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{\bf Question}

A gambler with capital $\pounds\!z$ bets against an opponent with
capital $\pounds(a-z)$ using \pounds1 stakes.  On each bet the
gambler wins \pounds1 with probability $\frac{1}{3}$, loses
\pounds1 with probability $\frac{1}{2}$ and retains his stake in
the event of a draw.  Show that the probability, $P_z$, that the
game ends after an odd number of bets satisfies the second order
difference equation $$2P_{z+1} + 7P_z +  3P_{z-1} = 6
\hspace{.2in} {\rm for \ \ } z = 1, 2, ...., a-1.$$  By solving
the equation, or otherwise, calculate the probability that the
game will end after an odd number of bets when the gambler and his
opponent both start with \pounds2.

Investigate the value of $P_z$ if the gambler plays against an
infinitely rich opponent.



\vspace{.25in}

{\bf Answer}

The Gambler has $\pounds\, z.$  The opponent has $\pounds (z-a)$
pounds.

$P_z = $Prob(game ends after as odd number of bets)

Consider the first bet.  There are 3 outcomes.
\begin{description}
\item[(i)] gambler wins.  Then he has $\pounds (z+1)$ and the game must not
end in a further odd number of bets.  Probability $\frac{1}{3}(1 -
P_{z+1})$
\item[(ii)] gambler loses.  Then he has $\pounds (z-1)$ and the games must
not end in a further odd number of bets.  Probability
$\frac{1}{2}(1 - P_{z-1})$
\item[(iii)] draw.  Then the gambler still has $\pounds\, z$ and the game
mustn't end in an odd number of bets.  Probability
$\frac{1}{6}(1-P_z)$
\end{description}

So $P_z = \frac{1}{3}(1 - P_{z+1}) + \frac{1}{2}(1 - P_{z-1}) +
\frac{1}{6}(1 - P_z)$ for $0&lt;z&lt;a$.  Also $P_0 = 0$ and $P_a = 0$

The equation when simplified becomes$$2P_{z+1} + 7P_z + 3P_{z-1} =
6 \hspace{.2in} 0 &lt; z &lt; a$$  Putting $P_z =\lambda^z$ in the
homogeneous equation gives $$2 \lambda^2 + 7\lambda + 3 = 0
\Rightarrow (2\lambda +1)(\lambda +3) = 0 \Rightarrow \lambda =
-\frac{1}{2}, -3$$A particular solution of the inhomogeneous
equation is $P_z = \frac{1}{2}$ (constant).

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

$P_0=0$ so $\ds A + B + \frac{1}{2} = 0$

$P_a=0$ so $\ds A\left( -\frac{1}{2} \right)^a + B(-3)^a +
\frac{1}{2} =0$

Solving gives $\ds B = \frac{1 - (-2)^a}{2(6^a -1)} \Rightarrow A
= -B - \frac{1}{2}$

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

When $z = 2$ and $a = 4,$  $\ds P_z = \frac{1 - 2^4}{2(6^4 -
1)}\left(9 - \frac{1}{4}\right) + \frac{1}{2} \left(1 -
\frac{1}{4}\right) = \frac{12}{37} = 0.\overline{324}$

In general, when $\ds a \to \infty, \ \ P_z = \frac{1}{2}\left(1 -
\left(- \frac{1}{2}\right)^z\right)$

So $\ds P_z \to \frac{3}{8}$



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