8. (a) $[\mathbf{5}]$ Find $a_{1}, a_{2}$, and $a_{3}$, so that the following equation holds for all $m \geq 1$ :
$$
p_{m}(0)=a_{1} p_{m-1}(0)+a_{2} p_{m-1}(2)+a_{3} p_{m-1}(4)
$$
(b) [5] Find $b_{1}, b_{2}, b_{3}$, and $b_{4}$, so that the following equation holds for all $m \geq 1$ :
$$
p_{m}(2)=b_{1} p_{m-1}(0)+b_{2} p_{m-1}(2)+b_{3} p_{m-1}(4)+b_{4} p_{m-1}(6)
$$
(c) [5] Find $c_{1}, c_{2}, c_{3}$, and $c_{4}$, so that the following equation holds for all $m \geq 1$ and $j \geq 2$ :
$$
p_{m}(2 j)=c_{1} p_{m-1}(2 j-2)+c_{2} p_{m-1}(2 j)+c_{3} p_{m-1}(2 j+2)+c_{4} p_{m-1}(2 j+4)
$$
Solution: We will show how to find $a_{1}$ in the first equation; other coefficients can be evaluated in the same way. $a_{1}$ is the probability that, beginning with $|T-H|=0$, the value of $|T-H|$ remains 0 after using the optimal strategy for one round. This only happens when the first three coins are not all heads or all tails. Therefore, $a_{1}=\frac{3}{4}$. It follows that $b_{1}=\frac{1}{4}$. The rest of the terms are the binomial coefficients of the expansion of $\left(\frac{1+1}{2}\right)^{3}$.
(a) $a_{1}=\frac{3}{4}, a_{2}=\frac{1}{2}, a_{3}=\frac{1}{8}$.
(b) $b_{1}=\frac{1}{4}, b_{2}=\frac{3}{8}, b_{3}=\frac{3}{8}, b_{4}=\frac{1}{8}$.
(c) $c_{1}=\frac{1}{8}, c_{2}=\frac{3}{8}, c_{3}=\frac{3}{8}, c_{4}=\frac{1}{8}$.