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Putting a value on implied pot odds
Implied pot odds are the pot odds, factoring in how much you think you'll get paid off if you hit, right? I want to be able to put a value on how much you need to be paid off to make it worth the draw.
So a made up hand, assuming infinite stacks.
Heads up, the flop is 9h, 5h, 2c and the pot is $3.
You hold Ah, 8h - the nut flush draw.
Villain bets $2. You have to invest $2 to make it a $7 pot. Ignoring other draws you have about a 1/6 chance of hitting, so the pot should be 6 x $2 = $12 for you to have the odds, but you think you'll get paid off so you make the call.
If you hit on the turn you need to win a pot of at least $17 to effectively win on the draw (they need to invest $5 more to make it a $12 pot - the $5 you invest on the turn doesn't count towards that).
If you don't hit on the turn. The villain bets another $4 and you call again without odds. Pot is now $8 +$7 = $15, and you hit on the river.
How much do I need to win now to cover my draw? First thing that comes to mind is that it's the same as on the turn: $4 x 6 - $15 = $9. $9 x 2 + $15 = $33. This covers the turn bet, but not the flop bet. To cover the flop bet as well I think we just have to cover the difference between the cost of a good draw and our $2 (if that makes sense). That would be exactly $1 (5 + 1 = 6 x 1). So our final pot needs to be $35.
I guess that means that the implied odds are only useful if you're slightly priced out of a draw.
If anyone can do better than my calculations I'd be glad to hear.
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