standard error formula for Bernoulli distribution

I know that standard error of the sample average Y_bar should be an estimator of the standard deviation of the sampling distribution $$\bar Y$$. This should be based on the case where the population std is unknown. The formula should be SE[$$\bar Y$$] = sample standard deviation / sqrt(n).

However, i am considering the case where i know i am drawing samples from a Bernoulli distribution; this has two cases.

1. I know it is Bernoulli, but I don't know its real p(probability of success). Then I shouldn't know its variance. Then the methods to compute SE should be using the formula above. For example, Y1 = 0, Y2 = 1, Y3 = 1. Then sample variance is 1/2 * [(0 - 2/3) ** 2 + 2*(1-2/3)**2) = A. so the sample SE should be $$\sqrt{\frac{A}{n}}$$, which is different from $$\sqrt{\frac{\bar Y*(1 - \bar Y) }{ n} }$$. (which formula is correct?)

2. I know it is Bernoulli and I know its real p. Then I should know the sample variance to be p * (1-p) / n, the sample standard deviation is the square root of sample variance. Then we probably don't even need to estimate sample standard error. As it is the same as sample standard deviation.

However, the following statement makes me confusing: When Yi are iid draws from a Bernoulli distribution with success probability p, the variance of $$\bar Y$$ should be p * (1-p)/n, and SE[$$\bar Y$$] is $$\sqrt{\frac{\bar Y * (1-\bar Y) }{ n}}$$. I am not sure, why when computing SE, we don't use p, but use $$\bar Y$$ instead.

The standard error of the mean $$SE(\bar Y)$$ is calculated using the standard deviation of the population($$\sigma$$) using $$\sigma/n$$, in this case, the standard deviation of the population ($$\sigma$$) is $$p(1-p)$$. However, we don't know the value for p therefore, we can't calculate the standard error of the mean $$SE(\bar Y)$$. However, we can estimate the standard error of the mean. In order to estimate, we use the standard deviation of the sample ($$\sigma_x$$) first and the estimator for the standard error of mean will be calculate by ($$\sigma_x/n$$). So, the only missing piece is to calculate the standard deviation of the sample($$\sigma_x$$). We use:$$\sigma_x = \frac{1}{n-1} \sum_{i=0}^n (x_i - \bar Y)^2$$ and $$\bar Y$$ is the sample mean. To sum up, if you knew the actual value of p then the $$SE(\bar Y) = \sqrt{ p(1-p)/n}$$ and when you don't know the p, $$SE(\bar Y)$$ is estimated using ($$\sigma_x/n$$)