[BruceZ]

In your note you state:
"But if you could go all-in on the flop, then this is 6 outs with 2 cards to come, and you would only need pot odds of 1-(41/47)(40/46) = 4.1-1. "

This calculation is actually only true if the 6 outs were six outs on the turn and on the flop. The backdoor out is a pseudo out as it requires the first card to be one of the backdoor out cards. THat is if you are not all-in the number of outs on the turn could reduce to five if the allin outs dont hit.

This calculation is valid for the case of the backdoor flush plus middle pair too, and it is no coincidence that this came out the same as the exact calculation in the above post. The reason is that when you hold two suited cards, it is impossible to make your backdoor draw AND hit one of your other 5 outs since the two cards of that suit are already on the flop. Since these events are mutually exclusive, the probability of getting one of them is the sum of the probabilities for each one, and this is the same as summing the outs (5+1=6). So it IS the same as if you had 6 outs on both the flop and the turn. The only reason it's even off in the 2nd decimal place is because the backdoor draw isn't precisely 1 out.