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{\bf Question}

In the Gambler's ruin problem, changing the stakes from 1 unit to
$\frac{1}{2}$ unit is equivalent to doubling the initial capitals.
What effect does this have on the game?

\vspace{.25in}

{\bf Answer}

In the gambler's ruin problem, to investigate the effect of
doubling the initial capital.
\begin{description}
\item[(i)] $p=q: \hspace{.2in} q_z=1-\frac{z}{a}$

So if $z$ is replaced by $2z$ and $a$ by $2a$ $q_z$ is unchanged.

$\ds E_z = z(a-z), $ so if $z \rightarrow 2z$ and $a \rightarrow
2a$

$E_{2z} = 2z(2a - 2z) = 4E_z$
\item[(ii)] $\ds p \not= q: \hspace{.2in} q_z = \frac{\left( \frac{q}{p}
\right)^z - \left( \frac{q}{p}\right)^a}{1- \left(
\frac{q}{p}\right)^a}$

$$E_z = \frac{z}{q-p} - \frac{a}{q-p} \left(  \frac{1-\left(
\frac{q}{p}\right)^z}{1-\left( \frac{q}{p}\right)^a} \right)$$

Replacing $z$ by $2z$ and $a$ by $2a$

\begin{eqnarray*} q_z & \rightarrow & \frac{\left( \frac{q}{p} \right)^{2z} -
\left( \frac{q}{p}\right)^{2a}}{1- \left( \frac{q}{p}\right)^{2a}}
= \frac{\left( \frac{q}{p} \right)^z - \left(
\frac{q}{p}\right)^a}{1- \left( \frac{q}{p}\right)^a} \cdot
\frac{\left( \frac{q}{p} \right)^z + \left(
\frac{q}{p}\right)^a}{1+ \left( \frac{q}{p}\right)^a} \\ & = & q_z
\cdot K \end{eqnarray*}

Where $K>1$ if $q>p$ and $K<1$ if $q<p.$

or: $$q_z  \rightarrow  \frac{\left[ \left( \frac{q}{p}
\right)^{2} \right]^z - \left[ \left(
\frac{q}{p}\right)^{2}\right]^a}{1- \left[ \left(
\frac{q}{p}\right)^{2}\right]^a}$$  So the game is \lq \lq
equivalent \rq \rq to the original with the odds changed from
$p:q$ to $p^2:q^2$ with draws allowed since $p^2 + q^2 < 1$

$$ E_z \rightarrow \frac{2z}{q-p} - \frac{2a}{q-p} \left( \frac{1
- \left( \frac{q}{p} \right)^{2z}} {1 - \left( \frac{q}{p}
\right)^{2a}} \right)$$

\end{description}



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