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Calculating the entropy of a password
Let's say I'm generating a random string of length $n>0$ from a finite, nonempty alphabet $A$. The formula for entropy is, according to Wikipedia:
$$\sum_{i=1}^n P(x_i) \log_b P(x_i)$$
$P$ is the probability mass function; we assume randomness is uniformly distributed over A, so $P = \frac{1}{A}$. Is it therefore the case that the entropy of my whole string is:
$$ \frac{n}{A} \log_b A$$
That means, if I have a string that is 16 characters long, selected uniformly from lower and uppercase Latin characters and numbers (i.e., $A=62$), then my string only has an entropy of around 1.5 bits... Is that right? It seems low, but I may be misinterpreting how this is all supposed to work.
EDIT I had originally assumed that entropy was $n \log_b A$, so my 16 character password would have an entropy of around 95 bits, but the above formula has this additional divisor. Which is correct?
The $n$ in the formula isn't the $n$ you're working with. What you are supp

The $n$ in the formula isn't the $n$ you're working with. What you are supposed to do is sum over all possible outcomes $x$ of the random variable $P(x)\log_2 P(x)$. Here there are $62^{16}$ possibilities, all with probability $62^{16}$, so you get $62^{16}62^{16}\log_2(62^{16})=16\log_2 62\approx 95$.
In fact whenever all possibilities are equally likely the entropy is just $\log_2$ of the number of possibilities.
(The formula is for a random variable with $n$ possibilities $x_1,\ldots,x_n$, not a random variable $x$ with $n$ components.)
20170804 16:31:42