Probability of gaining a game series

Let’s say I have a team, and we are playing a game against another team, where our probability of winning each game is $p$. We get the offer to play either $2k-1$ games or $2k+1$ games, where $k$ is fixed. For which $p$ is better one or another?

Intuitively, I’m sure that for $p > 1/2 $ it’s better to play $2k+1$ games while for $p<1/2$ is better to play $2k-1$. I’ve tried to write this idea. If I’m not mistaken, this is like a $BN$ distribution. What I’ve done is the following:

$X text{~} BN(k, p)$, $Y text{~} BN(k+1, p)$, so $F_X(2k-1)$ is my probability of winning in the case that $2k-1$ games are played, while $F_Y(2k+1)$ is my probability of winning in the case that $2k+1$ games are played. So I tried reducing each one and operating them but couldn’t find a way to clear $p$.

What I’ve found is that (I’m ommiting the process, but if someone wants it, ask for it) $F_X(2k-1)=binom{2k-2}{k-1}p^{k-1}(1-(1-p)^k)$ and $F_Y(2k+1)=binom{2k}{k}p^{k}(1-(1-p)^{k+1})$. Since there are combinatorial numbers, I tried to divide them to compare with $1$ and got:

$$frac{F_Y(2k+1)}{F_X(2k-1)} = frac{2p(2k-1)(1-(1-p)^{k+1})}{k(1-(1-p)^{k})}$$

But after comparing to $1$ I couldn’t go on