Who gets the first ace

Idea. The answer is not $\frac{1}{2}$. The first ace is more likely to come early than late, and A always receives the earlier card of each pair, so A has a small edge.

Step 1: the first ace at position $k$. The positions of the four aces are a uniform random set of 4 of the 52 positions. The first ace is at position $k$ when one ace is at $k$ and the other three are among the $52-k$ positions after it: $$\mathbb{P}(\text{first ace at }k)=\frac{\left(\genfrac{}{}{0ex}{}{52-k}{3}\right)}{\left(\genfrac{}{}{0ex}{}{52}{4}\right)}.$$

Step 2: why A has the edge. These probabilities decrease with $k$. A gets positions $1,3,5,\dots$ and B gets $2,4,6,\dots$, so pairing position $2j-1$ with $2j$, A's position always has the larger probability. So A's total is more than $\frac{1}{2}$.

Step 3: the exact value. A wins when $k$ is odd. Writing $m=52-k$, which runs over the odd numbers from 3 to 51, $$\begin{array}{rl}\mathbb{P}(A)& =\frac{{\sum }_{m\text{ odd}}\left(\genfrac{}{}{0ex}{}{m}{3}\right)}{\left(\genfrac{}{}{0ex}{}{52}{4}\right)}=\frac{140,725}{270,725}\\ & =\frac{433}{833}\approx \mathrm{0.520.}\end{array}$$ (The 25 terms of the sum can be added by hand; their total, together with that of the even terms, is $\left(\genfrac{}{}{0ex}{}{52}{4}\right)$.)

The trap. “By symmetry, $\frac{1}{2}$.” The situation is not symmetric: A is always dealt first in each round.