Backdoor flush = 2 outs.

[Mike Haven]

MicroBob was asking about equivalencies a day or two ago.

Here's the proof of the "outs value" of a backdoor flush, for your notebook:

with a two-outer on the flop you bet 0.5BB 45 times out of 47 and miss, so you have to bet another 1BB to hit twice out of 46

if you play 47x46 = 2162 hands, you hit 2x46 = 92 times and miss 45x46 = 2070 on first card, and on the 2070 misses you hit 2070/46x2 = 90 times

or, 182 hits out of 2162 hands = odds of 10.9 to 1, or 8.4% success rate

cost to hit two-outer 182 times per 2162 hands = 1081BB + 2070BB = 3151BB = 17.3BB per hit

***

with a backdoor flush you bet 0.5BB to hit the four-flush 10 times out of 47 and then have to bet another 1BB to hit the full flush 9 times out of 46

or, if you play 47x46 = 2162 hands, you hit 10x46 = 460 and miss 37x46 = 1702 on first card, and on the initial 460 four-flush hits you then hit the full monte 90 times, giving 90 flushes out of 2162 hands = odds of 23 to 1, or 4.2% success rate

the cost to hit a backdoor flush 90 times per 2162 hands = 1081BB + 460BB = 1541BB = 17.1BB per hit

hence, as the cost is exactly the same, you can count a backdoor flush draw to be exactly equivalent to a two-outer, ON THE FLOP (as soon as the turn is dealt, you either have a new nine-outs or a new zero-outs, as far as the flush draw part of your hand is concerned)

[sthief09]

[slavic]

Darn Mike I didn't need you tell me I had more outs than I was counting a BD flush for. I'm going to have to call more now.

However because of the duel bet nature of the out I may leave my 1.5 figure in place.

[Nemesis]

what about a back door straight?

[slavic]

They are a little weaker than a BD flush. But I normally throw in 1 to 1.5 outs for my in game figuring. It doesn't have to be perfect just close.

In either case you are rarely calling on just the flush or straight alone. However 2 overcards and a BD flush can suddenly make a fold a call.

[me454555]

I'm not sure if my math is wrong but I think its worth more than 2 outs.

W/a runner runner flush assuming your going to fold if the turn is a blank and your opponent will check call the river regardless of what card falls. here's your outcomes.

Lose $.5 79 times = -$39.5 (Folded on turn)
Lose $1.5 16 times = -$24 (Folded on river)
Win $2.5 5 times = $12.5 (Hit Draw)

Grand total for 100 tries = -$56

With 6 outs on the river, your stats look like this.

Lose $1.5 76 times = -$114 (Folded on river)
Win $2.5 24 times = $60 (Hit Draw)

Grand total for 100 tries = -$54

Again I'm assuming your opponent will call no matter what card falls on the river.

Could this be right? B/c of the fact that you'd fold a runner, runner draw if the turn card doesnt hit combined with the fact that the flop bet is half the size of the turn and river bets, a runner runner draw is about equal to six outs on the flop???

Someone show me where I'm going wrong

[Ed Miller]

This is not right. The flaw is that you are automatically making what is usually a bad turn call with a two-outter. This is essentially what you are saying:

"If you badly misplay a two-outter by always calling a bet on the turn, then a backdoor flush draw becomes worth as much as that two-outter."

Of course, if you play the two-outter correctly, it will be better than the backdoor flush draw.

[Ed Miller]

BTW, I discuss this topic (equivalent outs for backdoor flush and straight draws) in my book.

So many people have sent me emails asking what will be in the book that I have decided to mention, when the topic comes up on the forum, if it is treated specifically in the book.

[sthief09]

you aren't considering the fact that you're going to win more than $2.50

you have to only compare what you lose, not what you win. like this:

(all calculations assuming your chances of hitting a draw is .04x. this makes it a lot easier to follow)

Backdoor:
Lose .50 79 times = -$39.50
Lose 1.50 16 times = -$24
Win y 5 times
= -63.50 + 5y

x Outer:
Lose 1.50 (100-4x) times = -150 + 6x
Win y 4x times = 4yx
= 6x + 4xy - 150

-63.50 + 5y = 6x + 4xy - 150

now I'm getting 2 variables. It seems as though the size of the pot affects which draw is worthwhile. Taking an extreme case, if we're in a y = $.00001 pot, then it's not worth even drawing, but since you're drawing to such a small pot, you're better off folding more. that is why in small pots, where you have no right drawing, it's better to draw to a backdoor flush, because the draw won't cost as much.

but now let's suppose it's a real pot, where you are getting odds to draw. that is, the pot, y, you win will be greater than or equal to your effective odds. mathematically we are making a second equation so we can solve it as a system...

