Thanks Ginger,
NOW our answers almost match.

So there are 2304 possible outcomes (for 2 consecutive throws) but we only care about the ones inside this Venn diagram
So there are 36 ways to get six heads
and 64 ways to get two 6s.
and one way to get 2 sixes AND six heads.
So there is 35 ways out of 99 relevant outcomes where 6 heads is thrown BEFORE 2 sixes.
P(HHH,HHH before 6,6) = \(\frac{35}{99} = 0.\bar{35}\)
The logic I used last time was more complicated but similar. I think I just made careless errors.