Rolling Right Along

The process is a Markov chain with transition matrix:

$P = \begin{bmatrix} \frac{125}{216} & \frac{25}{72} & \frac{5}{72} & \frac{1}{216} \\[0.3em] 0 & \frac{25}{36} & \frac{5}{18} & \frac{1}{36} \\[0.3em] 0 & 0 & \frac{5}{6} & \frac{1}{6} \\[0.3em] 0 & 0 & 0 & 1 \\[0.3em] \end{bmatrix}$

Finding the fundamental matrix:

$N = \left( \begin{bmatrix} 1 & 0 & 0 \\[0.3em] 0 & 1 & 0 \\[0.3em] 0 & 0 & 1 \\[0.3em] \end{bmatrix} - \begin{bmatrix} \frac{125}{216} & \frac{25}{72} & \frac{5}{72} \\[0.3em] 0 & \frac{25}{36} & \frac{5}{18} \\[0.3em] 0 & 0 & \frac{5}{6} \\[0.3em] \end{bmatrix} \right)^{-1} = \begin{bmatrix} \frac{216}{91} & \frac{2700}{1001} & \frac{5490}{1001} \\[0.3em] 0 & \frac{36}{11} & \frac{60}{11} \\[0.3em] 0 & 0 & 6 \\[0.3em] \end{bmatrix}$

Finally the expected number of steps to reach the absorbing state is:

$\begin{bmatrix} \frac{216}{91} & \frac{2700}{1001} & \frac{5490}{1001} \\[0.3em] 0 & \frac{36}{11} & \frac{60}{11} \\[0.3em] 0 & 0 & 6 \\[0.3em] \end{bmatrix} \begin{bmatrix} 1 \\[0.3em] 1 \\[0.3em] 1 \\[0.3em] \end{bmatrix} = \begin{bmatrix} \frac{10566}{1001} \\[0.3em] \frac{96}{11} \\[0.3em] 6 \\[0.3em] \end{bmatrix}$

and since we started from the first state our expected number of rolls is $\frac{10566}{1001}$ .

Let $E(n)$ be the expected number of rolls of $n$ dice needed before each of the dice has shown a $6$ and been removed according to the description in the text of the question.

It is well-known that $E(1) = 6.$ Now for two dice, there is a probability of $\frac{1}{36}$ that both dice show $6$ on the first roll, a probability of $(\frac{5}{6})^{2} = \frac{25}{36}$ that neither do and a probability of $\frac{10}{36}$ that precisely one of the dice shows a $6.$ Thus

$E(2) = \dfrac{1}{36}*1 + \dfrac{10}{36}*(1 + E(1)) + \dfrac{25}{36}*(1 + E(2))$

$\Longrightarrow \dfrac{11}{36}*E(2) = \dfrac{96}{36} \Longrightarrow E(2) = \dfrac{96}{11}.$

Next, for $3$ dice, there is a probability of $\frac{1}{216}$ that all $3$ dice show $6$ on the first throw, $3*(\frac{5}{6})^{2}*\frac{1}{6} = \frac{75}{216}$ that precisely one dice does, $3*(\frac{1}{6})^{2}*\frac{5}{6} = \frac{15}{216}$ that precisely $2$ do and $(\frac{5}{6})^{3} = \frac{125}{216}$ that none do. We then have that

$E(3) = \dfrac{1}{216}*1 + \dfrac{75}{216}(E(2) + 1) + \dfrac{15}{216}(E(1) + 1) + \dfrac{125}{216}(E(3) + 1)$

$\Longrightarrow \dfrac{91}{216}E(3) = \dfrac{306}{216} + (\dfrac{75}{216})(\dfrac{96}{11}) \Longrightarrow E(3) = \dfrac{10566}{1001}.$

Thus $a - b = 10566 - 1001 = \boxed{9565}.$