\begin{equation*} \binom{n}{2t-1}\cdot\sum_{k=0}^{t-1}\binom{2k}{k} \binom{2(t-1-k)}{t-1-k} =\binom{n}{2t-1}\cdot 4^{t-1}, \end{equation*} \begin{align*} \EE(d_{te}(a,b)) &= (4/\sigma)^{t-1} \cdot r+ O(2t\sigma/n \cdot \sigma^{-t} r) + \sum_{q=2}^t O((2t\sigma/n)^{q-1} \cdot \sigma^{1-t} r)\\ &= (4/\sigma)^{t-1} \cdot r + O(2t\sigma^{2-t}/n) \cdot r. \end{align*} As in the proof of the expected value, the total number of tuples \((I_1, J_1, \dots, I_m, J_m)\) is \(\binom{n}{m(2t-1)} f(t)\) for some function \(f\) independent of \(n\), and the contribution of each tuple will also be independent of \(n\).