To solve this problem, we first identify the relative strengths of the horses based on the given information. We have three horses from each side, King Qi and Tian Ji, with different tiers: top, middle, and bottom. Let's denote King Qi's horses as $A$ (top-tier), $B$ (middle-tier), and $C$ (bottom-tier), and Tian Ji's horses as $a$ (top-tier), $b$ (middle-tier), and $c$ (bottom-tier).

Given the conditions:
- Tian Ji's top-tier horse ($a$) is better than King Qi's middle-tier horse ($B$), worse than King Qi's top-tier horse ($A$).
- Tian Ji's middle-tier horse ($b$) is better than King Qi's bottom-tier horse ($C$), worse than King Qi's middle-tier horse ($B$).
- Tian Ji's bottom-tier horse ($c$) is worse than King Qi's bottom-tier horse ($C$).

When a horse is randomly selected from each side for a race, we have the following possible matchups: $Aa$, $Ab$, $Ac$, $Ba$, $Bb$, $Bc$, $Ca$, $Cb$, $Cc$. This gives us a total of $9$ combinations.

Next, we determine which matchups result in Tian Ji's horse winning:
- Tian Ji's top-tier horse ($a$) wins against King Qi's middle-tier ($B$) and bottom-tier ($C$) horses, so $Ba$ and $Ca$ are winning combinations.
- Tian Ji's middle-tier horse ($b$) wins against King Qi's bottom-tier horse ($C$), so $Cb$ is a winning combination.

Thus, Tian Ji's horses win in $3$ out of the $9$ possible matchups: $Ba$, $Ca$, $Cb$.

Therefore, the probability that Tian Ji's horse wins is calculated as the number of winning combinations divided by the total number of combinations:
$$\frac{3}{9} = \frac{1}{3}$$

So, the correct answer is $\boxed{\text{A: }\frac{1}{3}}$.