WOHEP: Backdoor flush outs estimates

[SlantNGo]

On pages 66 and 67 where Yao covers the case of runner-runner flush and provides out estimates, there is no distinction made between what are, in my opinion, drastically different situations.

In one situation, you have suited hole cards, and one of your suit on the flop gives you a backdoor draw. For example, you hold J /images/graemlins/spade.gif 2 /images/graemlins/spade.gif on a K /images/graemlins/spade.gif 5 /images/graemlins/heart.gif 9 /images/graemlins/diamond.gif flop.

In the other, you have unsuited hole cards, and two of your suit on the flop gives you a backdoor draw. For example, you hold J /images/graemlins/spade.gif 2 /images/graemlins/heart.gif on a K /images/graemlins/spade.gif 5 /images/graemlins/spade.gif 9 /images/graemlins/heart.gif flop.

According to Yao, we are drawing to a 2nd or 3rd highest flush, and we should assign 1.5 outs to the backdoor. Is this the case for both situations? Shouldn't we be assigning more outs (maybe close to the full 2) for the first situation, while deducting for the second situation (maybe to just 1)?

[cwsiggy]

Why would they be different? You are not drawing to the nut flush obviously, but each still has 9 outs. Always assume they are in the deck.

Are you saying that because there are 2 on the board, and you only have one, therefore more people could be drawing to the flush also and therefore you have to discount it? I don't think that matters.

[Mr Mojo Risin]

The first one is a much better hand to hit the flush on. A jack high flush is good when there are only three on the board, but when there are four on the board, any higher spade, by itself beats you. I don't know if i would count more outs or not, but i would much prefer the first hand to hit than the second.

[binions]

[ QUOTE ]
According to Yao, we are drawing to a 2nd or 3rd highest flush, and we should assign 1.5 outs to the backdoor. Is this the case for both situations? Shouldn't we be assigning more outs (maybe close to the full 2) for the first situation, while deducting for the second situation (maybe to just 1)?

[/ QUOTE ]

Backdoor flush draws are 23:1 to hit, which is equal to 1 out with 2 cards to come. So, I don't know where you get "the full 2" idea.

1.5 outs is already generous.

[bravos1]

[ QUOTE ]
The first one is a much better hand to hit the flush on. A jack high flush is good when there are only three on the board, but when there are four on the board, any higher spade, by itself beats you. I don't know if i would count more outs or not, but i would much prefer the first hand to hit than the second.

[/ QUOTE ]

Indeed, the first is better to have because it is less likely that someone else will also have 2 spades, BUT just speaking about odds, it does not matter a bit if you have 2 in your hand and one on the board, or 1 in the hole and 2 on the board, or even none in your hand and 3 on the board (not a backdoor draw though /images/graemlins/grin.gif)! The odds of YOU getting a flush are the same in all of these cases. The odds of you having a WINNING hand will greatly vary though, with the first case being the best, then the second, and then the 3rd.

[King Yao]

[ QUOTE ]
Backdoor flush draws are 23:1 to hit, which is equal to 1 out with 2 cards to come.

[/ QUOTE ]

Can you show your derivation?

I used:
10/47 you hit a flush draw on the Turn.
if you hit the flush draw, then you will have 9 outs.

10/47 x 9 outs = 1.91 expected outs

[King Yao]

I think you have a good point. With one flush card in your hand and two on the board, I wouldn't bother to count any flush outs at all. The number of outs are the same in the two situations, but the other players will play the hand, and react to your hand differently. You could reduce the expected pot size, but for me, it would be easier not to count the flush outs at all.

With two flush cards in your hand, if you do hit the flush on the River, it will be hidden and a surprise. You are more likely to be paid off. On the other hand, with four flush on the board, you are less likely to be paid off, and also more likely to be trumped by a higher flush.

[SlantNGo]

That's what I was wondering... I discussed it with a friend, and he thought maybe you were talking only about the case where you have suited hole cards and one of your suit appears on the flop.

