Picking the safest full house before a random deal

From a standard deck you may choose any five cards that form a full house (three of one rank plus two of another). Your opponent then gets five random cards from the remaining $47$, and standard poker rankings apply. Card values are $2$–$10$ at face value and J, Q, K, A $=11,12,13,14$; a full house's value is the sum of its five cards (e.g. $22299$ is worth $24$). Consider the full houses, identified by their ranks only, that maximise your probability of winning. What is the sum of their values?

$228$. Three aces mean no opponent full house can win, leaving only quads (equally likely whatever pair you hold) and straight flushes. Holding aces already blocks the ace-low and ace-high straight flushes in three suits, so the pair should be a rank in the most remaining straight flushes, namely $6,7,8$ or $9$ (a brute-force count confirms these tie). The hands $AAA66$ through $AAA99$ are worth $54+56+58+60=228$.