\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}
{\bf Question}
A gambler with initial capital $£z$ plays against an opponent with
initial capital $£(a-z)$, where $a$ and $z$ are integers with $0
\lq z \leq a$. At each play the gambler wins $£1$ with probability
$p$ and loses $£1$ wit probability $q$. A draw occurs with
probability $r<1$, where $p+q+r-1$, and in the case of a draw both
players retain their $£1$ stake.
${}$
Let $q_z$ denote the probability that the gambler will eventually
be ruined. Write down a difference equation for $q_z$ and solve it
to obtain an explicit formula for $q_z$ in the case $p \ne q$.
${}$
The two players initially have $£10$ between them. If $p=0.4$ and
$q=0.3$ how much of the $£10$ must the gambler have in order to
have a better than even chance of not being ruined?
${}$
With these values of $p$ and $q$, how much initial capital does
the gambler need in order to have a better than even chance of not
being ruined when playing against an infinitely rich opponent?
\vspace{.25in}
{\bf Answer}
${}$
$\begin{array}{rcl}\mbox{P(ruin)} & = & \mbox{P(ruin and win 1st
bet)}\\ & & + \mbox{P(ruin and lose 1st bet)}\\ & & +\mbox{P(ruin
and draw 1st bet)}\\ & = & \mbox{P(ruin | wins 1st bet)P(win 1st
bet)}\\ & & +\mbox{P(ruin | loses 1st bet)P(lose 1st bet)}\\ & &
+\mbox{P(ruin | draws 1st bet)P(draw 1st bet)}\end{array}$
${}$
so
\begin{eqnarray*} q_z & = & q_{z+1} \cdot p+q_{z-1}\cdot q+q_z \cdot
r\\ q_z(1-r) & = & p \cdot q_{z+1}+q \cdot q_{z-1}\\ q_z & = & \ds
\frac{p}{p+q}q_{z+1}+\ds \frac{q}{p+q}q_{z-1}\ \ (r \ne 1)
\end{eqnarray*}
with boundary conditions $q_0=1;\ q_a=0$
Substituting $q_z=\lambda^z$ gives
$$\lambda^z=\ds \frac{p}{p+q}\lambda^{z+1}+\ds
\frac{q}{p+q}\lambda^{z-1}$$
$$\begin{array}{rcl} \rm{i.e.}\ \ds
\frac{p}{p+q}\lambda^2-\lambda+\ds \frac{q}{p+q} & = & 0\\
(\lambda-1)\left(\ds \frac{p}{p+q}\lambda-\ds \frac{q}{p+q}\right)
& = & 0
\end{array}$$
so $\lambda=1$ or $\lambda=\ds \frac{p}{q} \ne 1$ when $p \ne q$.
So the general solution is
$$q_z=A+B\left(\ds \frac{q}{p}\right)^z$$
The boundary conditions give
$$\begin{array}{rcl} 1 & = & A+B\\ 0 & = & A+B\left(\ds
\frac{q}{p}\right)^a \end{array}$$
so $1=B\left(1-\left(\ds \frac{q}{p}\right)^a\right)$
${}$
Thus $B=\ds \frac{1}{1-(\frac{q}{p})^a}$ and $A=1-B=\ds
\frac{-(\frac{q}{p})^a}{1-(\frac{q}{p})^a}$
Hence $q_z = \ds
\frac{(\frac{q}{p})^z-(\frac{q}{p})^a}{1-(\frac{q}{p})^a}$
Now with $p=0.4,\ q=0.3$, and $a=10$ we have
$$q_z=\ds
\frac{(\frac{3}{4})^z-(\frac{3}{4})^{10}}{1-(\frac{3}{4})^{10}}$$
so $q_z<\ds \frac{1}{2}$ if and only if $\left(\ds
\frac{3}{4}\right)^z-\left(\ds \frac{3}{4}\right)^{10}<\ds
\frac{1}{z} \left(1-\left(\ds \frac{3}{4}\right)^{10}\right)$
i.e. $2 \left(\ds \frac{3}{4}\right)^z<1+\left(\ds
\frac{3}{4}\right)^{10}$
$$z>\ds \frac{\ln \left(1+\left(\ds
\frac{3}{4}\right)^{10}\right)-\ln 2}{\ln\left(\ds
\frac{3}{4}\right)}=2.219\cdots$$
${}$
Playing against an infinitely rich opponent is analysed by letting
$a \to \infty$, giving the probability of ruin as $\left(\ds
\frac{3}{4}\right)^z$.
Now $\left(\ds \frac{3}{4}\right)^z<\ds \frac{1}{2}$ if and only
if $z>\ds \frac{\ln(\frac{1}{2})}{\ln(\frac{3}{4})}=2.41\cdots$
so in fact with $£3$ the gambler has a better than even chance of
not being ruined against an infinitely rich opponent.
\end{document}