your effective odds are 1.50*(1-.04x), and you're winning y*(.04x)

set them equal and you get: y >= 37.5/x - 1.5

meaning the pot has to be y or greater for it to even be worth chasing an x outer. the more outs you have, the smaller the pot you need. x = 14 should yield y = 0 since you're getting even odds, but it doesn't since the 4x approximation starts to fall apart when your outs increase above 11.

go back to the original equation:
-63.50 + 5y = 6x + 4xy - 150

combine it with:
y >= 37.5/x - 1.5

yields x ~ 2.7 outs

this number might be high since I used the stupid 4x approximation

as the pot gets bigger, x decreases, as your backdoor draw becomes less valuable in comparison to the actual x-outer

I have no idea whether this is right. I had a choice between doing this or reading Crime and Punishment for class, so I did this.

please comment

[me454555]

[ QUOTE ]
Backdoor:
Lose .50 79 times = -$39.50
Lose 1.50 16 times = -$24
Win y 5 times
= -63.50 + 5y


[/ QUOTE ]

If Y is the size of the pot, then the minimum pot size you'd want to win is $12.7 or 8.5 times your investment of 1 sb (.5) and 1 bb (1).

So... If the minimum size pot you need to break even on a runner runner is 12.7. Then Y=12.7

[ QUOTE ]
x Outer:
Lose 1.50 (100-4x) times = -150 + 6x
Win y 4x times = 4yx
= 6x + 4xy - 150


[/ QUOTE ]

Then 0=6x+50.8x-150 (The minimum number of outs you'd need to draw to the same size pot)
and x=2.64085

The number of outs youd need to draw to the same size pot is 2.65. So does this mean that a runner runner is worth 2.65 outs?

[sthief09]

Y is dependent upon X

and as for the results, they are somewhat meaningless, as I made a lot of esimations. I just wanted to show you where your mistakes were

[Mike Haven]

With the greatest of respect, Ed, in the context of my post, what I have said is correct.

I have spent some time pondering how you could possibly say that what I have proved in mathematical form "is not right", and I have come to the conclusion that you must be comparing value in terms of money won, rather than using my comparison in terms of cost to proceed. Presumably your thinking is along the lines of looking at S&M Hand Groupings - everyone knows that an eight-high hand is hardly anything to write home about, but an 87s can win much more money than a hand that might appear to be a better starter when the right flop falls with it?

I am in no way saying, "If you badly misplay a two-outter by always calling a bet on the turn, then a backdoor flush draw becomes worth as much as that two-outter."

I am saying that in the case of having a backdoor flush, where this unique state can exist only in the few seconds before the turn card is dealt, it is obvious that there is some value in having it, but as soon as the turn card is dealt, that value either disappears entirely, or, becomes a new and very important part of the drawing value of the hand.

It has always been a puzzle as to how to give the backdoor flush an easily useable and accurate value. As slavic has said, many of us have used one and a half outs as its value for a long time, without really knowing why. My calculations have shown that we should be using the added value of a backdoor flush by counting it as an additional two outs at that single specific point of decision in a relevant hand of comparing a bet to the pot odds available.

If a four-person pot is $95 and we have As9s on Kd7s3h we may decide to call for $10, whereas probably we should never call with As9s on Kd7c3h.

As I am positive you realise, it is absolutely irrelevant if someone with 88 does or doesn't play their hand with their two outs on the flop or the turn, but, yes, at the single, specific point of decision on the flop, they would have as much chance of finding a third eight by the end of the hand as you have of filling your flush, IF they played out the hand.

Please feel free to pm me, in total confidence, of course, the section of your book that comes to a different conclusion of the "outs value" of a backdoor flush and I will be delighted to give you private comments, prior to publication.

Good luck with your book.

[Ed Miller]

...at the single, specific point of decision on the flop, they would have as much chance of finding a third eight by the end of the hand as you have of filling your flush, IF they played out the hand.

That's not right, though. You have a (10/47)(9/46) = 4.2% chance to complete a backdoor flush draw. You have a 1 - (45/47)(44/46) = 8.4% chance to complete a two outter by the river.

And I think you are missing an important point. The value of a draw is determined not only by how often it comes in, but also in how EARLY IN THE HAND you get information about whether it's going to come in.

Say you are playing a variant of hold 'em where the turn card is always black (a spade or club) and the river card is always red (a diamond or heart). A spade flush draw would be worth more than a diamond flush draw in such a game.

A backdoor flush draw differs from a one or two out draw because you get MORE information on the turn about whether you are going to get there than you have with the normal draw.

EDIT: I apologize. I reread your original post, and it seems you have tried to take this into account. I'll go back through it in more detail.