I think that in the case where two of your suit appear on the flop, if you hold the nut or 2nd nut backdoor flush draw, that we should still count those as outs, however, certainly not as much as for a nut or 2nd nut draw where you have suited hole cards. I'm gonna take a stab and estimate that they are worth about 1 out.

[binions]

[ QUOTE ]
[ QUOTE ]
Backdoor flush draws are 23:1 to hit, which is equal to 1 out with 2 cards to come.

[/ QUOTE ]

Can you show your derivation?

I used:
10/47 you hit a flush draw on the Turn.
if you hit the flush draw, then you will have 9 outs.

10/47 x 9 outs = 1.91 expected outs

[/ QUOTE ]

Do you agree a backdoor flush draw is 23:1 to hit? 10/47 * 9/46 = 4.16% which is 23:1 against.

If so, look at any standard odds chart. 1 out with one card to come is 45:1 assuming 46 unseen cards. 1 out with 2 cards to come is 22.5:1 with 47 unseen cards (or 21.5:1 if there are 45 unseen cards).

Bingo. See Ed Miller's SSHE at p. 102 for confirmation on the math.

Anyway, I have seen Ed Miller (and I think Abdul) say a backdoor flush draw is worth 1.5 outs. But this has to do with:

1. The fact you can fold on the turn when you don't pick up a draw;

2. The fact that you get paid off relatively more often than a flopped 4 flush when you hit.

So the math "out" is inflated by 50% to 1.5 outs given these implied odds/game considerations.

But the math is a backdoor flush draw is 23:1 against which is equivalent to 1 out.

To say you have 1.91 outs on the flop means you have an 8% chance to hit your hand by the river. That's simply false with respect to backdoor flush draws. You don't have an 8% chance. You have a 4.16% chance.

[King Yao]

Hi Binion, thanks for responding. I hope you don’t think I am questioning you. I just want to get to the bottom of possible differences, and I want to make sure I didn’t screw up something in my book. That wouldn’t be good. I would appreciate it if you could please take a look at this and let me know what you think.

[ QUOTE ]
Do you agree a backdoor flush draw is 23:1 to hit? 10/47 * 9/46 = 4.16% which is 23:1 against.


[/ QUOTE ]

Yes, I agree with this.

I'm going to go through an example using DIPO because I think it will illustrate how the way I count outs works in accordance with DIPO. It is possible the number of outs may not work using other methods (maybe it is a case of mixing and matching different definitions). I haven't thought about using other methods since I don't use them, so if there is an inconsistency after the definitions are adjusted accordingly, please let me know.

Just to summarize the DIPO method for those that are unaware of it :
Good Number: Expected Outs x Expected Pot Size
Bad Number: Expected Non-Outs (also known as Unknown Cards minus Expected Outs)

If Good Number is greater than Bad Number, then you have pot odds to call.

Let's assume you have suited cards in your hand, and the Flop has one card of your suit. Let's assume you know for a fact that your backdoor flush will be good if you hit, but you have no other outs.

You estimate 10/47 x 9 as your expected outs and use 1.91 as the expected outs. And then you try to see how big you need the expected pot size to be in order for you to call. (note in DIPO, I suggest that you only count the bets accumulated on the Turn, but not all the way to the River as a way of estimation – and it is just that, just a quick estimation at the table)

Expected Outs = 1.91
Expected Non-Outs = 45.09

This means the Expected Pot Size must be greater than 23.6 small bets in order for a call to be correct.

1.91 x 23.6 = 45.1

So DIPO is saying you can call if you have greater than 23.5 to 1. That sounds fairly close to what you are claiming (22.5 to 1). I recommend rounding to either 2, 1.5 or 1 out. 2 for the highest backdoor flush draw (a little bit of the possibility of getting a ton of bets if someone else also has a backdoor flush draw – but more because it is easier to use 2 than 1.9 in your head). 1.5 for the 2nd and 3rd highest flush draws (lower than 1.9) and 1 for all other backdoor flush draws (to account for the times you get smacked down against another flush draw).

