The 6-sided dice case can be encoded as a Markov chain on the state space $\{0,1,2,3,\partial\}$ where each state $0\leqslant i\leqslant3$ means that $i$ pairs from the set $\{\{1,6\},\{2,5\},\{3,4\}\}$ have already been visited but none is completed yet, and state $\partial$ means that at least one of these pairs has been completed. For example, after the results 26622 the state of the chain is $2$, after the results 266224246 the state of the chain is $3$, and after the results 2662242465 the state of the chain is $\partial$.
The probability that any two dice add to 7 on a throw of $n$ is $P_0[T\leqslant n]$ where $T$ is the hitting time of $\partial$ and the subscript $0$ refers to the starting point of the chain.
To characterize $P_0[T\leqslant n]$ for every $n$, one usually computes the generating functions $u_i=E_i[s^T]$ for $0\leqslant i\leqslant3$, where $|s|\leqslant1$. Writing down carefully the transition probabilities of the chain, one sees that the Markov property after one step yields the relations
$$
u_0=su_1,\quad u_1=s(\tfrac16u_1+\tfrac46u_2+\tfrac16),\quad u_2=s(\tfrac26u_2+\tfrac26u_3+\tfrac26),\quad u_3=s(\tfrac36u_3+\tfrac36).
$$
Solving this Cramér system, one gets (something similar to)
$$
u_0=\frac{s^2(6+3s+s^2)}{(2-s)(3-s)(6-s)}.
$$
Thus, $u_0$ is a rational fraction with respect to $s$, which can be decomposed as
$$
u_0=\frac{c_2}{2-s}+\frac{c_3}{3-s}+\frac{c_6}{6-s}+as+b,
$$
for some suitable constants $a$, $b$, $c_2$, $c_3$ and $c_6$. Expanding each fraction as a power series in $s$ and collecting all the terms of degree at least $n$ yields, for every $n\geqslant2$,
$$
P[T\geqslant n]=\sum_k\frac{c_k}{k-1}\frac1{k^n},
$$
where the sum runs over $k$ in $\{2,3,6\}$.
If one uses some $2z$-sided dice instead, one finds similarly
$$
P[T\geqslant n]=\sum_k\frac{c_k}{k-1}\frac1{k^n},
$$
for some suitable constants $c_k$, where the sum runs over $k=2z/\ell$ with $1\leqslant\ell\leqslant z$. In particular, when $n\to\infty$,
$$
P[T\geqslant n]\sim\frac{c_2}{2^n}.
$$
Going back to the 6-sided case, $c_2=\lim\limits_{s\to2}(2-s)u_0=16$ hence
$$
P[T\geqslant n]\sim\frac{16}{2^{n}}.
$$
Well, if you have all the individual probabilities figured out, your answer would be:
$$
P(winning)=P(A>36\mid B=36)+P(A>35\mid B=35)+\cdots+P(A>4\mid B=4)
$$
Since your events are independent, we have
$$
P(winning)=P(A>36)P(B=36)+P(A>35)P(B=35)+\cdots+P(A>4)P(B=4)
$$
Then $P(A>n)=\sum_{k=n+1}^{36} P(A=k)$ so
$$
P(winning)=0\cdot P(B=36)+P(A=36)P(B=35)+\cdots+[P(A=36)+\cdots+P(A=4)]P(B=4)\\
$$
Finally,
$$
P(winning)=\sum_{n=4}^{36}\left(P(B=n)\sum_{k=n+1}^{36} P(A=k)\right).
$$
Best Answer
Probability (in this case) is the ratio of desired outcomes to all outcomes. Let's count desired outcomes in your particular case (and by the way we derive a general formula to compute it).
Suppose you roll 4 dice and you have a total success - on every die you get "7 to 10". How many quadruplets of numbers "1 to 10" may cause it? It is not so difficult: $4 \cdot 4 \cdot 4 \cdot 4$ or $4 ^ 4$.
Then you roll the 4 remaining dice - but now you need a total failure - numbers 1 to 6 on every die. How many quadruplets is able to cause it? The answer is similar: $6 ^ 4$.
So for the first group of four dice to be totally successful and in the same time the second group of four dice to be totally unsuccessful - e. g. $(8, 8, 8, 8, 1, 1, 1, 1)$ - there is $4^4 \cdot 6^4$ possibilities.
But where is written that the first four dice have be successful (and in the same time the four others not)? We may select four successful dices by $_8 \mathop{C}_4 = 70$ ways. So all successful possibilities count $70 \cdot 4^4 \cdot 6^4$ which gives the result $23{.}224{.}320$.
So the general formula for the number of desired outcomes is
$$_n {\mathop{C}}_k \cdot p^k \cdot q^{n-k}$$
Note: Please don't forget to divide it by the number of all possibilities, i. e. by $(p + q)^n$ to compute the probability.