[Chris Daddy Cool]

When can I pre-order this moneymaker?

[maurile]

Abdul's "Theory of Sucking Out" counts a backdoor flush draw as 1.5 outs for purposes of deciding whether to chase on the flop. (I'm putting aside his comment that it should really count as one out since a lot can go wrong with it. For now, I don't care what can go wrong with it -- I'm just interested in the likelihood of making the flush, as measured in terms of third-street outs.)

Abdul didn't show his work on this point, but I agree with his result of 1.5, at least as an approximation. Here's why.

To nail the backdoor flush, you need to hit one of your 10 outs on the turn, followed by one of your 9 outs on the river. If you get to a four-flush on the turn, you'll complete the flush 9/46 = 19.6% of the time. Since hitting one of our 10 outs on the turn will only give us the flush 19.6% of the time, we have to discount each of those 10 outs by 19.6%, meaning that we really have the equivalent of 1.96 outs before the turn. (Another way to think about this is that you'll hit your runner-runner 4.16% of the time, which is just slightly less than the 4.26% of the time you'll hit a two-outer on the turn. It is no coincidence that 1.96 is just slightly less than 2.)

So where does the 1.5 come from? It comes from the doubling of the bet size on the turn. You'll have worse pot odds on fourth street than you did on third street, so it will take a larger number of outs to make a call on fourth street worthwhile. Accordingly, each out on fourth street is worth less than an out on third street. And since we're measuring the value of a backdoor draw in terms of third street outs, we have to convert our 9 fourth street outs into their equivalent value in third street outs before we figure them into our equation.

Doing so is tricky because it depends on the number of players in the pot at each stage in the hand, how many raises there have been (if any) on each round of betting, etc. For example, if there are raises before the flop and on the flop, but not on the turn, the pot odds will often be better on the turn than they were on the flop. In a typical hand, however, the pot odds on fourth street will generally be 65% to 75% of what they were on third street. (E.g., 6:1 on third street and 4.5:1 on fourth street with no raises, 4 people seeing the flop, 3 seeing the turn, and 2 seeing the river. Or 10:1 on third street and 6.5:1 on fourth street under that same scenario except with one pre-flop raise.)

The value of an out is directly proportional to our pot odds. So if our pot odds on fourth street are typically only 75% of our pot odds on third street, each out on fourth street is worth about 0.75 third street outs. Therefore, the value of our 9 outs on the river as part of our backdoor draw is really more like 6.75 outs if we keep our units constant and measure everything in terms of third street outs. So our 10 outs on third street toward our backdoor flush should be discounted not by 19.6%, but by 14.6%, meaning that a backdoor flush draw is worth approximately 1.5 third street outs under normal circumstances.

[uuDevil]

Mike,

Hope this isn't throwing sand in the gears, but as I'm having trouble reconciling the various arguments....

It looks like you've come full circle from this thread. (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=43984&page=&view=&sb= 5&o=&vc=1)

I had used a value of 2 til I saw a post referring to this old Dynasty post (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=105679&page=&view=&sb =5&o=&vc=1) quoting Sklansky, so I now discount the value to "a little more than" 1.

Now you're saying 2 again?

/images/graemlins/crazy.gif

[MicroBob]

just thought i'd chime in here.
thanks to mike for informing me of the post.

i feel, in a small way, responsbile for this little debate taking place.
kind of amusing that, even though it was my question that sparked this whole thing, i don't really understand the arguments.

will look over the posts a bit more closely and see if i can decipher the little nuances everyone is talking about.

[Al_Capone_Junior]

During the "olden daze," before 2+2 went to the new software, this topic was discussed in great detail. The conclusion finally came to pass that a backdoor flush draw could be considered as one extra out on the flop. It is unfortunate that the older archives are in digest form only, because that makes it nearly impossible to search out the old thread.

[ QUOTE ]
That's not right, though. You have a (10/47)(9/46) = 4.2% chance to complete a backdoor flush draw. You have a 1 - (45/47)(44/46) = 8.4% chance to complete a two outter by the river.


[/ QUOTE ]

In the table "probability of completing a hand" found in HPFAP, one out is listed as 4.4% chance of hitting by the river. This is comparable to Ed's calculation of a 4.2% chance of hitting a backdoor flush, which is also comparable with the general estimation of about 24:1 odds.

I respect your opinions Mike, as I always have, but I think I am going to take Ed's side here, at least until shown the error of my ways, in which case I'll revise my opinion gladly. /images/graemlins/grin.gif

al / dave

[sthief09]

[ QUOTE ]
During the "olden daze," before 2+2 went to the new software, this topic was discussed in great detail. The conclusion finally came to pass that a backdoor flush draw could be considered as one extra out on the flop. It is unfortunate that the older archives are in digest form only, because that makes it nearly impossible to search out the old thread.

