= Solution
A <probability space> consists of a sample space $\Omega$, a <sigma-algebra> $\mathcal F$ of events, and a <probability measure> $\mathbb P$ satisfying:
* $\mathbb P(A)\geq0$ for every $A\in\mathcal F$.
* $\mathbb P(\Omega)=1$.
* For pairwise disjoint events, $\mathbb P(\bigcup_{j\ge1}A_j)=\sum_{j\ge1}\mathbb P(A_j)$.
For arbitrary events, \b[<Boole's inequality> is $\mathbb P(\bigcup_jA_j)\leq\sum_j\mathbb P(A_j)$]. To prove it, replace each $A_j$ by $D_j=A_j\setminus\bigcup_{i<j}A_i$. The $D_j$ are disjoint, have the same union, and satisfy $D_j\subseteq A_j$. Countable additivity and monotonicity give $\mathbb P(\bigcup_jA_j)=\sum_j\mathbb P(D_j)\leq\sum_j\mathbb P(A_j)$. Finite unions are included by taking subsequent events empty.
Let $H_i$ be the $i$th heads event. For every $m$, the event of infinitely many heads lies inside $\bigcup_{i\ge m}H_i$. By the <union bound>, its probability is at most $\sum_{i\ge m}p_i$, which tends to zero. Thus \b[the probability of infinitely many heads is $0$]. This is the <First Borel-Cantelli lemma>, and requires no <independence> of the coin tosses.
For the dice experiment there are $4$ ordered outcomes giving a sum of $5$ and $6$ giving $7$, out of $36$. Each trial therefore has $p_5=1/9$, $p_7=1/6$ and irrelevant-outcome probability $13/18$. By <independence> between trials, summing over the number of initial irrelevant outcomes gives
$$
\boxed{\mathbb P(5\text{ before }7)=\sum_{k\ge0}\left(\frac{13}{18}\right)^k\frac19=\frac{p_5}{p_5+p_7}=\frac25.}
$$
The geometric series also shows that one of the two relevant sums eventually appears almost surely. Equivalently, <first-step analysis> gives $p=p_5+(13/18)p$.
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