Yeah, I ran it through the odds calculator a little while ago and came up with 2.8:1. For $740, that means you need close to $2,000 in the pot. (also of note is that it doesn't matter whether he has 3d or not, since that's not an out anyway).
I guess this is pretty close either way. I really agree with you that his hand range is VERY heavily weighted toward set (and at least at FR, I'd view it as pretty ballsy with bottom set, which I'd personally very seriously consider laying down in this situation).
Ok, so SB is a LAG, which makes 64s more plausible. But that's the only hand I can really think of except for a set (and diamonds only). I just can't see any overpair at all making this move reasonably (I actually do see QQ-AA do it, but I don't consider the players where I see it done very good). And if it were me with 6d4d, I'd probably call rather than raise given that action--with the intention of going away if the board pairs the turn.
But anyhow, supposing the range is 64s or any set. Then there are 9 hands where you're 2.8:1 dog and 1 hand where you're favorite.
And 64s in diamonds has 12 outs, so should be better than 2:1 to catch up.
Ok, so how big would the pot have to be given that exact range to make the call, if we assume that you have to call $1,000 (just to make it easy)?
X = pot-size
1) Set scenario (9 cases): You win X once and lose $1,000 2.8 times. Equity is thus (X - 2,800)/3.8
2) 64s scenario: You win X twice and lose $1,000 once. Equity is thus (2*X - 1,000)/3
Total equity times 10 (for number of cases) is thus X*((9/3.8)+(2/3)) - (9*737+333) = X*(2.7) - (6,633 + 333) = X*2.7 - 6,966.
It works if the pot is bigger than $2,457.
So, basically, if the hand is that heavily weighted toward a set (and I really think it is--maybe even more), then you'd need odds of about 2.5:1 to make the call.
lol, well, ok, I'm back to folding it!