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n <- 3instead of100andr <- 0instead of0.8. There are several problems, one of which is that the formula of the question is a reasonable standard error only when $r$ is close to zero. Another is that the sampling distribution can be highly skewed, rendering any SE less than useful: that's why better analyses re-express $r$ with the Fisher transformation. $\endgroup$n=3is a too special case to judge. I don't know, mayben-2is not the good divisor, maybe it isn-3. Beyond that, the two formulae give the same result forr=0. $\endgroup$