Cracking a key using a combination of partial keys
If you know the probabilities $P(k_1), P(k_2), \dots, P(k_n)$ of candidate keys $k_1, k_2, \dots, k_n$, the fact that the probability mass function $P$ factors into a product for independent subkeys is irrelevant; if you sort the $k_i$ by descending $P(k_i)$ so that the most probable answer is $k_{\pi(1)}$ with probability $P(k_{\pi(1)})$, the second-most probable answer is $k_{\pi(2)}$ with probability $P(k_{\pi(2)}) \leq P(k_{\pi(1)})$, etc., then this minimizes the expected number of guesses because each successive guess covers the maximal fraction of probability mass left unguessed so far.

Specifically, the expected number of guesses is $$\sum_{i=1}^n i\cdot P(k_{\pi(i)}).$$  Any transposition of the weights that puts a higher weight $P(k_{\pi(i)})$ on a larger number $i$ will lead to a higher weighted sum.