This recommendation is a quick and dirty estimation to add to the outs….I feel if using DIPO, it is fairly accurate.

I've got a head-ache now. Can't believe I did this on a Friday night, I guess that shows my lifestyle, eh? /images/graemlins/ooo.gif

[binions]

I don't think DIPO works that way when you are trying to figure runner runner. DIPO is usually applied trying to figure outs for the next card.

Put another way, you need 23.5:1 under DIPO to see both cards, not just the turn.

So really, you should divide 1.91 by 2 to get 0.96 outs for a backdoor flush draw. This is accurate. I have rounded to 1.

[ninjia3x]

OMG, this is so simple to find, I don't get what you ppl are writting about. Just run all possible combinationas and find the EV, here is my crude and straightfoward method.

odds of hitting the cards we need on the turn is 10/47, river is 9/46

so to get through all the possible cases, i'm going to run it 2162 times

so first, u'll lose -2162 on the flop to see the turn card,

next, on the turn, 460 times u'll get your four flush draw. so u'll have to pay to see it, so -2*460

now, for u'r pay off, it will be final pot times 90. cause u'll hit 90/460 times.


so u hae FinalPot*90 = 2162 + 2*460 for 0EV


Final Pot = 34.24

so working backward, -2(him) for his call on the river, -2(you) -2(him) for the turn, -1(you) -1(him) for the flop bet, u'r left with 26.24

so pot needs to be 26.24 on the flop. with the pot odd on the flop, u can easily figure out the number of outs it equals.. add 1 for his bet, and u get the pot odd you need.

which is 27.24

so u need to get 27.24:1, so that is 1/27.24 * 47 = 1.73 outs. (out/47* pot = 1 ...out = 1/pot *47 )

Q.E.D.

[binions]

[ QUOTE ]
OMG, this is so simple to find, I don't get what you ppl are writting about. Just run all possible combinationas and find the EV, here is my crude and straightfoward method.

odds of hitting the cards we need on the turn is 10/47, river is 9/46

so to get through all the possible cases, i'm going to run it 2162 times

so first, u'll lose -2162 on the flop to see the turn card,

next, on the turn, 460 times u'll get your four flush draw. so u'll have to pay to see it, so -2*460

now, for u'r pay off, it will be final pot times 90. cause u'll hit 90/460 times.


so u hae FinalPot*90 = 2162 + 2*460 for 0EV


Final Pot = 34.24

so working backward, -2(him) for his call on the river, -2(you) -2(him) for the turn, -1(you) -1(him) for the flop bet, u'r left with 26.24

so pot needs to be 26.24 on the flop. with the pot odd on the flop, u can easily figure out the number of outs it equals.. add 1 for his bet, and u get the pot odd you need.

which is 27.24

so u need to get 27.24:1, so that is 1/27.24 * 47 = 1.73 outs. (out/47* pot = 1 ...out = 1/pot *47 )

Q.E.D.

[/ QUOTE ]

You are figuring implied odds, and the savings you make when you miss your draw on the turn.

That is not an answer to the math problem. This gets into why Ed says to value the backdoor flush draw as 1.5 outs instead of its true math "out value."

The fact remains that you have a 4.16% chance to hit your hand with 2 to come. This equals one out.