[ QUOTE ]
That's not right, though. You have a (10/47)(9/46) = 4.2% chance to complete a backdoor flush draw. You have a 1 - (45/47)(44/46) = 8.4% chance to complete a two outter by the river.


[/ QUOTE ]

In the table "probability of completing a hand" found in HPFAP, one out is listed as 4.4% chance of hitting by the river. This is comparable to Ed's calculation of a 4.2% chance of hitting a backdoor flush, which is also comparable with the general estimation of about 24:1 odds.

I respect your opinions Mike, as I always have, but I think I am going to take Ed's side here, at least until shown the error of my ways, in which case I'll revise my opinion gladly. /images/graemlins/grin.gif

al / dave

[/ QUOTE ]


a backdoor flush is cheaper than one out. if you miss on the turn, you can fold. if you miss a 1 outer you call again (unless you don't have odds to call)

[Al_Capone_Junior]

all decisions are implied to be taken on the correctness of pot odds.

al

[Bob T.]

In Mike's 'proof' he made the assumption that you were going to chase your two outer to the river. If that is true, then the value of chasing a backdoor flush is the same as the value of chasing a two outer to the river.

Hopefully we don't spend to much time chasing either two outers or only a backdoor flush to the river. When you have some other kind of hand, and have to make a decision about whether or not to see the turn, this kind of information is useful.

Good luck,
play well,

Bob T.

[Ed Miller]

...that David wrote a four page essay on the value of a backdoor flush draw in Poker, Gaming, and Life. I believe it's on p.106 (though I don't have the book in front of me).

In any event, I am almost certain that Mike's estimate of 2 outs is too generous.

I will discuss this with Mason and David before it's finalized in the book. So the book will contain an answer to this question that all three of us agree with.

[Mike Haven]

every time we look at our hand we compare it with the hands we deduce our opponents to have

normally we can see that we have nine outs, or eight outs, or five outs, or whatever, for our hands to catch up and overtake the leading opponent's hand

when we have a backdoor flush or straight we can't use the conventional outs-counting method because the ten outs or eight outs we need to improve our hand to a four-flush or an OESD still leave us well behind even if we catch, and still leave us with only a draw

therefore we need to establish a sensible and realistic comparison between drawing to a standard finite draw and to one of these rather imprecise situations

looking at my previous calculations we can see that AT THE MOMENT OF STARTING TO DRAW TO A BACKDOOR FLUSH OUR EXPECTATION OF SUCCESS IS ALMOST EXACTLY THE SAME, IN MONETARY EXPENSE TERMS, AS THAT OF STARTING TO DRAW TO A TWO-OUTTER

whether we finish drawing to improve to the flush, as with whether we finish trying to hit our two-outter, is irrelevant

the backdoor flush's outs-value of 2 applies only at the moment we have to decide whether or not we have enough outs to warrant drawing to our inferior hand when comparing the bet to be made with the pot odds available

finally, let's use a specific example to prove that the 2 outs is correct (we have to wonder why the pot is so large, but, whatever!)

we have Ac2c and our opponent has AdKd on a flop of Kc9h6s in a 2-4 game

ignoring runner-runner twos we have only our backdoor flush as a winner

the pot is 50

37x46 = 1702 times we lose our 2 = 3404 loss

10x46 = 460 times we have to call 4 more and we lose 370x4 = 1480 loss

3404+1480 = 4884 total loss

90 times we win 50+4 = 4860 win

this is proof that the pot odds needed to break even over 2162 hands were 50 to 2 or 25 to 1, which are the same odds as an equivalent 1.8 outter on the flop
QED

summary:

a backdoor flush is worth an additional 2 outs (actually 1.8 outs, but i have rounded up because of the reasonable likelihood of winning extra bets on the river) in terms of the cost of drawing

[Dylan Wade]

It would be more interesting to link us to the original post
at RGP (http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&oe=UTF-8&threadm=3a31e107.0303131615.c737df2%40posting.go ogle.com&rnum=2&prev=/groups%3Fq%3Douts%2Bto%2Bbackdoor%2Bflush%2Bgroup: rec.gambling.poker%26hl%3Den%26lr%3D%26ie%3DUTF-8%26oe%3DUTF-8%26group%3Drec.gambling.poker%26selm%3D3a31e107.0 303131615.c737df2%2540posting.google.com%26rnum%3D 2) than to copy it here. Mainly because of the Gary Carson factor. Interstingly enough Gary Carson published the equivalent outs of a backdoor flush as "2 outs" in his book. Later on RGP he admits his mistake. (http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&oe=UTF-8&threadm=99fd5b1.0207011545.7e4bf36e%40posting.go ogle.com&rnum=1&prev=/groups%3Fq%3Douts%2Bto%2Bbackdoor%2Bflush%2Bgroup: rec.gambling.poker%26hl%3Den%26lr%3D%26ie%3DUTF-8%26oe%3DUTF-8%26group%3Drec.gambling.poker%26selm%3D99fd5b1.02 07011545.7e4bf36e%2540posting.google.com%26rnum%3D 1)

I do not know whether it's 1, 1.5, or 2, but it does seem to confuse a lot of people.

