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{\bf Question}

\begin{description}
\item[(a)]
A gambler with initial capital $£z$ plays against an opponent with
initial capital $(a-z)$ where $a$ and $z$ are integers, and $0
\leq z \leq a$. At each play the gambler wins $£1$ with
probability $\df{1}{3}$,\ loses $£1$ with probability $\df{1}{2}$,
and retains his stake money in the event of a draw.

Formulate a difference equation for the probability $P_z$ that the
game ends in an even number of bets, with appropriate boundary
conditions involving $P_0$ and $P_a$. Solve the equation to find
an explicit formula for $P_z$ in terms of $z$ and $a$.

What is the value of $P_z$ if the gambler plays against an
infinitely rich opponent?

\item[(b)]
In the classical gambler's ruin problem one of the players will
eventually be ruined, with probability 1. Write a brief
explanation of what is meant by thus, giving particular attention
to the concept \lq\lq with probability 1" in this context.
\end{description}






\vspace{.25in}

{\bf Answer}

\begin{description}
\item[(a)]
Consider the first bet. There are 3 possible outcomes

\begin{description}
\item[(i)]
Gambler wins. Then he has $£z+1$ and the game must not end in a
further even number of bets. Prob: $\df{1}{3}(1-P_{z+1})$.

\item[(ii)]
Gambler loses. Then he has $£z-1$ and the game must not end in a
further even number of bets. Prob: $\df{1}{2}(1-P_{z-1})$.

\item[(iii)]
Draw. Then the gambler still has $£z$ and the game must not end in
a further even number of bets. Prob: $\df{1}{6}(1-P_z)$.
\end{description}

${}$

So
$P_z=\df{1}£{}(1-P_{z+1})+\df{1}{2}(1-P_{z-1})+\df{1}{6}(1-P_z)$
for $0<z<a$. $P_0=1,\ P_a=1$

${}$

The equation when simplified becomes

$$2P_{z+1}+7P_z+3P_{z-1}=6\ \ 0<z<a$$

Putting $P_z=\lambda ^2$ in the homogeneous equation gives

$$2\lambda^2+7\lambda+3=0\ \ (2\lambda+1)(\lambda+3)=0$$

so $\lambda=-\df{1}{2},\ -3$.

A particular solution of the inhomogeneous equation is

$P_z=\df{1}{z}(\cos t)$????????

${}$

So the general solution is
$P_z=A\left(-\df{1}{2}\right)^2+B(-3)^2+\df{1}{2}$

${}$

The boundary conditions give
$\begin{array}{l}P_0=A+B+\df{1}{2}=1\\
p_a=A\left(-\df{1}{2}\right)^a+B(-3)^a+\df{1}{2}=1 \end{array}$

\newpage
Solving these gives

$$B=\df{1}{2}\df{((-2)^a-1)}{6^a-1}\ \ A=-B+\df{1}{2}$$

$$P_z=\df{1}{2}\df{((-2)^a-1)}{6^a-1}\left[(-3)^z
-\left(-\df{1}{2}\right)^2\right]+\df{1}{2}\left[1+\left(-\df{1}{2}\right)^z\right]$$

As $a \to \infty\ \ P_z \to
\df{1}{2}\left(+\left(-\df{1}{2}\right)^z\right)$

\item[(b)]
Consider a large number $N$ of realisations of the procedure, and
let $R(N)$ denote the number of these which result in one of other
of the players being ruined. Let $D(N)=N-R(N)$. Then to say that
one of the players will be ruined, with probability 1, means that
$\df{R(N)}{N}\to 1$ as $N \to \infty$.

${}$

It also implies $\df{D(N)}{N}\to 0$ as $N\to \infty$ so that
neither player is ruined, with probability zero. This does not
mean impossibility, as clearly the two players could win
alternately and the game would continue for ever. Thus \lq\lq with
probability 1" does not mean certainty.
\end{description}



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