Hitting probabilities for conditioned oriented random walk monotonic? 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!
 A: I have a brute force calculation that establishes this. Your formula for $q_i$ is not completely correct; it should read $q_i=\binom{2m}{m+i}/A_i$ with
$$
A_i = 4^m - \binom{2m}{m} - 2\sum_{j=1}^{i-1} \binom{2m}{m+j} .
$$
To see that $q_1\ge p^*$, let's just write
$$
\binom{2m}{m+1}=\binom{2m}{m}\frac{m}{m+1} ,
$$
so after rearranging the claim becomes
$$
\binom{2m}{m}4^{-m} \ge \frac{1}{m+1} ,
$$
and this is true (induction on $m$ if all else fails).
Similarly, I can use that $\binom{2m}{m+i+1}=\binom{2m}{m+i}\frac{m-i}{m+i+1}$ to rewrite the other claim $q_{i+1}\ge q_i$ as
$$
\frac{m-i}{m+i+1}A_i \ge A_i-2\binom{2m}{m+i} .
$$
This I can rearrange some more to obtain
$$
A_i \le \frac{2(m+i+1)}{2i+1}\binom{2m}{m+i} . \quad\quad\quad\quad (1)
$$
Now $2\binom{2m}{m+i}$ is the new summand that gets added on when going from $A_i$ to $A_{i+1}$, so an equivalent form of (1) is
$$
A_{i+1}\le \frac{2(m-i)}{2i+1}\binom{2m}{m+i} = \frac{2(m+i+1)}{2i+1}\binom{2m}{m+i+1}  .\quad\quad\quad\quad (2)
$$
We can now establish (2) by induction on $i$, starting at the maximum value $i=m$. Since $A_{m+1}=0$, (2) certainly holds for $i=m$. Now if I assume (2) for $i+1$ taking the role of $i$ (and with $i\le m-1$), then I have (1) for $i+1$, so to obtain (2) for $i$, it suffices to check that
$$
\frac{2(m+(i+1)+1)}{2(i+1)+1}\le  \frac{2(m+i+1)}{2i+1} .
$$
This is straightforward.
