dang so many outs

I like the raise on the flop to push out the third player. However, on the turn there's 570 in the pot and you have 1210. If you push and get called, and we assume that any of your fifteen straight and flush outs are good, then you're going to win 570 + 1210 = 1780, 15 out of 46 times. So about 1/3 of the time, you win 1780, giving you (1/3)*1780 = +593 from winning when you get called. Two out of three times, you lose 1210, for a loss of (2/3)*1210 = -807. So, you need to get him to fold a good percentage of the time for the push on the turn to be a profitable play. How often? Well, there is 570 in the pot and you need to recover 807-593 = 214 chips. This is about 38% of the pot. So, if the guy who flat called a bet and a raise on that flop folds 38% of the time, you break even.

You really think he folds that often?

EDIT: I'm not sure this is right...the percentage of the time he folds changes the amount of chips you lose overall, because he doesn't always call. I'll try to work it out better, but this is sufficient for the basic premise I guess.

c/f the turn?

Yeah...you can probably cut your raise on the flop to like 150 as well, and save your money when this kind of thing happens, and then check/call turn if the cold caller gives you odds, check/fold otherwise.

EDIT: I was still wrong. Here we go...

If he folds a proportion x of the time, then we need that proportion multiplied by the size of the current pot to be equal to (1-x) times the average outcome when he does call. On average, you lose 213 chips when he calls. So, we have

(1-x)[213] = x(570)

So

213 - x(213) = x(570)
213 = x(570+213)
x = about 27%