Let us consider the positions $1,2, \ldots, 52$, from left to right, of the fifty-two cards at the beginning of the game. Whenever Aws removes two adjacent cards, we subtract $2$ from the positions of all the cards on the right of these adjacent removed cards.

We notice that, at each operation, if a card is not removed, even if its position changes, it will keep the same parity. We notice also that whenever two adjacent cards of the same color are removed, the positions of these two cards have different parity.

Define $R$ to be the number of red cards with even positions at the beginning subtracted from the number of red cards with odd positions at the beginning. Clearly, $R$ is an invariant under Aws' operations. Hence, if at the beginning, $R \neq 0$, Aws can never remove all his cards.

Now assume that $R=0$ and that Aws has started playing. If there are no more red cards, then Aws is left with only black cards, so he can keep removing adjacent cards until he ends up with no card. If there is at least one red card, because $R=0$, there is at least one red card with even position $e_{1}$ and one red card with odd position $o_{1}$. Assume, without loss of generality, that $e_{1}<o_{1}$.

Consider $o=\min \mathcal{O}$, where $\mathcal{O}$ is the set of odd positions $o'$ of red cards with $o'>e_{1}$. Number $o$ exists since $o_{1} \in \mathcal{O}$. Consider $e=\max \mathcal{E}$, where $\mathcal{E}$ is the set of even positions $e'$ of red cards with $o>e'$. Number $e$ exists since $e_{1} \in \mathcal{E}$. If $o=e+1$, then there are two adjacent red cards that Aws can remove. If $o>e+1$ then $o \geq e+3$ and the positions between $e$ and $o$ are all occupied by adjacent black cards. In this case, Aws can remove adjacent black cards. This proves that if $R=0$, Aws can keep removing cards until he removes all cards.

Hence, Aws can win if and only if $R=0$. In this situation, there must be $13$ red cards with even positions and $13$ red cards with odd positions. There are precisely $\binom{26}{13}^{2} \cdot 26!^{2}$ possible such starting positions and the probability for Aws to win is

$$\frac{\binom{26}{13}^{2}}{\binom{52}{26}}.$$