Tossing a coin $n$ times, probability for a palindrome?

Question

Tossing a coin $n$ times, calculate the probability for a palindrome result?

So I said

We have $2 \choose 1$ options for throws $1,n$, $2 \choose 1$ options for throws $2, n-1$.... thus: $P(A) = \frac{ \frac{n}{2} \cdot {2 \choose 1} }{2^n}$

Is that correct?

Answer 1

if n is even the probability is $(\frac{1}{2})^{n/2}$.

if n is odd the probability is $(\frac{1}{2})^{(n-1)/2}$.

I think the above is the correct answer as each flip in the first half sets up a requirement for the second half. I don't think there is anything I am missing to make the problem more complicated.