Tied chess matches and the monotonicity of $\sum_{k=0}^n \binom{2n}{k,k,2n-2k} (pq)^k (1-p-q)^{2n-2k}$

For convenience, define out-of-range multinomial coefficients as $0$. Let

\begin{gather*} r = 1 - p - q, \quad s = p + q, \\ a_k = 1 - \frac{1}{2k + 1} + \frac{1}{2n - 2k + 1}, \\ b_k = 1 + \frac{1}{2k + 1} - \frac{1}{2n - 2k + 1}, \\ c_k = r^2\frac{f(p, q, n; k, k)}{f(p, q, n + 1; k, k)} = \frac{(2n - 2k + 2)(2n - 2k + 1)}{(2n + 2)(2n + 1)}, \\ d_k = 4pq\frac{f(p, q, n; k - 1, k - 1)}{f(p, q, n + 1; k, k)} = \frac{4k^2}{(2n + 2)(2n + 1)}. \end{gather*}

Since $r^2 + a_krs + b_krs + s^2 = r^2 + 2rs + s^2 = 1$, we have

\begin{split} t(p, q, n) &= \sum_{k=0}^n f(p, q, n; k, k) \\ &= \sum_{k=0}^{n} (r^2 + a_krs) f(p, q, n; k, k) + \sum_{k=0}^{n} (b_krs + s^2) f(p, q, n; k, k) \\ &= \sum_{k=0}^{n + 1} ((r^2 + a_krs) f(p, q, n; k, k) + (b_{k-1}rs + s^2) f(p, q, n; k - 1, k - 1)) \\ &= \sum_{k=0}^{n + 1} \left((r^2 + a_krs)\frac{c_k}{r^2} + (b_{k-1}rs + s^2)\frac{d_k}{4pq}\right)f(p, q, n + 1; k, k) \\ &≥ \sum_{k=0}^{n + 1} \left((r^2 + a_krs)\frac{c_k}{r^2} + (b_{k-1}rs + s^2)\frac{d_k}{s^2}\right)f(p, q, n + 1; k, k) \\ &= \sum_{k=0}^{n + 1} \left(c_k + a_kc_k\frac sr + b_{k-1}d_k\frac rs + d_k\right)f(p, q, n + 1; k, k) \\ &≥ \sum_{k=0}^{n + 1} (c_k + 2\sqrt{a_kb_{k-1}c_kd_k} + d_k)f(p, q, n + 1; k, k) \\ \end{split}

We claim $c_k + 2\sqrt{a_kb_{k-1}c_kd_k} + d_k ≥ 1$; this is trivial if $k = 0$ ($c_k = 1, d_k = 0$) or $k = n + 1$ ($c_k = 0, d_k > 1$), and otherwise, it follows from

\begin{multline*} 4a_kb_{k-1}c_kd_k - (1 - c_k - d_k)^2 \\ = \frac{k^2(2n + 2(n - k) + 5)(4(2n + k + 1)(n - k) + 10n + 7)}{(2n + 1)^2(n + 1)^2(2(n - k) + 3)(2k + 1)(2k - 1)} ≥ 0, \end{multline*}

since every factor on the right is nonnegative. Therefore,

\begin{split} t(p, q, n) &≥ \sum_{k=0}^{n + 1} f(p, q, n + 1; k, k) = t(p, q, n + 1). \end{split}

Not an answer but a reformulation.

The problem can be stated as a random walk $X_n$ , starting at $X_0=0$ and with $X_n = X_{n-1} + Z_n$ where $Z_n$ is iid taking values over $\{-1,0,1\}$ with given probabilities $p_{-1}=q>0$, $p_{0}=1-(p+q)$, $p_1 = p>0$.

We want to prove the conjecture $P(X_{2(n+1)} = 0) < P(X_{2n} = 0)$

This seems particularly non-obvious for small $n$. For large $n$, one expects (by the CLT) that

$$P(X_{2n}=0) \approx \frac{1}{\sqrt{4\pi n\sigma^2}} \exp \left( - \frac{2n\mu^2}{\sigma^2}\right)$$

where $\mu = p-q$ and $\sigma^2=p+q-(p-q)^2$, provided that $p+q<1$. Elsewhere, for $p+q=1$ it should be multiplied by $2$.

Granted, this says little about the conjecture in itself.