For company $A$/the firm/giant corporation/"big pharma"/"THE MAN", the strategy does not change from the symmetric version:
Consider a round where the probability of seeing only lesser candidates thereafter is $> .5$. If company $A$ keeps the candidate, then it has a chance of winning $> .5$. If $A$ does not keep the candidate, then company $B$ can hire the candidate and company $A$ has a chance of winning $< .5$. So, obviously, company $A$ would hire (and company $B$ would attempt to hire) in this situation.
For a candidate with winning odds of exactly $.5$, $A$ may or may not choose to hire, but $B$ would choose to hire because $B$ can never get odds better than $.5$.
If company $A$ hired before it saw a candidate with winning chance $>= .5$, then the odds of a better future candidate existing (and hence $B$ winning) would be $> .5$. So $A$ will not hire until it sees a candidate of winning odds $>= .5$.
Therefore, $A$'s strategy is identical to the symmetric case: hire the first candidate that yields winning odds of $> .5$.
$B$'s strategy, then, is formed with $A$'s strategy in mind. Obviously, if $A$ hires (at or) before $B$, then $B$'s strategy is to hire the next candidate that is better than $A$'s, if any. Also, if a candidate comes by with winning odds $> .5$, $B$ should try to hire, even though $A$ will also try to hire (and force $B$ to keep looking).
The only question left is: is it ever beneficial for $B$ to hire when the odds of winning is $<= .5$. The answer is: yes.
Intuitively, say there is a round where the odds of winning with the candidate is $.5 - \epsilon$. Also, there is "likely to be" (explained later) a future candidate with winning odds $> .5 + \epsilon$. Then it would benefit $B$ to choose the earlier candidate.
Let $d_r$ be the candidate interviewing in round $r$ for all $1 <= r <= N$.
Officially, $B$'s strategy is: "hire $d_r$ if doing so yields better odds of winning than if not". The following is how we calculate such a decision.
Let $p_{r,i}$ be the probability of winning after interviewing and hiring $d_r$ given $d_r$ is the $i$th best interviewed candidate. Then:
$p_{r,i} = $ probability that $d_s < d_r$ for $s > r$
$ = (1-\frac{i}{r+1})(1-\frac{i}{r+2})\times ... \times (1-\frac{i}{N})$
...
$ = \frac{(N-i)!r!}{(r-i)!N!}$
Notably, $p_{r,i}$ is easily computable to constant accuracy.
Let $P_{B,r}$ be the probability that $B$ wins given that neither company hired in rounds $1$ through $r-1$.
Then $B$ would hire $d_r$ if the probability of winning after hiring $d_r$ is better than $P_{B,r+1}$.
Note that $P_{B,N} = 0$, because if it is the last round, then $A$ is guaranteed to hire and $B$ will not hire anyone and loose.
Then, in round $N-1$, $B$ is guaranteed to try to hire and will succeed unless $A$ hires as well. So:
$$
P_{B,N-1} = \sum_{i=1}^{N-1}\frac{1}{N-1}
\left\{
\begin{array}{lcl}
p_{N-1,i} & : & p_{N-1,i} < .5 \\
1-p_{N-1,i} & : & p_{N-1,i} >= .5
\end{array}
\right.
$$
Which leads to the recursive function:
$$
P_{B,r} = \sum_{i=1}^{r}\frac{1}{r}
\left\{
\begin{array}{lcl}
1-p_{r,i} & : & p_{r,i} >= .5 \\
p_{r,i} & : & P_{B,r+1} < p_{r,i} < .5 \\
P_{B,r+1} & : & \text{else}
\end{array}
\right.
$$
It's pretty obvious that $P_{B,r}$ can be calculated to constant accuracy in polynomial time. The final question is: "what is the probability of $B$ winning?" The answer is $P_{B,1}$ and varies with $N$.
As to the question of how often does $B$ win? I have not calculated exactly, but looking at $N$ from 1 to 100, it appears that as $N$ grows, that $B$'s winning rate approaches .4 or so. This result may be off as I just did a quick python script to check and did not pay close attention to rounding errors with floating numbers. It may very well end up that the real hard limit is .5.
