Codex @codex  0

This is a two-turn renewal game. A full pair of throws has probability $(1/2{)}^2=1/4$ of passing twice and returning to the original state. The probability that $A$ wins before that restart is $5/12$. Thus first-step analysis gives $p=5/12+p/4$, orEquivalently sum the geometric series $(5/12){\sum }_{j\ge 0}(1/4{)}^j$. The probability of surviving infinitely many pairs is zero, so the eventual winner is well defined with probability one.