Counting The Sixes

There is $1$ way to make a $6$ with a $1$ -digit subsequence (with a $6$ ), out of a total of $6$ $1$ -digit combinations, and there are $20$ $1$ -digit subsequences in the $20$ -digit sequence, for an expected value of $\frac{1}{6}\cdot 20 = \frac{10}{3}$ .

There are $5$ different ways to make a $6$ with a $2$ -digit subsequence (with a $15$ , $24$ , $33$ , $42$ , or $51$ ), out of a total of $6^2$ $2$ -digit combinations, and there are $19$ $2$ -digit subsequences in the $20$ -digit sequence, for an expected value of $\frac{5}{6^2}\cdot 19 = \frac{95}{36}$ .

There are $10$ different ways to make a $6$ with a $3$ -digit subsequence (with the digits of $114$ there are $\frac{3!}{2!} = 3$ ways, with the digits of $123$ there are $3! = 6$ ways, and with the digits of $222$ there is $\frac{3!}{3!} = 1$ way, for a total of $3 + 6 + 1 = 10$ different ways), out of a total of $6^3$ $3$ -digit combinations, and there are $18$ $3$ -digit subsequences in the $20$ -digit sequence, for an expected value of $\frac{10}{6^3}\cdot 18 = \frac{5}{6}$ .

There are $10$ different ways to make a $6$ with a $4$ -digit subsequence (with the digits of $1113$ there are $\frac{4!}{3!} = 4$ ways, with the digits of $1122$ there are $\frac{4!}{2!2!} = 6$ ways, for a total of $4 + 6 = 10$ different ways), out of a total of $6^4$ $4$ -digit combinations, and there are $17$ $4$ -digit subsequences in the $20$ -digit sequence, for an expected value of $\frac{10}{6^4}\cdot 17 = \frac{85}{648}$ .

There are $5$ different ways to make a $6$ with a $5$ -digit subsequence (with the digits of $11112$ there are $\frac{5!}{4!} = 5$ ways), out of a total of $6^5$ $5$ -digit combinations, and there are $16$ $5$ -digit subsequences in the $20$ -digit sequence, for an expected value of $\frac{5}{6^5}\cdot 16 = \frac{5}{486}$ .

Finally, there is $1$ way to make a $6$ with a $6$ -digit subsequence (with a $111111$ ), out of a total of $6^6$ $6$ -digit combinations, and there are $15$ $6$ -digit subsequences in the $20$ -digit sequence, for an expected value of $\frac{1}{6^6}\cdot 15 = \frac{5}{15552}$ .

The combined expected value is $\frac{10}{3} + \frac{95}{36} + \frac{5}{6} + \frac{85}{648} + \frac{5}{486} + \frac{5}{15552} = \frac{12005}{1728}$ , so $a = 12005$ and $b = 1728$ , and $a + b = \boxed{13733}$ .

Suppose that a total of $N \ge 6$ dice are rolled. For integers $1 \le u \le 6$ and $u \le v \le N$ , let $X_{uv}$ be the indicator random variable that is equal to $1$ if a consecutive subsequence of $u$ dice rolls (ending with the $v$ th in the main sequence) adds to $6$ , and which is equal to $0$ otherwise. Thus $X_{uv} \; = \; \left\{ \begin{array}{lll} 1 & \hspace{1cm} & d_v + d_{v-1} + ... + d_{v+1-u} = 6 \\ 0 & & \mathrm{ow} \end{array}\right.$ where $d_k$ is the value of the $k$ th die roll for $1 \le k \le N$ . Then $C \; = \; \sum_{u=1}^6 \sum_{v=u}^N X_{uv}$ and hence, by the linearity of expectation, $\mathbb{E}[C] \; = \; \sum_{u=1}^6 \sum_{v=u}^N \mathbb{E}[X_{uv}] \; = \; \sum_{u=1}^6 \sum_{v=u}^N \mathbb{P}[X_{uv} = 1]$ Now $\mathbb{P}[X_{uv} = 1]$ is the probability that the sum of $u$ dice rolls is equal to $6$ , and hence is $\frac{n_u}{6^u}$ , where $n_u$ is the number of ways in which $u$ dice rolls can total $6$ , which is the coefficient of $x^6$ in the expansion of $(x + x^2 + x^3 + x^4 + x^5 + x^6)^u$ . Since $x + x^2 + x^3 + x^4 + x^5 + x^6 \; = \; \frac{x(1-x^6)}{1-x}$ we see that $n_u$ is the coefficient of $x^6$ in $x^u(1-x)^{-u}$ , and hence is the coefficient of $x^{6-u}$ in $(1-x)^{-u}$ . Thus we deduce that $n_u \; = \; {u + (6-u) - 1 \choose u-1} \; =\; {5 \choose u-1} \hspace{2cm} 1 \le u \le 6$ Therefore $\begin{aligned} \mathbb{E}[C] & = \sum_{u=1}^6 \sum_{v=u}^N \frac{n_u}{6^u} \; = \; \sum_{u=1}^6 \frac{n_u(N+1-u)}{6^u} \; = \; \sum_{u=1}^6 \frac{N+1-u}{6^u}{5 \choose u-1} \\ & = \sum_{u=0}^5 \frac{N-u}{6^{u+1}}{5 \choose u} \; = \; \tfrac16N\left(1 + \tfrac16\right)^5 - \tfrac{5}{36}\left(1 + \tfrac16\right)^4 \\ & = \frac{2401(7N-5)}{46656} \end{aligned}$ In particular, when $N=20$ we have $\mathbb{E}[C] = \frac{12005}{1728}$ , making the answer $12005 + 1728 = \boxed{13733}$ .

We can find the expected value of C by making a list of all possible subsequences that add to 6 and their expected values, then adding up all of the expected values. $\begin{array}{l} \\ 111111 && 11112 && 1113 && 1122 && 114 && 123 && 222 && 15 && 24 && 33 && 6 \\ \hline 15(\frac 16) ^6 && 16(\frac 16) ^5 \cdot 5 && 17(\frac 16) ^4 \cdot 4 && 17(\frac 16) ^4 \cdot 6 && 18(\frac 16) ^3 \cdot 3 && 18(\frac 16) ^3 \cdot 6 && 18(\frac 16) ^3 && 19(\frac 16) ^2 \cdot 2 && 19(\frac 16) ^2 \cdot 2 && 19(\frac 16) ^2 && 20(\frac 16) \end{array}$ For the expected value part of this table, we have the number of places the subsequence could start within the 20 length sequence times the probability of rolling exactly that sequence times the number of ways to rearrange the sequence. For example, the 11112 subsequence could start in 16 different places, the probability of rolling all five dice perfectly is $(\frac 16) ^5$ and it could be rearranged 5 different ways. If we add up all of the expected values, we get $\boxed{\dfrac{12005}{1728}}$