Consider an oriented random walk on $\mathbb Z^2$ (i.e. only steps $\rightarrow$ and $\uparrow$ with equal probability.) Say we let the walk go $2m$ steps then start guessing sites at distance $2m$ from the origin until we pick up its trail. The best first guess is at $(m,m)$ which has probability $p^* = \binom{2m}{m} 4^{-m}$ of being included in the walker's trail. Say the walker was not at $(m,m)$, the next natural guess is $(m+1,m-1)$. Let $q_1$ be the probability the walker crossed $(m+1,m-1)$ given it didn't cross at $(m,m)$. I am looking for a proof that $$q_1 \geq p^*\text{.}$$ In general, take $A_i$ to be the event the walk crossed at $(m+i,m-i)$ and $B_{i-1}$ that the walk did not cross at $\{(m\pm j,m \mp j) \colon j=0,1,\cdots,i-1\}$. Define $q_i = \mathbf P[A_i \mid B_{i-1}]$. Pretty sure that $q_i$ can be written as $$q_i = \frac{\mathbf P[A_i \cap B_{i-1} ]}{\mathbf P[ B_{i-1}]}=\frac{ \binom{2m}{m+i} } { 4^m - \binom{2m}{m} - 2\sum_{j=1}^{i-1} \binom{2m}{m+j} }.$$ I'd like to show $q_{i+1} \geq q_i.$ Either via being good with binomial coefficients or with a probabilistic proof about conditioned random walk.

I'm also interested in the analogue for oriented walk on $\mathbb Z^d$. Thanks!