Let us consider only odd number of repetitions and $p < q$ where $p$ is the probability of a bit flip and $q=1-p$. Let $E_n$ be the event of an error when repeating a bit $2n+1$ times. Let us evaluate $P(E_{n+1})$. Here, error happens if there are more than $(n+1)$ flips. To find $P(E_{n+1}$), consider what happens in the first $2n+1$ transmissions out of the total $(2n+3)$ transmissions. There are 4 possibilities for the number of flips in the first $(2n+1)$ transmissions: (a) $\leq n-1$ flips, (b) exactly $n$ flips, (c) exactly $n+1$ flips, (d) $\geq n+2$ flips. The probability of event $E_{n+1}$ in these 4 cases are $0, p^2, 1-q^2$ and $1$ respectively.
For convenience, define $\alpha_n = {2n+1 \choose n} (pq)^n$. Then
$P(E_{n+1}) = p^2 \alpha_n q + (1-q^2) \alpha_n p + \sum_{k=n+2}^{2n+1} {2n+1 \choose k} p^k q^{n-k}$
The summation is the expression for $P(E_n)$ with the first term missing, i.e.
$\sum_{k=n+2}^{2n+1} {2n+1 \choose k} p^k q^{n-k} = P(E_n) - \alpha_n p$. Putting these together gives us
$P(E_{n+1}) - P(E_n) = \alpha_n pq (p-q) = {2n+1 \choose n} (pq)^{n+1}(p-q) < 0$.
Edit: Other answers beat me to it but I will leave this up anyway.