You repeatedly roll two dice and take their sum (like in Craps) until one of the following happens:
A) You roll five 7's
B) You roll each of these at least once: 4,5,6,8,9,10

What's the probability that B happens before A?

You should get an answer of

Let's modify it a bit so that it's less busywork: what's the probability of [at least one 5 + at least one 9] before three 7's?

You should get an answer of

1- [(6/14 x 6/14 x 6/14) + (6/14 x 6/14 x 8/14 x 6/10) + (6/14 x 8/14 x 6/10 x 6/10) + (8/14) x 6/10 x 6/10 x 6/10)] = .353

In English, three sevens win if they all hit before a five or nine or if its success, success, failure, success, or success, failure, success, success, or failure, success, success, success.

Hopefully no one needs to be explained why those words are reflected by the fractions. I didn't even reduce them to make the correlation more obvious.

Yes to your math, but you forgot to actually subtract from 1.

There's a shorter solution, and there's also an interesting one (borderline sorcery) that Bruce showed years ago to a similar problem. For now here's the latter: https://mathb.in/77766

That's the integral of P(<3 sevens in time t)*P(>0 fives)*P(0 nines)*P(nine on next roll), the product of which is the probability of the game ending with a nine in time t. Multiply by 2 because the game can equally likely end with a five. Evaluate to infinity because the game can potentially go on forever if we keep rolling irrelevant sums. We're able to multiply the probabilities like that because getting <3 sevens in time t, getting >0 fives in time t, etc are independent events.

How are they independent, why are those the probabilities, and why are we integrating over continuous time when talking about discrete dice rolls?

If we only had time for N rolls, rolling a seven on the next roll would remove an opportunity to roll a five, so rolling 3 sevens wouldn't be an independent event from rolling >0 fives. However, there is no limit on the number of rolls, and rolling a seven next roll doesn't interfere with rolling a five in eternity.

Furthermore, it makes no difference when in time the dice are rolled. This means we're free to imagine that the rolls occur at times determined by a Poisson point process with rate 1, making for an average of one roll per unit of time. In that case, the number of occurrences of a given sum within time t is Poisson distributed with a rate matching its roll probability. The events within a given time are independent because there aren't collisions in continuous time—if a number gets rolled at a given moment, another number can still get rolled at an infinitesimally later moment.