[MicroBob]

"(actually 1.8 outs, but i have rounded up because of the reasonable likelihood of winning extra bets on the river)"


regardless of the calculations and debate...i don't like rounding up here.

probably off-topic, but of the 10 cards that you need to hit, 2 of them will pair the board unless the two non-flush cards are already paired. and if the board is already paired then the turn and river could 2-pair it or put 3 of fthe same rank out there.

i'm way way too lazy to try doing the calculations...and i would get them wrong anyway,....but i wonder if the extra bets you win when your flush is good (not even counting the slim chance of losing to a bigger flush) are counter-balanced by the occasional full-house that burns you.

the full-houses are rare obviously, but not rare enough to let you get away with rounding the 1.8 figure up to 2.0.
jmho.

if anything, i might consider rounding down to 1.5 for this reason.

[Mike Haven]

i take your point, MicroBob, but in the proof above i have used the completed backdoor flush winning only one BB on the turn, and none on the river, to break even

changing the figures only slightly from winning one BB on the turn and river combined to two and a half BB on the turn and river combined gives a profit if the 2 outs equivalency is used instead of the more exact 1.8 outs:


"finally, let's use a specific example to prove that the 2 outs is correct (we have to wonder why the pot is so large, but, whatever!)

we have Ac2c and our opponent has AdKd on a flop of Kc9h6s in a 2-4 game

ignoring runner-runner twos we have only our backdoor flush as a winner

the pot is 45

37x46 = 1702 times we lose our 2 = 3404 loss

10x46 = 460 times we have to call 4 more and we lose 370x4 = 1480 loss

3404+1480 = 4884 total loss

90 times we win 45+4+6 = 4950 win"

[bernie]

1 exception is that you have to factor in the cost of going 2 streets against your odds instead of just 1 street. You cant hit your 2 out hand on the turn the way you can say a set. You still dont have a made hand. The effective odds are a little different i think.

b

[John Feeney]

I bumped into this thread in searching for something else. It prompted me to go looking for an old thread in the archives. An hour later, I can't even remember what I was originally searching for. /images/graemlins/crazy.gif But I did find the old thread.

I don't have time to read all of the present thread, so I'm not sure what y'all have concluded. Nor am I sure if the old thread will shed any light. But I seem to remember slightly adjusting my own thinking about bdf outs as a result of it. So, FWIW, here it is:

http://www.twoplustwo.com/digests/aug99_msg.html#27088

[maurile]

[ QUOTE ]
It would be more interesting to link us to the original post
at RGP (http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&oe=UTF-8&threadm=3a31e107.0303131615.c737df2%40posting.go ogle.com&rnum=2&prev=/groups%3Fq%3Douts%2Bto%2Bbackdoor%2Bflush%2Bgroup: rec.gambling.poker%26hl%3Den%26lr%3D%26ie%3DUTF-8%26oe%3DUTF-8%26group%3Drec.gambling.poker%26selm%3D3a31e107.0 303131615.c737df2%2540posting.google.com%26rnum%3D 2) than to copy it here. Mainly because of the Gary Carson factor.

[/ QUOTE ]
It was my post to r.g.p. I just wanted to point that out so it didn't look like I was plagiarizing.

[maurile]

[ QUOTE ]
So, FWIW, here it is:

http://www.twoplustwo.com/digests/aug99_msg.html#27088

[/ QUOTE ]
Thanks. That was a good thread.

Here are some of the highlights:

Re: How many outs is a runner-runner flush draw?
Posted by: Abdul Jalib (AbdulJ@PosEV.com)
Posted on: Thursday, 26 August 1999, at 4:43 p.m.

Louie, the stuff you write is great. However, I am confused about one thing. Can you explain how counting a pure backdoor flush draw (where you plan to fold if you don't get your first card on the turn) should be considered about 1 out? If you work through the expected value (EV) calculations, I fear you'll find that you're not taking implied odds into account properly with that method.

Here's how I would do it, using a trick from Zagie on rec.gambling.poker:

First, assume that you will get your 4-flush. Then your EV on the for a 15 big bet profit by the river is (9/46)(15+1.5)-1.5 = 1.7 big bets. (That's the chance you'll turn your four flush into a five flush times the amount you'll win including your investment, quantity, minus your investment.)

