A fair coin is tossed multiple times.
Every time Heads comes up, Team A wins a random score ranging from 1 to 101
Every time Tails comes up, Team B wins a random score ranging from 1 to 100
The game/series of tosses ends 'as soon as' and 'only' when Team A has more score than Team B. If Team B has a higher score then the coin is tossed again and again till the time team A has a higher score.
Mathematically, can we prove that team A has 100% probability of winning all games/series of tosses eventually?
Intuitively yes? But what if someone argues that there exists a sample set where Team B keeps on getting more scores than A on an average basis randomly over an infinitely long sample series of coin tosses.
Yes, team $A$ has $100\%$ probability of winning. Let $S_n$ be team $A$'s score minus team $B$'s score after the $n$-th round, we will show $\mathbb{P}(S_n \le 0 \text{ for all }n) = 0$. If we let $X_n := S_{n}-S_{n-1}$ be the amount the score difference changes in one round, then $\mathbb{E}[X_n] = \frac 12 (51 - 50.5) = .25$ so by the law of large numbers $\lim_{n \rightarrow \infty} \frac 1n \sum_{k=1}^n X_k = .25$ almost surely. But since $S_n = \sum_{k=1}^n X_k$, this implies that $S_n > 0$ for $n$ sufficiently large. More precisely, given any $\omega \in \left\{\omega : \lim_{n \rightarrow \infty} \frac 1n \sum_{k=1}^n X_k(\omega) = .25 \right\}$ we have that there exists $N = N(\omega)$ such that $\frac 1N \sum_{k=1}^N X_k(\omega) > .2$ so $S_N(\omega) = \sum_{k=1}^N X_k(\omega) > .2 \cdot N > 0$ and therefore team $A$ has won by the $N$-th round.