 Solution

A probability space consists of a sample space $\Omega$, a sigma-algebra $\mathcal{F}$ of events, and a probability measure $\mathbb{P}$ satisfying:

For arbitrary events, Boole's inequality is $\mathbb{P}({\bigcup }_j{A}_j)\le {\sum }_j\mathbb{P}(A_j)$. 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 $\mathbb{P}({\bigcup }_j{A}_j)={\sum }_j\mathbb{P}(D_j)\le {\sum }_j\mathbb{P}(A_j)$. 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 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 givesThe geometric series also shows that one of the two relevant sums eventually appears almost surely. Equivalently, first-step analysis gives $p=p_5+(13/18)p$.