Let denote the number of such sequences of H and T only where we don’t have any two consecutive H.
Let there be cases which end with H and cases with end with T.

With a case that ends with a T, we can make a case for by putting a T or a H after the T.
With a case that ends with a H, we can make a case for by putting a T after the H.
So we have cases with T in the end and cases with H in the end.
Hence

Arguing similarly, we see that

And so,



Hence forms the Fibonacci sequence.
Where and so on.
you need to show the bijection discussed above is correct though.
the probability is simply the number of favourable cases divided by .