I'd started preparing a response, but the OP seems to have lost interest. In any case for the benefit of future Readers here are my initial thoughts. If there are Comments I'll be glad to fill in more details.
The variant of the Secretary Problem considered here appears in the literature as early as 1966 with publications by:
- Gusein-Zade, S. M. "The problem of choice and the optimal stopping rule for a sequence of random trials," Teor. Veroyatnost. i Primenen., 1966, [Vol. 11,Issue 3 (1966), pp. 534-537] (English version: Theory of Probability and its Applications, 1966, 11:3, 472–476)
- Gilbert, John P. and Mosteller, Frederick "Recognizing the Maximum of a Sequence," Journal of the American Statistical Association Vol. 61, No. 313 (Mar., 1966), pp. 35-73
Although both of these links lead to papers behind paywalls, as far as I can tell, a good summary of the results by Gusein-Zade can be gleaned from this open access technical report by Frank and Samuels (1979) . In particular as the number of secretaries $N \to \infty$, the asymptotic chance of success for the classic problem $1/e \approx 0.3679$ rises to roughly $0.5736$ when choosing either the best or second-best candidate counts as a success.
The general rule that maximizes the chance of getting one of the $b$-best secretaries is shown by Gusein-Zade to have the form of $b+1$ consecutive intervals subdividing the interviews $1,\ldots,N$.
In the present case $b=2$ and there are three phases of the rule. In the first phase the interviews $1 \le i \le N\cdot m_1(N)$ are conducted and the relative rankings observed without stopping, In the second phase the interviews $N\cdot m_1(N) \lt i \le N\cdot m_2(N)$ are conducted and only stop if one interview results in a relative "top" ranking (better than all previous interviews) happens, in which event the interviews are ended with a job offer being made. If the second phase finishes without a job offer, the third and final phase of interviews $N\cdot m_2(N) \lt i \le N$ applies the rule that a job offer is made if and only if an interview results in a relative ranking of "top" or "next to top" (better than all but at most one of the previous interviews).
It is useful to express the boundaries between phases as coefficients $m_1(N),m_2(N)$ because these multipliers have fairly simple limits as $N\to \infty$, namely as Gusein-Zade showed:
$$ \lim_{N\to \infty} m_2(N) = \frac{2}{3} $$
and:
$$ \lim_{N\to \infty} m_1(N) = \varphi $$
where $\varphi$ is the unique solution in $(0,1)$ of:
$$ \varphi - \ln \varphi = 1 - \ln(2/3) $$
Now $\varphi \approx 0.3470$, and the exact limiting probability of success for large $N$ is $\varphi (2-\varphi)$, whose approximate value $0.5736$ was mentioned before.
$A$ is antisymmetric so the game itself is symmetric. Thus if player 1 can guarantee at least $v$, then player 2 can guarantee at most $-v$. Therefore, $v=0$ so you just have to find strategies $x$ and $y$ such that $x^TA=(0,0,0)$ and $Ay=(0,0,0)^T$. You get $x=y=(1/4,1/4,1/2)^T$.
Best Answer
Since this is all or nothing, there is no point in selecting a candidate who is not best so far.
If there are $n$ candidates in total and the $k$th candidate is the best so far then the probability that the $k$th candidate is best overall is $\frac{k}{n}$, which is an increasing function of $k$.
So if you have a decision method which selects the $k$th candidate when best so far but not the $m$th candidate when best so far, with $k \lt m$, then a better (or, in an extreme case, not worse) decision method would be not to select the $k$th candidate when best so far but to select the $m$th candidate when best so far.
So in an optimal method, if at any stage when you are willing to select a best so far candidate, you should be willing to select any subsequent best so far candidates. That gives the strategy in your question of not selecting up to a point and then selecting any best so far candidates after that point.
There is an extreme case: for example if there are two candidates, it does not make any difference whether you accept the first candidate or wait to see the second candidate.