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{\bf Question}

A gambler with initial capital $£z$ plays against an opponent with
capital $£(z-1)$, where $a$ and $z$ are integers and $0 \leq z
\leq a$. At each play the gambler wins $£1$ with probability $p$
and loses $£1$ with probability $q=1-p$.

Let $q_z$ denote the probability that the gambler will eventually
be ruined. Write down a recurrent relation for $q_z$ and solve it
to obtain explicit formulae for $q_z$ in terms of $z,\ a,$ and
$p$, in both cases $p=\df{1}{2}$ and $p\ne \df{1}{2}$.

Two players begin a game of dice with $10$ each. At each play they
both stake $£1$ and each of them throws a fair cubical die. If
player $A$ has a higher score than player $B$ he wins, otherwise
he loses. Player $B$ says that if player $A$ gets down to his last
$£5$ he will give him a chance by changing the game to one of
tossing a fair coin until one of the players us ruined.  In this
game $A$ wins if the coin lands heads and $B$ wins if it lands
tails again with $£1$ stake.

Calculate the probability that player $A$ will eventually be
ruined.



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{\bf Answer}

We argue conditionally on the result of the first play to obtain

$q_z=pq_{z+1}+qp_{z-1}$ where $q=1-p$

The auxiliary equation is

$$p\lambda^2-\lambda+q=0$$

i.e., $(p\lambda-q)(\lambda-1)$ since $p+q=1$.

so $\lambda=\df{q}{p}\ \ \lambda=1.$

We have unequal roots of $q \ne p$.

Then $q_z=A+B\left(\df{q}{p}\right)^z$\ \ \ for $0<z<a$.

Boundary conditions are $q_0=1m\ q_a=0$

and these give $A$ and $B$

$$q_z=\df{(\frac{q}{p})^a-(\frac{q}{p})^z}{(\frac{q}{p})^a-1}$$

For $p=1=\df{1}{2},\ q_z=Az+B$, and then boundary conditions give

$$q_z=1-\df{z}{a}$$

${}$

For player $A$ to be ruined, he must get down to $£5$ and then be
ruined on the coin-tossing game.

To find the probability that he gets down to $£5$ is equivalent to
playing the dice game where player $A$ has $£5$ and player $B$ has
$£10$.

P(player $A$ has a higher score than player
$B$)=$\df{1}{2}\left(\df{36-6}{36}\right)=\df{15}{36}$

so $p=\df{15}{36},\ q=\df{21}{£6},\ z=5,\ a=15,\
\df{q}{p}=\df{21}{5}=\df{7}{5}$

so P(player $A$ gets down to
$£5$)=$\df{(\df{7}{5})^{15}-(\frac{7}{5})^5}{(\frac{7}{5})^{15}-1}=0.97167
\cdots$

The game now becomes a fair game with $z=5$ and $a=20$

so P(player $A$ ruined in fair game)=$1-\df{5}{20}=\df{3}{4}$

The plays throughout are independent so overall

P($A$ ruined)=$0.97167 \cdots \times \df{3}{4}=.728755771\cdots$



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