[ninjia3x]

[ QUOTE ]
[ QUOTE ]
OMG, this is so simple to find, I don't get what you ppl are writting about. Just run all possible combinationas and find the EV, here is my crude and straightfoward method.

odds of hitting the cards we need on the turn is 10/47, river is 9/46

so to get through all the possible cases, i'm going to run it 2162 times

so first, u'll lose -2162 on the flop to see the turn card,

next, on the turn, 460 times u'll get your four flush draw. so u'll have to pay to see it, so -2*460

now, for u'r pay off, it will be final pot times 90. cause u'll hit 90/460 times.


so u hae FinalPot*90 = 2162 + 2*460 for 0EV


Final Pot = 34.24

so working backward, -2(him) for his call on the river, -2(you) -2(him) for the turn, -1(you) -1(him) for the flop bet, u'r left with 26.24

so pot needs to be 26.24 on the flop. with the pot odd on the flop, u can easily figure out the number of outs it equals.. add 1 for his bet, and u get the pot odd you need.

which is 27.24

so u need to get 27.24:1, so that is 1/27.24 * 47 = 1.73 outs. (out/47* pot = 1 ...out = 1/pot *47 )

Q.E.D.

[/ QUOTE ]

You are figuring implied odds, and the savings you make when you miss your draw on the turn.

That is not an answer to the math problem. This gets into why Ed says to value the backdoor flush draw as 1.5 outs instead of its true math "out value."

The fact remains that you have a 4.16% chance to hit your hand with 2 to come. This equals one out.

[/ QUOTE ]

The only implied odd that may be the last 2 dollars u make off of him. which he is getting 15:1.

other wise, there is no implied odd included in that, I think it is pretty clear, please go read again.

[King Yao]

[ QUOTE ]
I don't think DIPO works that way when you are trying to figure runner runner. DIPO is usually applied trying to figure outs for the next card.

[/ QUOTE ]

DIPO is used to figure out if you have pot odds to call a bet. You need to know your expected outs in order to use DIPO correctly. Figuring out the expected outs on a runner-runner draw fits in with this idea. I think it works good (or is that 'works well'?)

[binions]

[ QUOTE ]
[ QUOTE ]
I don't think DIPO works that way when you are trying to figure runner runner. DIPO is usually applied trying to figure outs for the next card.

[/ QUOTE ]

DIPO is used to figure out if you have pot odds to call a bet. You need to know your expected outs in order to use DIPO correctly. Figuring out the expected outs on a runner-runner draw fits in with this idea. I think it works good (or is that 'works well'?)

[/ QUOTE ]

I agree you can call with a backdoor flush draw if you get 23:1 to see BOTH cards, ie you are all in on the flop and will not be faced with another bet.

But if you are only getting 23:1 to see the turn card, you are overpaying if you will also be faced with calling a bet to see the river.

[King Yao]

very nice.

[SlantNGo]

Alright, here's my take on it. Binion is arguing that a backdoor flush draw is 2 outs for 2 streets, which should be reduced to 1 out for 1 street, since most of what we do in limit hold'em is using street to street odds.

By definition, a 1 outer hits:
1 / 47 + 46 / 47 * 1 / 46 = 4.26% of the time by the river.
Similarly, a 2 outer hits:
2 / 47 + 45 / 47 * 2 / 46 = 8.42% of the time by the river.
A backdoor flush draw, mathematically, hits 4.16% of the time by the river; hence, it is equivalent to about 1 out, mathematically.

The second argument is about how much we should add to the mathematical out due to other factors, such as the savings on the turn when we do not pick up our flush draw. SSH says approximate it as 1.5 outs. Well, ninjia and I had this discussion many months ago, and he feels that 1.5 is a bit low of an approximation, and that it's closer to 2. That's what he's trying to show by his calculation here.

King, your calculation showing that the backdoor flush draw is 1.91 outs was as follows:
10 / 47 * 9 / 46 = 4.16%
4.16% * 46 = 1.91 outs

However, that is a 2-street out calculation. Using that logic, a regular flush draw is:
9 / 47 + 38 / 47 * 9 / 46 = 34.9%
34.9% * 46 = 16.1 outs which is clearly incorrect because we know a flush draw is 9 outs. The 16.1 outs is a 2-street figure, similar to the 1.91 outs you arrived at for the backdoor flush draw. Hence, Binion says to divide by 2 to get the actual mathematical outs, which is approx. 1.

