If $\mu^{m+n} + \nu^{m+n} = 2\mu^m\nu^n$ then either "side" of dice flips can be achieved equally

Question

Here's yet another problem from my probability textbook:

A die of any number of faces has $\mu$ of its faces marked $1$, and $\nu$ marked $-1$, the rest being zero. It is repeatedly thrown until the sum of the numbers turned up no longer lies between the limits $m$ and $-n$. Show that either extreme of these limits is equally likely to be reached if$$\mu^{m+n} + \nu^{m+n} = 2\mu^m\nu^n.$$

Here's what I got so far. We want to solve the following equations:$$(\mu + \nu)p_{-n} = \mu p_{-n+1} + {1\over2}\nu, \quad (\mu + \nu)p_{-n+1} = \mu p_{-n+2} + \nu p_{-n}, \quad (\mu+\nu)p_{-n+2} = \mu p_{-n+3} + \nu p_{-n+1}, \quad \ldots\quad , \quad (\mu + \nu)p_0 = \mu p_1 + \nu p_{-1}, \quad \ldots\quad ,\quad (\mu+\nu)p_{m-2} = \mu p_{m-1} + \nu p_{m-3}, \quad (\mu + \nu)p_{m-1} = \mu p_{m} + \nu p_{m-2}, \quad (\mu + \nu)p_{m} = {1\over2}\mu + \nu p_{m-1}.$$However, I don't see how to use the condition that $$\mu^{m+n} + \nu^{m+n} = 2\mu^m\nu^n.$$I'm also not sure if the coefficients in front of $\nu$ and $\mu$ in the first and last equations should be ${1\over2}$. Should they be $1$? Or $0$? And why?

Any help would be well-appreciated.

Answer 1

There is a mistake in your setup. If $p_k$ is the probability of reaching $m$ before reaching $-n$, then $$ p_{-n}=0,\qquad p_m=1 $$ because you have already certainly won or certainly lost.

You have a system of equations:$$ p_k=\frac{\mu}{\mu+\nu}p_{k+1}+\frac{\nu}{\mu+\nu}p_{k-1} \qquad -n<k<m \\p_{-n}=0,\qquad p_m=1\tag{$*$} $$ The goal is to solve this system of equations for $p_0$, that is to get $p_0$ in terms of $\nu$ and $\mu$, and then to use the given equation involving $\nu$ and $\mu$ to show that $p_0=\frac12$.

$(*)$ can be rewritten as $$ p_{k+1}-p_k=(\nu/\mu)(p_{k}-p_{k-1})\tag1 $$ Let $\delta_k=p_{k+1}-p_k$, so that $(1)$ becomes $$\delta_{k}=(\nu/\mu)\delta_{k-1}\tag2$$This obviously implies by induction that $$ \delta_k=(\nu/\mu)^k\delta_0\tag3 $$ Now, we have expressed all of the $\delta_k$'s in terms of $\delta_0$, and can proceed to solve for $\delta_0$ as follows: \begin{align} 1&=1-0 \\&=p_m-p_{-n} \\&=\delta_{-n}+\delta_{-n+1}+\dots+\delta_{m-2}+\delta_{m-1} \\&=\sum_{k=-n}^{m-1} \def\r{\nu/\mu}(\r)^k \delta_0 \\&=\delta_0\cdot \frac{(\r)^{m}-(\r)^{-n}}{\r-1} \end{align} The last line of reasoning implies $\delta_0=\frac{\r-1}{(\r)^m-(\r)^{-n}}$. Therefore, we have solved for $\delta_0$, and therefore all of the $\delta_k$'s, and we just need to use this to find $p_0$. Fortunately, since $\delta_k$ is the difference of the $p_k$'s, $p_0$ is the cumulative sum of the $\delta_k$'s: $$ p_0=p_{-n}+\sum_{k=-n}^{-1}\delta_k=0+\sum_{k=-n}^{-1}\frac{(\r)^{k+1}-(\r)^{k}}{(\r)^m-(\r)^{-n}}=\frac{1-(\r)^{-n}}{(\r)^m-(\r)^{-n}} $$ The sum evaluates nicely since it is telescoping. All that remains is to show $$ \mu^{m+n} + \nu^{m+n} = 2\mu^m\nu^n\qquad \text{implies}\qquad \frac{1-(\r)^{-n}}{(\r)^m-(\r)^{-n}}=\frac12, $$ which I leave to you.

Answer 2

Let $X_n$ be your running total after $n$ die rolls. Furthermore, let $p=\mu/(\mu+\nu)$ be the probability that the next nonzero roll will be positive, and let $q=\nu/(\mu+\nu)$ be the probability that it is negative. You can show that $$ (q/p)^{X_n}\tag{!} $$ is a martingale. This means that $E[(q/p)^{X_{n+1}}\mid X_n]=(q/p)^{X_n}$, which implies the random variable $(q/p)^{X_n}$ does not change on average over time. In particular, if we let $\tau$ be the smallest integer for which $X_\tau\in\{m,-n\}$, then $$ E[(q/p)^{X_\tau}]=E[(q/p)^{X_0}]=E[(q/p)^0]=1\tag1 $$ On the other hand, if we let $r$ be the probability that we exit at $m$ before $-n$, then $$ E[(q/p)^{X_\tau}]=r\cdot (q/p)^m+(1-r)\cdot (q/p)^{-n}\tag2 $$ Combing $(1)$ and $(2)$, and solving for $r$, we get $$ r = \frac{1-(q/p)^{-n}}{(q/p)^m-(q/p)^{-n}}\tag3 $$ Now, using the equality $q/p=\nu/\mu$, and the given condition $\mu^{m+n} + \nu^{m+n} = 2\mu^m\nu^n$, I think you should be able to prove that $r=1/2$.

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I labeled the first equation $(!)$ because it is not at all obvious where it comes from. I happened to remember this trick was applicable from previously knowing about the gambler's ruin problem. I do not know if there is a different way to solve this without that prior knowledge.