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{\bf Question}

A gambler with initial capital \pounds z (where z is a positive
integer) plays a coin tossing game against an infinity rich
opponent.  Two fair coins are tossed: if both show heads the
gambler wins \pounds 2; if both show tails the gambler wins
\pounds 1; otherwise the gambler loosed \pounds 1.

${}$


Letting $q_z$ denote the probability the the gambler is
eventuality ruined, show that $$q_{z+2} + q_{z+1} - 4q_z +
2q_{z-1} = 0$$

${}$

Find the general solution of this difference equation.  Using the
assumption that $q_z \to 0$ as $4z \to \infty$, show that $$q_z =
(\sqrt 3 - 1)^z.$$

${}$

What is the minimum initial capital the gambler needs in order
that he has a better than even chance of not being ruined?

${}$

Suppose the gambler starts with \pounds 1.  Considering the
various possible outcomes, show that the probability of ruin in 5
or fewer steps is $\ds \frac{78}{128}$



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{\bf Answer}

Let $q_z$ denote the probability of ruin, with initial capitial
\pounds z.  Arguing conditionally on the result of the first bet
gives $$q_z = \frac{1}{4} \cdot q_{z+2} + \frac{1}{4} q_{z+1} +
\frac{1}{2}q_{z-1}$$ Rearranging this gives $$ q_{z+2} + q_{z+1}
-4q_z + 2q_{z-1}=0$$To find the general solution put $q_z =
\lambda ^z$.  This gives  the auxillary equation \begin{eqnarray*}
\lambda^3 + \lambda^2 - 4\lambda + 2 & = & 0 \\
(\lambda-1)(\lambda^+2\lambda-2) & = & 0 \\ {\rm So \ \ } \lambda
& = & 1, \, -1-\sqrt3, \, -1+\sqrt 3 \end{eqnarray*} Thus $\ds q_z
= A + B(-1 -\sqrt3)^z + C(-1+\sqrt3)^z$

Now $q_z$ is a probability  and since $-1-\sqrt3<-1$ and
$0<-1+\sqrt3<1$, we cannot have the righthand side between the )
and 1 unless B=0.  Now $\ds (-1-\sqrt3)^z\to0$ as $z\to \infty$
and so assuming $q_z \to 0$ as $z\to \infty$ we conclude that A=0.
Hence $$q_z = C(-1+\sqrt 3)^z, $$ and finally $q_0 = 1$ gives
$C=1.$

${}$

Now in order to have $q_z < \frac{1}{2}$ we must have
$\ds(-1+\sqrt3)^z < \frac{1}{2}$ i.e., $z \ln (-1+\sqrt3)<-\ln 2$
giving $\ds z>\frac{-\ln 2}{\ln(-1+\sqrt3)} = 2.22...$ so $\pounds
z \geq \pounds 3$.

Starting with \pounds 1, the paths leading to ruin in 5 or fewer
steps are:

\begin{tabular}{llr}

  1 step & L & prob $\ds \frac{1}{2}$ \\
  2 steps & IMPOSSIBLE &  \\
  3 steps & W(1) L L &  $\ds\frac{1}{2^4}$ \\
  4 steps & W(2) L L L  & $\ds\frac{1}{2^5}$ \\
  5 steps & W(1) W(1) L L L  & $\ds\frac{1}{2^7}$ \\
   & W(1) L W(1) L L  & $\ds\frac{1}{2^7}$ \\
\end{tabular}

So $\ds p= \frac{1}{2^7} (64 8 4 1 1) = \frac{78}{128}$

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