Flush Draw...Calculating Outs

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Ranger390

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I've been meaning to ask this for some time regarding counting outs to a draw, particularly a flush draw. Here's the situation: Texas Hold'em, Full ring game or tournament table of 10 players. You look down at two suited cards and call the preflop bet. The flop shows two additional cards of your suit, so you have 9 outs to your flush. The 2/4 rules (remember, there is a slight error due to rounding) says that you will hit your flush 36% of the time if you see both the Turn and River cards. The odds of hitting on the Turn is 18%. But, if you miss on the Turn, the odds are still 18% that you will hit on the River.

Here's my question, when figuring outs, ALL unseen cards are counted as potential outs and there are 9 remaining cards of your suit that are unseen. However, ALL unseen cards are not currently in the deck, as each player at the table has already been dealt two cards each. That means that there are 18 cards that have been dealt to the to the players that you can not see and that can not be dealt of the Turn or River. In a thoroughly suffled deck, that means that the odds are that 4.5 of the cards of your suit were dealt to other players and are unavailable to be dealt on the Turn or River. So, out of the remaining 29 cards in the deck, only 4.5 of your suit remain. That's an 18% chance of completing your flush on the Turn and River instead of a 36% chance. OR, a 9% chance of hitting on the Turn and 9% to hit on the River. Under these circumstances, the odds of hitting your flush are not as attractive.

While it would be more difficult to figure odds in this fashion for other types of drawing hands, why are the odds of other players already being dealt the cards that you need not figured into odds calcualtions? Why is this reasoning off base? Is my math really fuzzy???
 
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Ben_Dover

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(# players don't matter ;-) If the 9 outs were truly shuffled evenly across the 47 unknown cards then that only burns 3.44 outs in 18 cards (one of your 'outs' every 5.22 cards) and leaves 5.55 outs available in the remaining 29 cards.... So a single draw in that case is:

5.55/29 = 19.1489% !

(same as 9/47=19.1489%)
 
qwerrk

qwerrk

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Did not check your math but if you are going to start calculating flush odds this way, then you need to rethink all the other reference point odds you have gotten familiar with like open-ended and gut shot straight draws.

One thing about the flop you describe is odds are high you are the only one with 4 of the suit but if you are playing, say, 7-6, and the flush comes on the turn, your odds are now shaded by anyone holding an over-card in the suit, they might hit on the river; should your winning odds be adjusted? It gets a little fuzzy :p
 
smokedu

smokedu

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(# players don't matter ;-) If the 9 outs were truly shuffled evenly across the 47 unknown cards then that only burns 3.44 outs in 18 cards (one of your 'outs' every 5.22 cards) and leaves 5.55 outs available in the remaining 29 cards.... So a single draw in that case is:

5.55/29 = 19.1489% !

(same as 9/47=19.1489%)
wow i am so not good at math
 
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Ben_Dover

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why are the odds of other players already being dealt the cards that you need not figured into odds calcualtions?

Same story for two card draws: the other 9 players don't matter because even if they hold a few outs, they also hold a bunch of cards that *aren't* your outs.

For completeness, in two clicks you can find that the regular two card flush draw (9 outs with 47 unknowns) is calc'd as:

(38/47*37/46)=65.03% of not hitting, or 34.97% of hitting

In your scenario, if you exclude the cards held by the other players (with 5.55 outs on average that remain available), then that leaves 29.55 unknown cards so hopefully that gives the same EXACT answer:

(24/29.55)*(23/28.55)=65.03% of not hitting, or 34.97% of hitting.

Exactamundo!
 
Stu_Ungar

Stu_Ungar

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Actually this is pretty interesting and something I'm going to do a bit of work on. I thought of this a long time ago and was told it had no bearing on anything so I lost interest and eventually forgot about the idea.

Traditionally we count all unseen cards as equal. But they are not. There are two types of unseen card. Those we could be dealt and those we couldn't.

So with 9 other players they hold roughly 2/5 of the unseen deck and therefore we should assume they hold 2/5 of the outs.

So in that situation the outs should be discounted 1/3 because to make table maths easier the discount should be a simple number favoring on the optimistic side of rounding so if the table held 0.4 of the deck, you would round down to 1/3 rather than down to 1/2 because you just want to take into account there are less odds but don't want to starve yourself of odds.

