[MarkGritter]

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Is my analysis flawed here? I'm not sure how to do the math, if villian only bets busted draw 1/3 of the time and the SB only calls with QQ unimproved 1/2 the time.

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Probability problems can usually be solved via brute force. Just list all the cases.

Button:
A) Flush draw comes in (1/4.25) and villian bets = 0.235
B) Flush draw misses (3.25/4.25) and villian checks (2/3) = 0.510
C) Flush draw misses (3.25/4.25) and villian bets (1/3) = 0.255

We only care about cases A and C, because in B there is no bet to call.

SB:
1) QQ improves (1/2.5) and calls = 0.4
2) QQ doesn't improve (1.5/2.5) and calls (1/2) = 0.3
3) QQ doesn't improve and folds = 0.3

So, we have 7 cases (A1, A2, A3, B, B1, B2, B3) covering all the possibilities. But, we know that Button bet and SB called, so we need to find the probability that they both missed given that action.

Cases where there is a bet/call are
A1 = 0.235 * 0.4 = 0.094
A2 = 0.235 * 0.3 = 0.0705
C1 = 0.255 * 0.4 = 0.102
C2 = 0.255 * 0.3 = 0.0765
Total = 0.343

(Adding in the remaining cases should total to 1.0 if I got the math right. B = 0.510, A3 = 0.235 * 0.3 = 0.0705, C3 = 0.255 * 0.3 = 0.0765, total = 1.0. So far, so good.)

34% of the time that the action goes bet/call. But you have the best hand only in case C2. C2 is 0.0765/0.343 = 22% of the these cases.

So, given the assumptions you state, you are getting about 3.5:1 odds on having the best hand.