $n$ at time $T_n$. The particles cannot cross each other without meeting hence, at time $T_n$, they already met. Thus, the meeting time $M_n$ is almost surely finite and $$E(M_n)\leqslant E(T_n)=n^2. $$ Note that $M_n$ is also at most the hitting time $T'_n$ of $0$ by the particle starting from $n$, hence $$E(M_n)\leqslant E(\min(T_n,T'_n)),$$ where $T'_n$ is an independent copy of $T_n$. Conversely, the particles cannot meet before at least one of them hits $n/2$ hence $$E(M_n)\geqslant E(\min(T_{n/2},T'_{n/2})). $$ This suggests that there might exist some finite positive $\mu$ such that $E(M_n)\sim\mu n^2$ when $n\to\infty$. If this is so, one knows that $\frac14\theta\leqslant\mu\leqslant\theta$ where $\theta$ in $(0,1)$ is defined by the equivalent $E(\min(T_n,T'_n))\sim\theta n^2$.

Note finally that $\min(T_n,T'_n)$ is the first hitting time of the boundary of the square $[-n,n]^2$ by a random walk on $\mathbb Z^2$ starting from $(0,0)$ with steps $(\pm1,\pm1)$.

Did
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  • This looks like it's along the right track. Are you sure about your parity argument though? It may just be a difference in the way we are measuring the alley, but I think even is the case where they never meet. They certainly need an odd number of spaces inbetween them. – Max Jan 20 '14 at 10:42
  • To meet, they need an even number of spaces hence an odd number of states. Consider n=2, then the state space is {0,1,2} and they meet at 1 after one step. – Did Jan 20 '14 at 17:09