Consider two games with the same number $n$ of players.
Let $\mathcal{A}:=\{Y_1,Y_2\}$ be the set of equilibria of game 1. $Y_i$ is an $n\times 1$ vector reporting the action of each player, for $i=1,2$.
Let $\mathcal{B}:=\{X_1,X_2\}$ be the set of equilibria of game 2. $X_i$ is an $n\times 1$ vector reporting the action of each player, for $i=1,2$.
The selection rule is defined as "the rule according to which players pick the outcome to play among the predicted equilibria".
(*) Suppose that "the selection rule of game 1 is independent of the selection rule of game 2". In other words, how players pick the outcome to play in game 1 does not affect how players pick the outcome to play in equilibrium 2.
Now, let's attach some probabilities (denoted by "Pr") to these objects and consider
$$ Pr(\text{players pick $Y_1$ in game 1}| \text{the set of equilibria of game 1 is $\mathcal{A}$}) $$ and $$ Pr(\text{players pick $X_1$ in game 2}| \text{the set of equilibria of game 2 is $\mathcal{B}$}) $$ Does (*) imply that $$ Pr(\text{players pick $Y_1$ in game 1}| \text{the set of equilibria of game 1 is $\mathcal{A}$})\times Pr(\text{players pick $X_1$ in game 2}| \text{the set of equilibria of game 2 is $\mathcal{B}$})= Pr(\text{players pick $Y_1$ in game 1}, \text{players pick $X_1$ in game 2}| \text{the set of equilibria of game 1 is $\mathcal{A}$}, \text{the set of equilibria of game 2 is $\mathcal{B}$}) $$ ? Or do we need other conditions?
A random selection rule is not usual in game theory, but let's go with it.
I find your notation a tad confusing, because it suggests that "the set of equilibria of game 1" is a conditioning event and hence that it should be probabilised.
My guess (and of course I may be wrong) is that you mean to say that game $A$ has two equilibria and game $B$ has two equilibria. Let $p$ the probability of pick $Y_1$ in game $A$ and $q$ the probability to pick $X_1$ in game $B$. Then independent implies that $pq$ is the probability to pick $Y_1$ in $A$ *and$ $X_1$ in $B$.