Say, $R_i$ is a binary variable, and $n$ is the sample size.

What will be the expectation, $\mathbb E[R_i=0]$?

Is it $\mathbb E[R_i=0]=\mathbb E[1-R_i]=\sum_{i=1}^{n}(1-R_i)P(R_i=0)$?

But if anyone show that $n^{-1}\sum_{i=1}^{n}(1-R_i)$ converges in probability to $P(R_i=0)$, does he assume $P(R_i=0)=\frac{1}{n}$ ?

That is, to show $n^{-1}\sum_{i=1}^{n}(1-R_i)$ converges in probability to $P(R_i=0)$, is the following way correct:

$n^{-1}\sum_{i=1}^{n}(1-R_i)=\mathbb E[1-R_i]=\mathbb E[R_i=0]=P(R_i=0)$.