Second, consider whether you would call with a regular flush draw with a shot at a pot of 1.7 big bets. The pot is offering you 1.7 to 0.5, or 3.4 to 1. (Converting from big bets to small bets doubles the pot size.) You have 10 outs to make your flush draw draw. Should you call on the flop? Your expected value is your chance of making a flush draw times the expected value of that flush draw including your bet (quantity) minus the bet it will cost you, i.e., (10/47)(1+3.4565)-1 = -0.0518 small bets, so you should fold.

Now, this was a best case scenario. The most common thing that can go wrong is that your flush cards pair or double pair the board and you lose to a full house. Another thing that can go wrong is if you're not drawing to the nut backdoor flush and lose to a higher backdoor flush. You also have to take into account that the turn could cost you two or more big bets. These hazards decrease both your effective outs and your effective pot odds (as you will have to pay through the nose when you get there and lose.) So, I think there is no question that a fold is correct here, assuming the 15 big bet profit estimate was correct.

Zagie was trying to make an exact EV calculation procedure for backdoor draws but was worried that it wasn't exact. I think he made a slight error, which I fixed. You can double check my answer by doing the calculation the more straightforward way:

(10/47)((9/46)(15)+(1-9/46)(-1.5))+ (1-10/47)(-.5) = -.0259 big bets or -.0518 small bets.

If you want me to show how to do the calculations a third way, a way that you can do in your head at the tables, I can. See also the link below:

-Abdul


Re: How many outs is a runner-runner flush draw?
Posted by: John Feeney (johnfeeney@home.com)
Posted on: Friday, 27 August 1999, at 4:55 a.m.

Abdul -- I'm not certain how to tie together what you're saying. You disagree with Louie's looking at a backdoor flush draw as equivalent to about 1 out. Then to make your point you provide calculations of EV for a bdf draw given a specific size pot. But I don't see how that speaks to the question of whether " like about 1 out" is reasonable estimate. (Though you were referring to having *only* the bdf draw, while Louie was really talking about it's significance in conjunction with other draws, I still wonder about what you were saying.) In fact, it appears that you're saying "Louie, it's not as big a draw as you're estimating it to be. See, it's -EV with this 15bb pot." But if you were to use the 1 out estimate for a lone bdf draw, and really treat it as you would a draw to 1 card, you would quickly conclude that that pot was far too small, agreeing with you're saying - that it's -EV in that situation.

Also, in your "Theory of Sucking Out " don't you equate a bdf draw with 2 outs?

I've long used the same 1 out estimate that Louie and Chris mention basing it on the reasoning Louie gives below. I do think it makes sense when you're dealing with a bdf draw along with other outs. For just the bdf draw alone, I've used the reasoning David provides in the essay I mention in my other post in this thread. Comments?

John Feeney

Re: How many outs is a runner-runner flush draw?
Posted by: Abdul Jalib (AbdulJ@PosEV.com)
Posted on: Friday, 27 August 1999, at 3:21 p.m.

There are two perspectives. One says that your outs with two cards to come are the same as your outs with one card to come, although you actually distinguish them as apples and oranges, 2-draw-outs and 1-draw-outs. The other perspective says that you think of your outs with 2 cards to come as if you only had 1 card to come - this is the perspective I usually use. I say a straight draw on the flop has 16 outs (actually 14 1/2 outs) on the flop, you say a straight draw on the flop has 8 outs. I say tomayto, you say tomahto.

The reason I say a backdoor flush draw is worth almost 2 outs if you're going to the river is that (10/47)(9/46)=x/47 means that x is 1.9565.

Feeney became confused when I started off bickering about whether the outs should be considered 1 or 2 and then I proceeded into a calculation that didn't really use either number. The thing is, if you are going to fold on the turn if you don't make your draw draw, then you cannot use the 2 out estimate without making an adjustment, but I don't see how you can use the 1 out estimate either.

Here's yet another way to do the calculation, but using 2 outs after making an adjustment to the amount you have to call. 10/47 of the time you take a card off on the flop, your backdoor flush draw will turn into a real flush draw on the turn and you'll wind up putting in a big bet there. So, you'll put in an average of 1+10/47=1.42 small bets; let's round up to 1.5. The pot profit is expected to be 15 big bets or 30 small bets, so you're getting 30 to 1.5 or 20 to 1 on a 47 cards to 2 outs or 23.5 to 1 shot. So, you should fold.