[binions]

[ QUOTE ]
OMG, this is so simple to find, I don't get what you ppl are writting about. Just run all possible combinationas and find the EV, here is my crude and straightfoward method.

odds of hitting the cards we need on the turn is 10/47, river is 9/46

so to get through all the possible cases, i'm going to run it 2162 times

so first, u'll lose -2162 on the flop to see the turn card,

next, on the turn, 460 times u'll get your four flush draw. so u'll have to pay to see it, so -2*460

now, for u'r pay off, it will be final pot times 90. cause u'll hit 90/460 times.


so u hae FinalPot*90 = 2162 + 2*460 for 0EV


Final Pot = 34.24

so working backward, -2(him) for his call on the river, -2(you) -2(him) for the turn, -1(you) -1(him) for the flop bet, u'r left with 26.24

so pot needs to be 26.24 on the flop. with the pot odd on the flop, u can easily figure out the number of outs it equals.. add 1 for his bet, and u get the pot odd you need.

which is 27.24

so u need to get 27.24:1, so that is 1/27.24 * 47 = 1.73 outs. (out/47* pot = 1 ...out = 1/pot *47 )

Q.E.D.

[/ QUOTE ]

47*46=2162 possible combinations.

Assuming one opponent:

37*46=1702 of those combinations, you lose 1 small bet.

10*37=370 of those combinations, you lose 1 small bet and 1 big bet, or 3 small bets.

So, when we miss, we lose 2812 small bets.

We hit 90 times. 2812/90= 31.2.

So to break even, we need to win 31.2 small bets if we hit.

We win 1 big bet (ie 2 small bets) on the turn and river from our opponent when we hit.

31.2 - 5 = 26.2 is the number of small bets we need in the pot on the flop in order to call 1 small bet.

[binions]

[ QUOTE ]
[ QUOTE ]
OMG, this is so simple to find, I don't get what you ppl are writting about. Just run all possible combinationas and find the EV, here is my crude and straightfoward method.

odds of hitting the cards we need on the turn is 10/47, river is 9/46

so to get through all the possible cases, i'm going to run it 2162 times

so first, u'll lose -2162 on the flop to see the turn card,

next, on the turn, 460 times u'll get your four flush draw. so u'll have to pay to see it, so -2*460

now, for u'r pay off, it will be final pot times 90. cause u'll hit 90/460 times.


so u hae FinalPot*90 = 2162 + 2*460 for 0EV


Final Pot = 34.24

so working backward, -2(him) for his call on the river, -2(you) -2(him) for the turn, -1(you) -1(him) for the flop bet, u'r left with 26.24

so pot needs to be 26.24 on the flop. with the pot odd on the flop, u can easily figure out the number of outs it equals.. add 1 for his bet, and u get the pot odd you need.

which is 27.24

so u need to get 27.24:1, so that is 1/27.24 * 47 = 1.73 outs. (out/47* pot = 1 ...out = 1/pot *47 )

Q.E.D.

[/ QUOTE ]

47*46 = 2162 possible combinations.

Assuming one opponent:

37*46 = 1702 of those combinations, you lose 1 small bet.

10*37 = 370 of those combinations, you lose 1 small bet and 1 big bet, or 3 small bets.

So, when we miss, we lose 2812 small bets.

We hit 90 times. 2812/90 = 31.2.

So to break even, we need to win 31.2 small bets if we hit.

We win 1 big bet (ie 2 small bets) on the turn and river from our opponent when we hit.

31.2 - 5 = 26.2 is the number of small bets we need in the pot on the flop in order to call 1 small bet.

[/ QUOTE ]

Correction: 31.2 small bets is the final pot we need to win to break even, so that means 31.2 - 2 big bets on the turn (ours and his) - 2 big bets on the river (ours and his) - 1 small bet on the flop (ours) is 22.2 small bets on the flop that we need to call 1 small bet on the backdoor flush draw against one opponent.