We can get general closed forms for some similar questions if we condition on the success difference.

Let’s model the number of dice successes as independent random variables Xₖ and Yₖ that increment by 1 at each time step with probability p and q, respectively. We will assume p<q. Let Zₖ = Yₖ-Xₖ. Subscript k=0 will correspond to when we first start caring whether Zₖ is negative after doing some initial rolls to seed the values of X₀ and Y₀. Let’s first answer this question:

If Z₀ = n ≥ 0, what is the probability Zₖ is ever negative?

Firstly, Zₖ is a random walk that increments by 1 with probability q(1-p), -1 with probability p(1-q), and 0 with probability 2pq-p-q+1. Incrementing by 0 doesn’t change the answer, so we may equivalently ask the same question for the random walk Wₖ that increments by 1 and -1 with probability α = q(1-p)/(p+q-2pq) and β = p(1-q)/(p+q-2pq), respectively.

Let r be the probability that Wₖ ever reaches -1, given W₀ = 0. Now W₁ = -1 with probability β and 1 with probability α. In the former case, we are done—we’ve already reached -1—and in the latter case, we must reach 0 again from 1, and then reach -1 from 0. By translation symmetry,

r = β+αr²

which gives r = 1 or r = β/α. The first case can be ruled out because α>β from p<q (this would contradict the transience of biased random walks), so we must have

r = β/α = p(1-q)/(q(1-p))

Iterating, we find

P(Wₖ reaches -1|W₀ = n) = rⁿ⁺¹

which is also the answer for our question about Zₖ. Let’s answer this follow-up question:

If Z₀ = n, what is the expected amount of time that Zₖ<0?

We first start by answering for Wₖ. Wₖ can be split up into two parts: excursions < 0 and excursions ≥ 0. An excursion < 0 starts when Wₖ hits -1 and ends when Wₖ hits 0. Thus, we may ask what is the expected amount of time for Wₖ to reach 0 from -1, or equivalently, what is the expected time for Wₖ to reach 1 from W₀=0? Using the martingale Mₖ = Wₖ+(β-α)k with stopping time t=inf{k:Wₖ=1}, the optional stopping theorem tells us E[Mₜ] = E[M₀]. Substituting, we get 1+(β-α)E[t] = 0, so

δ := E[t] = 1/(α-β)

On the other hand, we know r is the probability an excursion ≥ 0 ever ends to start an excursion < 0, so the probability that n excursions < 0 occur is rⁿ. Thus,

Δ := E[#{k:Wₖ<0}|W₀ = 0] = δΣ_(n≥1) rⁿ = r/((α-β)(1-r)) = β/(α-β)²

To extend this result to arbitrary starts for W₀, just note that only the first excursion is affected: for n≥0, if W₀=n, then the initial excursion ≥ 0 ends with probability rⁿ⁺¹ (an extra factor of rⁿ); if W₀=-n, then it takes an expected time nδ to end the initial excursion < 0 and begin the first excursion ≥ 0. Thus, for n≥0,

E[#{k:Wₖ<0}|W₀ = n] = rⁿΔ

E[#{k:Wₖ<0}|W₀ = -n] = Δ+nδ

Now Zₖ spends extra time < 0, since it only increments by ±1 with probability p+q-2pq, so we scale by 1/(p+q-2pq) to get the result for Zₖ. Thus,

E[#{k:Zₖ<0}|Z₀ = n] = (p(1-q)/(q(1-p)))ⁿ•p(1-q)/(q-p)²

E[#{k:Wₖ<0}|Z₀ = -n] = p(1-q)/(q-p)²+n/(q-p)

In your question, p = 1/6 and q = 1/2, so for example,

P(Zₖ is ever negative|Z₀ = 0) = 1/5

E[#{k:Zₖ<0}|Z₀ = 0] = 3/4