when to use odd to improve hand from flop to river in FLHE

Newbie question here about LIMIT Holdem:

I think I understand how & why to calculate odds on the flop to improve your hand on the turn when deciding whether to call a bet or fold. But what I'm not sure about is how/when you would use your odds to make your hand on by the river. I've been playing with the former calculations for situations like this:

I'm in late position against three opponents and make middle pair with a rainbow flop (one is the same suit as my suited hold cards). The first to act bets, and I put him on top pair or an overpair; one person folds, the other calls.

Normally, I'd think I have 5 outs at this point to either make trips or two pair of the turn to beat the villain's pair, which would require better than 8.5 pot odds to make the call. But I've also flopped a back door flush draw here, which means I'm a little less than 2:1 to hit my flush by the river. And that's a big difference.

So the question here is.... which odds to I use to determine what I do -- the 8.5:1 odds to hit trips or 2 pair on the turn, or the 2:1 odds to hit my back door flush by the river? I think people the latter sometimes, but perhaps it's different in FLHE vs NLHE. I've been almost exclusively using the odds for the turn. Is that a bad idea?

I wouldn't valuate my outs like that,

a) if you have 10 9 on a A 10 4 board and your opponents has A 9, you would not want to see a 9 as you would be drawing very thin.

b) I don't know what the exact odds for a runner runner flush are, but playing like that seems more like praying for a miracle than poker to me.

Post...hurt...brain...

Try counting ALL of your outs together (I think a backdoor flush counts as 1 out). Multiply by 4 with 2 cards to come, and 2 with 1. This will give you a rough approximation of your percentage chance to hit your outs.

In your example (I think), you have 5 outs to make two-pair or three of a kind. Counting your backdoor flush as 1 out, you have 6 outs. So, with 2 cards to come (we're on the flop) you multiply 6 outs times 4 for a product of 24. You have a 24% chance of hitting your outs.

Subtract your percentage from 100 and then divide by the percentage for odds. (100-25)/25=3. You need 3 to 1 to make the call on the flop.

It's not a bad idea to only calculate your odds for the turn if you know that the pot won't be big enough (on the turn) to continue drawing to the river.

I think that helps...right?

For runner-runner flush, you are looking at 35% chance with two cards to come, and ~ 20% with one card to come. Flushes can be rare and hard to chase (as I am trying to make a note of myself, as I also play limit, and am trying not to throw too much money away chasing....) so the turn concept may not be a bad idea. If you miss the flush AND the other draws, bail. At microstakes I play at, you cannot make anyone fold; they will always call or bet regardless of what they have.

Also, understand that regardless of odds; if only one card in your hand actually matches the flush on the board, and someone has a higher hand of the same suit, you are screwed. I have been burned by that a few times. such a flush draw is very weak and is almost not even worth going after. Fold.

rowhousepd said:

But I've also flopped a back door flush draw here, which means I'm a little less than 2:1 to hit my flush by the river. And that's a big difference.

Click to expand...

Sorry gang, you're totally right -- I meant 23:1. Typo.

But the question still holds: If I need to improve my hand in a scenario like this, are chances of hitting runner-runner just too slim to really take into account?

Double-A said:

Try counting ALL of your outs together (I think a backdoor flush counts as 1 out). Multiply by 4 with 2 cards to come, and 2 with 1. This will give you a rough approximation of your percentage chance to hit your outs.

Click to expand...

OK, wait a sec. Is this equation how you’d factor in a backdoor draws in general -– where you count it as ONE out in addition to your outs for the turn? Also, as you said above, why multiply 4 with 2 cards, and then 2 with one. (And likewise, if there essentioally 5+1 outs, why multiply it by 4? Now my brain hurts.

PokerMagpie said:

For runner-runner flush, you are looking at 35% chance with two cards to come, and ~ 20% with one card to come.

Click to expand...

Ok so, if you're saying 35% (unlike Daouble-A), how did you get that number, Magpie? And yes, at micros where I play as well, I'm realizing that no one gets scared out of pots. Gotta learn to call & fold more.

Btw, I do realize that the game involves more than a simple formula like this, but I’m just trying to get the math down first.

rowhousepd said:

But the question still holds: If I need to improve my hand in a scenario like this, are chances of hitting runner-runner just too slim to really take into account?

Click to expand...

Your chances of hitting a runner-runner draw are too slim to take into account by themselves. But sometimes you'll have more than one draw, like two over cards w/ backdoor straight and flush draws.

Focus on counting all of your outs.

rowhousepd said:

OK, wait a sec. Is this equation how you’d factor in a backdoor draws in general -– where you count it as ONE out in addition to your outs for the turn? Also, as you said above, why multiply 4 with 2 cards, and then 2 with one. (And likewise, if there essentioally 5+1 outs, why multiply it by 4? Now my brain hurts.

Click to expand...

I just googled it, and the consensus seems to be that a backdoor flush should be considered as 1.5 outs.

Multiplying your outs on the flop by four is a quick and dirty way to estimate your percentage chance of hitting your outs on the turn AND river. If your lone opponents bet on the flop puts him all in, meaning you won't have to call any more bets, and you know that you have to hit your flush draw to win then... You have nine outs (multiplied by four) or a 36% chance to hit your flush with two cards to come. If this situation occurred on the turn you'd be half as likely to hit your flush (nine outs times two for an 18% chance).