Multiple testing, alpha, and familywise error I am confused why it is that in multiple testing, the expected number of false positives is: 
$$
\alpha \times\text{number of tests}\tag{1}
$$
(That is, for $\alpha = 0.05$, you would expect $5\%$ false positives, regardless of the number of tests.)  
But if the percent of false positives stays the same regardless of the number of tests, why does the probability of at least one false positive increase according to
$$
1 - (1-\alpha)^\text{number of tests}\tag{2}
$$
What are formulas $1$ and $2$ saying?
 A: Okay, I am going to write separately how we get to eqs. (1) and (2).
Definitions: 


*

*$\alpha=$ type-I error 

*$n=$ the total number of tests.
Let's start with eq. (1)
The expected number of false positives FP can be calculated by knowing the probability of getting zero false positives, one false positive, two false positives etc., like so
\begin{equation}
\mathbb{E}(FP) = 0\times\mathbb{P}(FP=0) + 1\times\mathbb{P}(FP=1)+2\times\mathbb{P}(FP=2)\dots+N\times\mathbb{P}(FP=N)
\end{equation}
The probability of k false positives is 
$$\mathbb{P}(FP=k) = \binom nk\alpha^k (1-\alpha)^{n-k}$$
Does this formula ring a bell? It is the formula for the binomial distribution!
So the expected number of false positives is nothing more than the expected value of a binomial distribution with probability of success $\alpha$,
$$ 
\bbox[white,5px,border:2px solid red]{\mathbb{E}(FP)=n\alpha}
$$
Now, let's worry about eq. (2)
In general, given $n$ tests the probability of rejecting the null in any of the tests can be written as
\begin{equation}
\mathbb{P}(\mathrm{rejecting\ the\ null\  in \ any \ of \ the \ tests})=\mathbb{P}(r_1\lor r_2\lor\dots\lor r_n)
\label{eq:prob}
\end{equation}
in which $r_j$ denotes the event "the null is rejected at the j-th test".
While it is difficult to evaluate this equation in general, the expression greatly simplifies for independent tests. Indeed, under this assumption we can write
\begin{equation}
\mathbb{P}(r_1\lor r_2\lor\dots\lor r_n) = 1 - \mathbb{P}(r_1^* \land r_2^* \land\dots\land r_n^* ) = 1 - \prod_{j=1}^n \mathbb{P}(r_j^* ),
\end{equation}
where $ r_j^* $ denotes the event "the null is NOT rejected at the j-th test".
Now, note that $\mathbb{P}(r_j^*)=1-\alpha$, so we can rewrite the previous equation to find the family-wise type I error
\begin{equation}
\bbox[white,5px,border:2px solid red]{\mathrm{Type \ I \ error} = \mathbb{P}(r_1\lor r_2\lor\dots\lor r_n)= 1-(1-\alpha)^n} \ \ \ (\mathrm{independent \ tests})
\label{eq:typeI_independent}
\end{equation}
