Since other rolls do not matter, I thought I could consider there are only $7$ possible outcomes for each roll, 6 for 7 (1/6, 2/5, 3/4, 4/3, 5/2, 6/1) and 1 for 12 (6/6). So the odds of getting 3 7s before a 12 would simply be $(6/7)^3$ or $62.97\%$.
Is this correct?