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3:1 doesn't price in a flush draw. Or does it?
Say you have a flush draw.
The board is

You hold

The pot is $1000
Stacks are $12,000 each (currently, not before this pot)
Villain bets the pot, $1000, presumably with an ace.
You're getting ($1000 + 1000 +1000 = 3000 for 1000)
You get 3 to 1 odds.
Hey! a flush comes 1 in 3 so I'm in, right?
***
Why does everyone I know believe this? Isn't it completely wrong?
The odds of a flush coming (ON THE NEXT CARD) is 5:1. Thus, according to pure pot odds, you have been priced out.
Why should we calculate the odds of the flush hitting on the next card, instead of the next two, IMO? Because there is usually more betting done on the turn by the villain. The price is much higher than expected.
For our example hand, let's continue
You call the $1000 bet, having taken the 3:1 lay.
The turn makes the board

the pot is now $3000
Villain bets $2000.
(3k + 2k + 2k = 7k). You're getting 7:2 ("3.5" to 1)
You are priced out again. Odds of spade coming on the river are ~1:5 and so 7:2 is just no good.
Nevertheless, this hand always plays out the same. Hero with spade draw calls, hits, and gets little if any money on the river (nixing the much-needed implied odds that might make his call worthwhile [+EV]). He pats himself on the back later.
I've folded nut flush draws before, I'm just wondering if I've been wrong all this time. I understand that implied odds might be there. It's just that, usually, a three-suit board kills action. Thoughts? I think my problem is just not understanding how the pot odds work with two different bets on two different streets with cards to come.
Example: How big of a pot does the hero in this example need to build by the showdown to make his flop call profitable? His turn call? The whole hand (flop + turn) altogether? I know that if he doesn't get any extra on the river he spewed.
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