3.4.1 problem 46

If a failure is seen on the first trial, then there are $0$ successes and $1$ failure, so it is clearly possible that there are more than twice as many failures as successes.

(a)

If we think of the Bernoulli trial success as a win for player $A$, and the Bernoulli trial failure as a loss for player $A$, then have more than twice as many failures as successes is analogous to $A$ losing the Gambler’s Ruin starting with $1$ dollar. For instance, if $A$ wins the first gamble, then $A$ has $3$ dollars, and $B$ needs $2\ast 1+1$ gamble wins for $A$ to lose the entire game.

Thus, we need to find $p_1$.

(b)

$p_k=\frac{1}{2}{p}_{k+2}+\frac{1}{2}{p}_{k-1}$ with conditions $p_0=1$ and $\underset{k\to \infty }{\lim }{p}_k=0$

The characteristic equation is $\frac{1}{2}{t}^3-t+\frac{1}{2}=0$ with roots $1$ and $\frac{-1\pm \sqrt{5}}{2}$.

Thus,

$$p_k=c_1+c_2{\left(\frac{-1+\sqrt{5}}{2}\right)}^k+c_3{\left(\frac{-1-\sqrt{5}}{2}\right)}^k$$

Using the hint that $\underset{k\to \infty }{\lim }{p}_k=0$, $c_1$ and $c_3$ must be $0$. Thus,

$$p_k=c_2{\left(\frac{-1+\sqrt{5}}{2}\right)}^k$$

Using $p_0=0$, we get that $c_2=1$. Thus,

$$p_k={\left(\frac{-1+\sqrt{5}}{2}\right)}^k$$

(c)

$$p_1=\frac{-1+\sqrt{5}}{2}$$