Suppose $n>m$.
Person 1 samples $n$ times (i.i.d.)
Person 2 samples $m$ times (i.i.d.)
$x_{(n-2)}$ is the 3rd-highest number obtained from $n$ sampling from distribution $F$
$y_{(m-2)}$ is the 3rd-highest number obtained from $m$ sampling from distribution $F$
Question: Conditional upon the largest number is the same from both people, is the expected value of 3rd-best number of person 1 bigger than the expected value of 3rd-best number of person 2?
That is,
$$E(x_{(n-2)}-y_{(m-2)}|x_{(n)}=y_{(m)})>0 ?? $$
I guess this ``condition'' part is irrelevant because it's i.i.d. sampling. (correct?)
Then the problem reduces to
$$E(x_{(n-2)}-y_{(m-2)})>0 ??$$
This should be intuitively true. But should there be some regularity conditions imposed on $F$ to make this hold with "strict" inequality?
Intuitively, I would say the condition is relevant. As I put in a comment, first of all, were it not relevant, this would have nothing to do with the fact that each sampling is of i.i.d. r.v., since $X_{(n-2)}$ and $X_{(n)}$ are not independent (same for $Y_{(m-2)}$ and $Y_{(m)}$); at most, it would have to do with the (not explicitly stated) fact that each $X_i$ is independent from each $Y_j$.
Even more, given that $E\left(X_{(n)}\right)>E\left(Y_{(m)}\right)$, this seems to impose a condition.
Supose that the condition were $X_{(n)}=Y_{(m-3)}$, if you use the same reasoning you would conclude that this fact does not affect the sign of $E(X_{(n-2)}-Y_{(m-2)})$ while, in fact, is saying that for sure $X_{(n-2)}\le Y_{(m-2)}$ and so $$E\big(X_{(n-2)}-Y_{(m-2)}|X_{(n)}=Y_{(m-3)}\big)\le 0.$$
Try again not dismissing the condition. Intuition says to me that anyway the answer is affirmative.
Let's assume that both samples from the same distribution verify $x_{(n)}=y_{(m)}=K\in \mathbb R$, and for the sake of simplicity let's suppose that the distribution has support on $(0,+\infty)$, you have $n-1$ observations to accommodate in the interval $(0,K)$, and then another (fewer) $m-1$ observations to accommodate in the same interval. In the extreme case that $n>>m$ it is to be expected that $X_{(n-1)}>Y_{(m-1)}$ with great probability, and perhaps also that $X_{(n-2)}>Y_{(m-2)}$ (imagine, for instance, that $n=10$ or $n=100$, and that $m=3$ or that $m=4$).
I can't say this reasoning works over and over since $n-k$ eventually becomes small, although still $m-k$ is even smaller and eventually becomes $1$, but I guess is fine for $k=2$ (of course, this is not even close to a formal proof).