Anna stands at $A(0,0)$ and Eve at $B(5, 3)$. Every minut Anna move $1$ right (with probability $p$) or $1$ up and Eve move $1$ left (also with probabilty $p$) or $1$ down. What is the maximum probability they will meet, as $p$ varies from $0$ to $1$? Suppose they move independently.
Solution: Let $q=1-p$. They can meet only after 4 steps in $C(1,3)$, $D(2,2)$, $E(3,1)$ or $F(4,0)$. Probability that they meet at $T(k,4-k)$ where $k\in\{1,2,3,4\}$ is \begin{eqnarray*}P(Anna = T \cap Eve = T)&=& P(Anna = T)\cdot P(Eve = T)\\ & =& {4\choose k}p^kq^{4-k} \cdot {4\choose 5-k}p^{5-k}q^{k}\\ & =& {4\choose k}{4\choose 5-k}p^5q^3 \end{eqnarray*} So the probability they meet is $$P=p^5q^3\Big( {4\choose 1}{4\choose 4}+ {4\choose 2}{4\choose 3}+ {4\choose 3}{4\choose 2}+ {4\choose 4}{4\choose 1}\Big)$$ $$=56p^5q^3\leq 56{5^5\cdot 3^3\over 8^8}\approx 0,2816$$
This may depend on the interpretation of the wording of the question.
It is not clear to me that $p$ and $q$ are relevant to the answer if each individual is constrained to taking $5$ horizontal steps and $3$ vertical steps in a random order; there are ${8 \choose 5}=56$ such paths.
(Changed from my earlier answer of $\frac{56}{225}$). For Anna, ${4 \choose 3}{4 \choose 0} = 4$ of these paths pass through C $(1,3)$; ${4 \choose 2}{4 \choose 1} = 24$ pass through D $(2,2)$; ${4 \choose 1}{4 \choose 2} = 24$ pass through E $(3,1)$; and ${4 \choose 0}{4 \choose 3} = 4$ pass through E $(4,0)$. The calculations are similar but reversed for Eve.
I could then say the probability of them meeting after each has taken $4$ steps is $$\dfrac{{4 \choose 3}{4 \choose 0}{4 \choose 0}{4 \choose 3} +{4 \choose 2}{4 \choose 1}{4 \choose 1}{4 \choose 2}+{4 \choose 1}{4 \choose 2}{4 \choose 2}{4 \choose 1}+{4 \choose 0}{4 \choose 3}{4 \choose 3}{4 \choose 0}}{{8 \choose 5}^2} \\ = \frac{4^2+24^2+24^2+4^2}{56^2} = \frac{37}{98} \approx 0.37755$$