To do this in your head, I assume you have memorized that you're going to put in about 1.5 small bets by the start of the river in a best case scenario on average with your backdoor flush draw. Next estimate how big the pot will become not counting your future action: 30 small bets is what the original poster said. Next compute your effective pot odds: 30/1.5 is easier to think of as two thirds of 30, which is 20, 20:1. Finally see if you have odds to call: multiply your outs by one more than the effective pot odds and compare to the number of unseen cards, so that's 2(1+20)=2(21)=42<47. Since 42 is less than 47, you should fold. (See my Theory of Sucking Out webpage for a little more detail on that technique.) It's close in these terms, but not so close after you discount your outs and increase your expected cost to account for losing when you make your hand, getting raised on the turn, etc.

A heuristic I use based on past analyses is to not even consider calling on the flop with a backdoor flush draw unless the current pot is bigger than 10 big bets. (Exceptions: when heads up or with other outs.) Feeney said Sklansky put the minimum pot size at 15 big bets; consider 10-14 a gray area that depends on how many players will be hanging around on the turn.

Now, your turn. Show me how to do the math from the 1 out perspective.

-Abdul

[tripdad]

hey guys, i'm no mathematician, but to me this is fairly simple. on the flop, odds against hitting your 4 flush on the turn are 4.7 to 1. assuming you do, on the turn the odds to hit the flush on the river are 5.11 to 1. this makes the probability of hitting a runner runner flush 24.017 to 1.

the odds of hitting a 2 outer are 23.5 to 1.

so, a runner runner is aproximately equal to 2 outs.

this of course doesn't take into account how much it costs to draw, nor should it. just that you will hit the flush 1 in 24.017 times when you have a 3 flush on the flop.

(i think i remember reading in HPFAP where they approximate it as a 25 to 1 longshot, or approximately the same as spiking a set when you have a pair)

cheers!

[Lost Wages]

You cannot multiply odds like you can the probability of mutually exclusive events.

The probability of making a backdoor flush is 10/47 * 9/46 = 4.16% or 23:1.

Lost Wages

[tripdad]

[ QUOTE ]
The probability of making a backdoor flush is 10/47 * 9/46 = 4.16% or 23:1.

Lost Wages

[/ QUOTE ]

i believe if you divide 1 by 24.017, you get .0416371...4.16%

23:1 would be 4.35%

so, the probability of hitting a backdoor flush is 10/47 * 9/46 = 4.16% or 24.017:1

cheers!

[Lost Wages]

If the probability of an event is P then the odds against are (1-P)/P:1

10/47 * 9/46 = 0.4163 = 23.02:1

Edit:
The inverse operation, converting odds to probability. If the odds are X:Y then the probability is Y/(X+Y).

1/(23.02 + 1) = 0.4163

Lost Wages

[tripdad]

if the probability of an event=P, and the odds against are (1-P)/P:1, then the odds against me understanding any of these inversely operational proportionalactic math type posts are longer than hitting a backdoor flush....much longer.

cheers!

[Ed Miller]

this of course doesn't take into account how much it costs to draw, nor should it.

If you don't take into account how much it costs to draw, then the number is COMPLETELY worthless. Which draw is more valuable?

1. A six-outter that you must call a pot-sized bet to draw to?
2. A four-outter that you get to draw to for free?

EDIT: In fact, you can't even answer this question without knowing how much money you could potentially win AFTER you make your draw. Trying to assign value to a draw (and THAT is what you are trying to accomplish when you assign a backdoor draw a virtual number of outs) is meaningless without evaluating the monetary situation.

[Al_Capone_Junior]

I agree with your sentiments.

the thing I want to impart is that if you have a backdoor flush draw, your hand is only 4% better on the flop than if you didn't. That's like one extra out. It's virtually never worth chasing on its own, but if you are faced with a truly borderline decision, sometimes that extra 4% might turn a fold into a call. Such things are discussed in several 2+2 books, but I am too lazy to look up specific examples right now.

One thing that should be noted... if you pick up a flush draww on the turn, that makes your hand a LOT stronger equity wise. But you should not be chasing backdoor flushes just because you might gain equity on the turn. You must treat them for what they are worth WHEN YOU HAVE TO DECIDE. And here, we're talking about the flop. /images/graemlins/grin.gif

al

[Al_Capone_Junior]

The odds of hitting a two outer are in fact 23.5:1 ON THE NEXT CARD.

The odds of hitting a flush draw BY THE RIVER are about 24:1.

the odds of hitting a ONE outer IF YOU CALL TWO TIMES is about 22:1.

