We have a set of i.i.d. random variable $X_i$ with some discrete distribution.
Further we have a random variable Y, Independent from $X_i$ with a Binomial Distribution Bin(n,p).
Now we are interested in $$q_i := P[Y \geq X_i ]$$
Is it true that
$$q_i:=P[Y\geq X_i ] = \sum_{x} P[X_i =x] P[Y\geq x] =E_{X_i}[P[Y \geq X_i]].$$ Especially, $q_i$ is Independent from $i$ and for all $i$ we have that $q_i = q$ where $$q:=E_X[P[Y \geq X]]=\sum_{x} P[X=x] \sum_{j=x}^{n} { n \choose j} p^j (1-p)^{n-j}$$ where $X$ has the same Distribution as $X_i$.
It is true that $q_i$ does not depend on $i$ under the condition that $Y$ is independent of the family $(X_i)_i$ (otherwise anything can happen). Then $q=E(F(Y))$, where $F:x↦P(X⩽x)$ is the common CDF of the random variables $X_i$.