I understand that the expected value of a Bernoulli variable is $p\cdot1 + (1-p)\cdot0 = p$, since it provides the amount of outcomes that match a condition out of the total possibilities, on a single try.
But why is the mean of the distribution $p$ as well? Shouldn't it be a weighted average of the values of the outcomes (for example, 3.5 on a dice toss)? Or are there any definitions that I'm confusing?