If I choose a random alphabet from (a-z), $26$ characters, what is the entropy?
Shannon's formula:
$$H = - \sum p \log_2(p) = - (1/26)\log_2(1/26) = 0.18$$ bits.
However, other formulas on the Internet use:
$$H = \log_2 (N^L) = \log_2(26) = 4.7$$
bits.
Which one is correct?
The internet is correct.
The sum is over all possible outcomes, and not just one. So you should get:
$H = \sum_{x = 'a'}^{'z'} - p(x) \log_2(p(x)) = \sum_{x = 'a'}^{'z'} - \frac{1}{26} \log_2(\frac{1}{26} = 26 * (-\frac{1}{26} \log_2(\frac{1}{26})) = -\log_2(\frac{1}{26}) = \log_2(26) = 4.70043971814...$