Expected number of moves of a random walk inside a simplex

Question

Few days ago, Roger Blazey, retired head maths teacher, at LinkedIn group told about the following problem:

“This problem was raised as part of a lecture at The Biennial AAMT (Australian Association of Mathematics Teachers) in Sydney in 1972. In 1972 I was a young, beginning Mathematics Teacher. The lecture was given, I think, by a Professor RADE , and if my memory serves me well, the Title of his Lecture, in one of the many plenary sessions was, "Stochastic Processes."

The simplicity of the result for the basic problem, fascinated me then and nothing has changed for me today!

Here is the problem, in it's most basic form.

”On a table there are $2$ piles of disks... $4$ disks in one pile (pile $A$) and $3$ disks in the other pile (pile $B$). So based on the outcome resulting in the throw of a coin ($H$ or $T$) .....................................

when a head occurs, a disk is transferred from $A$ to $B$; otherwise (for a tail) a disk is transferred from $B$ to $A$. What is the expected number of moves to have either $A$ or $B$ void?

Now computer modelling will quickly suggest a result of $12$ with the strong suspicion that when $n(A)=x$ and $n(B)=y$ the expected number of moves is given by “$xy$”.

So how can this be proved?

The obvious extension (which, to my knowledge was not raised by anyone at the lecture) is, “what is the formula for $3$ piles of disks and the outcome on the fall of a die.... And further for $4$ piles of disks, $5$ piles,............ “$p$” piles of disks?"

The formula for $3$ piles of disk (numbering $x$, $y$ and $z$) is certainly not the product "$xyz$"

Hymn ... a tad too difficult for me I'm afraid! Over to you!”

Here is the link to the discussion at LinkedIn.

My try. It seems the following.

Let $x=n(A)$, $y=n(B)$, $x+y=N=\operatorname{const}$, and $E(x;N)$ be the expected number of moves to have either $A$ or $B$ void.

Then we obtain the following boundary and recurrent formulas for $E(x;N)$:

$E(0;N)=E(N;N)=0$ (there is no need to move) and if $0<x<N$ then $$E(x;N)=1+(E(x+1;N)+E(x-1;N))/2.$$

(By the symmetry, $E(N-x;N)=E(x;N)$, but we shall not need this in the following proof).

The recurrent formula can be transformed to

$$E(x+1;N)=2E(x;N)-E(x-1;N)-2$$

which solution for $x\ge 1$ is

$$E(x;N)=xE(1;N)-x^2+x.$$

Since $E(N;N)=0$ we obtain $NE(1;N)-N^2+N=0.$

So $E(1;N)=N-1$ and $E(x;N)=x(N-x)=xy.$

The general problem can be interpreted as a random motion inside a simplex

$$\{x_1,\dots,x_k\ge 0 : x_1+\dots+x_k=N\}$$

with respective boundary and recurrent conditions.

So, this is the problem. The two piles case was simple and we have a luck to obtain a nice little neat answer here (although Roger disagreed with me :-) ) . But the case of more than two piles seems to be much more hard and need to involve methods which are far beyond the simple observations which I used to solve two piles case. I googled for a general case and found none. But I found a couple of similar problems, which are hard and use complicated techniques to investigate: Bruce Hajek “Gambler's Ruin: A Random Walk on the Simplex” and Wikipedia remark on N-player gamblers ruin problem.

Roger wrote:

I would like to think there is a formula for 3 piles of disks and perhaps if Dr. Alan Turing were alive today he would derive a result for it!

But because an invocation of a spirit of Dr. Alan Turing is not an easy task, I hope that participants of MSE will found an other way to solve this problem. :-)

PS. There is already a lot of questions related to two piles case. For instance — Alex Ravsky, Jan 06 '15 at 16:39
Mircea Grecu just pointed to me a publication “A matrix-analytic approach to the N-player ruin problem” by Yvik C. Swan and F. Thomas Bruss — Alex Ravsky, Jan 06 '15 at 17:00
With 3 piles A, B, C, what is the move when the die selects pile A? — Did, Jan 06 '15 at 17:28
@Did I thought about it. :-) Roger did not provided a complete description of a generalization, so I think we may assume that each of six moves $A\to B$, $B\to A$, $A\to C$, $C\to A$, $B\to C$, and $C\to B$ has an equal probability $1/6$. — Alex Ravsky, Jan 06 '15 at 17:41

score 2 · Answer 1 · answered Jan 06 '15 at 20:38

Assume 3 piles of disks and that at each step a disk passes from one pile to another, the 6 possible choices of the "emitting" pile and of the "receiving" pile are equiprobable. Then the mean time for some pile to be empty, starting from piles of heights $x$, $y$, $z$, is $$\frac{3xyz}{x+y+z}.$$

To prove this, note that the function $$t:(x,y,z)\mapsto xyz,$$ defined on the nonnegative integers $(x,y,z)$ such that $x+y+z=n$ for some given $n$, is such that $t(x,y,z)=0$ if $x=0$ or $y=0$ or $z=0$ and, for every positive $(x,y,z)$, $$t(x,y,z)=\tfrac13n+\tfrac16(t(x,y+1,z-1)+t(x,y-1,z+1)+t(x+1,y,z-1)+t(x-1,y,z+1)+t(x+1,y-1,z)+t(x-1,y+1,z)).$$ The uniqueness of the solutions of Pascal equation makes that $\frac3nt(x,y,z)$ is indeed the mean exit time from the triangle, starting from $(x,y,z)$.

For more on the subject, see this.

Note that for 4 piles of disks the mean time for some pile to be empty is not a multiple of the product of their heights.

Expected number of moves of a random walk inside a simplex

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