The rest of the discussions on this thread should explain fully the value of a backdoor flush on the flop! /images/graemlins/grin.gif

al

[Mike Haven]

as always, bernie, you have gone straight to the nub of the problem

it is exactly the fact that a bdf cannot be completed in one move that has caused us to have difficulties in assessing its value, over the years, with different suggestions having been made

in no way have i been suggesting its strength is the same as a two-outter pocket pair, which draw can be completed on the turn, with possibly further strength to come from the fall of a miracle river card

i have to repeat once again that my calculations show that in terms of the overall average cost of successful completion drawing to a bdf draw is very close to the overall average cost of successful completion of a draw to a two-outter, whether or not the latter draw is completed on the turn, before the attempt to successfully complete the draw is commenced

(as with any betting sequence, once the attempt to to improve your hand is commenced, anything can happen, and the attempt to improve is not necessarily completed, depending on other players' actions in the various betting rounds

(for instance, if you held A4s on a J74r flop you might decide you have 5 outs to proceed and you might call a $10 bet in an $85 pot - if you are then raised, reraised and capped you might decide to fold - this does not mean that your call was not correct, in pot odds/outs terms, at the moment of making it even though it might be imprudent to continue trying to complete the draw)

[bernie]

The BD draw is just an add on to other draws you may have.
Almost a bonus.

I think the implied odds of a 2 outer -pair-are higher since you may collect on 2 streets rather than 1. Even though it may cost the same to get there if you went both streets to complete it. This, imo, is one reason the BD draw is weaker than the pair draw.

[ QUOTE ]
once the attempt to to improve your hand is commenced, anything can happen, and the attempt to improve is not necessarily completed, depending on other players' actions in the various betting rounds


[/ QUOTE ]

Also factoring in what you 'anticipate' your opponents may do based on prior hands theyve played or, if live, you have a read of possible raises coming behind. Which, of course, makes closing the action rather than being near the bettor most desirable.

b

[tripdad]

this thread, if i'm not mistaken, is about determining the LIKELYHOOD of completing a backdoor flush, and assigning a value in terms of outs, not dollars. the value in dollars, of course is determined by the size of the pot, how many players are involved,etc...

honestly, anyone chasing a bd flush without getting 25 to 1 pot odds (highly unlikely) is a bit nuts, assuming the flush is his only way to win the pot. if you do get those odds, and you make it, you WILL be paid off by all those who stuck around to the river. the pot will be so huge, they will all be spending the money in their heads before showdown.

the point of my posts were to (in plain english) back up the assumption that the original poster of this thread wrote....a bd flush= 2 outs .i think he's right, and noone has proven otherwise.

BTW, i don't think the odds of hitting a ONE outer if you call 2 times is 22:1. it think it is 46:1 against hitting on the turn, then 45:1 against hitting on the river. how do you get 22:1 out of that.

cheers!

[tripdad]

no...it is just the number we are looking for. you're right, the number is useless if you don't take into account how much it is to draw, but i think that is determined by the individual game/pot/# of opponents, type of opponents, etc... the poster here just gives the outs, thats all.

can you tell me the "value" of a royal flush? no. in fact i won more money with a pair of sixes one session than i did with a royal flush the very same session.

it is up to each player to determine the value when he comes across the hand and decides whether or not to play it.

cheers!

[slavic]

anyone chasing a bd flush without getting 25 to 1 pot odds (highly unlikely) is a bit nuts

Those odds would work if you didn't have to put in another bet or two on the turn. So if you are all in on the flop it should be OK, but you almost never will be.

[Al_Capone_Junior]

[ QUOTE ]
BTW, i don't think the odds of hitting a ONE outer if you call 2 times is 22:1. it think it is 46:1 against hitting on the turn, then 45:1 against hitting on the river. how do you get 22:1 out of that.


[/ QUOTE ]

Direct quote from the table in HPFAP, "Probability of completing a hand"

No. of outs 1 percentage 4.4

this table refers to your chance of making it by the river from the flop, calling twice.

convert that to odds and it's about 22:1. if you want the decimal places, figure it out yourself.

As much as I respect Mike and his posts, he did NOT prove that a backdoor flush is worth two outs on the flop. If you cannot gather why from the conversations already posted in this thread, then believe whatever you wish.

The point tho is that this is the small stakes forum, where presumably there are novice players who may very well overvalue backdoor flushes, i.e. count them as being worth two additional outs on the flop. There is a big difference between a four and a six outer on the flop.

al

[tripdad]

if you don't improve with the turn card, you don't call the extra bet. if you do improve, you now get huge pot odds for a 5.11 : 1 chance to make your hand. in fact, you'd be getting good enough odds to call multiple raises.

cheers!

[BugsBunny]

Then there's a small error in HPFAP /images/graemlins/smile.gif With 1 out I make the percentage to be 4.255% (4.3 if rounding to the nearest tenth of a percent) at 22.5 to 1 odds.

1 - 46/47 * 45/46 = 0.0425531915

-bugs