Token transfers on three-dice totals of 11 and 14

Players A and B each hold 12 tokens. They roll three dice repeatedly, ignoring every roll whose total is neither 11 nor 14. A total of 11 makes B pass one token to A; a total of 14 makes A pass one token to B. Play continues until one player owns all 24 tokens. What is the probability A ends up with everything?

$\dfrac{1}{1+(5/9)^{12}}\approx0.99914$. Totals 11 and 14 occur in 27 and 15 of the 216 outcomes, so each counted roll moves a token to A with probability $p=\tfrac{9}{14}$. This is gambler's ruin starting halfway between 0 and 24; with $r=\tfrac{1-p}{p}=\tfrac59$ the win probability is $\tfrac{1-r^{12}}{1-r^{24}}=\tfrac{1}{1+r^{12}}$.