[tczMUFlmoNk]

Mostly yes, but to your second paragraph:

> if instead one letter only 1/1000th of the time (aaa....aaa...aa..a.z.aaaa), then when the rare beast shows up, it is a big surprise, so the total amount of surprise available in the sequence is high

…when a Bernoulli distribution is skewed, the maximum surprise is high, yes, but the average surprise (= entropy) is low. The entropy of a Bernoulli distribution is maximized when p = 0.5 and falls off to either end:

https://en.wikipedia.org/wiki/Binary_entropy_function

For your examples, if the sequence is uniformly distributed (Bernoulli(1/2)), the entropy is log(2) ≈ 0.693 bits per symbol; if instead one letter occurs 1/1000th of the time, the entropy is about 0.0079 bits per symbol.