I know the recursion formula. It is $p\_n = p\_{n-1}q + p\_{n-2}pq + p\_{n-3}p^2q + p^3$, but I am having trouble finding $p\_6$.

I was thinking it would be 
$p\_3 = p^3$ since it would have to be $H$ on each roll which has a probability of $p$, and 
$p\_4 = p\_3 + P($1st roll is $T$ and $2, 3, 4$ are $H)$, but that is where I cannot figure out what is next.