So instead of 9 outs at a full ring table, there should be around 6.

Miller, Sklansky and Malmuth talk about discounting odds in Small Stakes hold em but I have never read a book that advocates discounting odds as a function of players sat at the table.

Essentially this means that the smaller the table, the closer your odds become to the true odds. So on a 6 max table the discount should be around 1/5 so the flush is worth about 7 outs

With 4 people the discount becomes about 1/8 and so the flush is worth 8 outs.

With less than that the maths isn't worth the bother as it gets so close 9 its not worth discounting.

So in summary

full ring discount outs by 1/3
6 max discount outs by 1/5
4 max discount outs by 1/8
heads up - No discount needed.

Always round the discounted figure up and not down.

This would apply to all outs and not simply flush draws.

I see that this goes against traditional logic, but It does fit in nicely with the concept that play loosens up as the table size reduces.. Is this partly because draws increase in strength? Not all individual draws are as effected as greatly due to the rounding up. But the total outs for a hand should first be calculated and then the total outs be discounted in this way.

I also hear a lot of 6 max players say the enjoy 6 max more because they get more action, their bets are more respected and they dont get sucked out as often.. Is this because they have more outs for the same draw than a 9 handed table?


Makes for an interesting discussion.

Feel free to look over the maths and correct it if necessary but the actual discount figures are incidental, at this point, to the arguments for and against whether the discount is needed.

Same story for two card draws: the other 9 players don't matter because even if they hold a few outs, they also hold a bunch of cards that *aren't* your outs.

For completeness, in two clicks you can find that the regular two card flush draw (9 outs with 47 unknowns) is calc'd as:

(38/47*37/46)=65.03% of not hitting, or 34.97% of hitting

In your scenario, if you exclude the cards held by the other players (with 5.55 outs on average that remain available), then that leaves 29.55 unknown cards so hopefully that gives the same EXACT answer:

(24/29.55)*(23/28.55)=65.03% of not hitting, or 34.97% of hitting.

Exactamundo!

Now I do like that arugument too.. can we get others who have a good knowlege of statistics to compare the two ideas? One is a case for the discount and the other against it. And discuss which is correct and why.

Thanks
 
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Stu_Ungar

Stu_Ungar

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OK Ill leave the above post up for posterity!! ^^^^^^^^^^^^

Put some more thought into it and there is NO NEED TO DISCOUNT ODDS due to number of players.

Here is why.. I think its what DR was saying.

Yes there are effectively 2 packs of cards, 1 that you can be dealt from and one you cant.

Yes some of the outs are contained in the pack you cannot be dealt from (the cards which other players have been dealt)

However if you discount the outs then you also have to reduce the size of the pack you can be dealt from by exactly the same ratio. So if you remove 1/3 of the outs, you must be working from a pack of cards which also has to be reduced by 1/3.

The numbers just cancel each other out and so the odds remain the same as they were prior to the discount.

So the number of players does in fact reduce the number of outs you have available BUT it also reduces the size of the pack they will be dealt from. Thus the odds remain the same.

The total effect is that the number of players has NO EFFECT on drawing odds.
 
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Ranger390

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Thanks for correcting my math and for showing me why the unseen cards in other players hands make no difference in calculating outs. I realized that I was probably not the first person to think about this, but I could not figure out how and why it did not matter, since I was assuming that it did. Thanks for setting me straight!
 
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baudib1

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I think you should begin discounting outs when there are multiple players continuing after the flop.

Let's say you have KhQh in a limped pot and the flop comes Jh9c4h.

If 5 people see the flop and there is a 3/4 PSB bet thrown out, let's say four people continue after the flop. You decided to call because you have a strong drawing hand: 9 outs to a king-high flush, 3 more outs to a nut straight. Plus you might count 3 Ks and 3 Qs as possible outs if someone is betting on TP.

However, in a multiway pot, chances are running extremely high that some of your outs are gone and many of them may be dirty. Eg., limped pots invite people who play ATSC. It would be a disaster if the Ahxh is out against you. Also, someone could be betting/calling with TP and a straight draw themselves, so you could be reverse-dominated by KJ/QJ. QT is a possibility here, so hitting your K or Q would be disastrous and could give someone else the nuts or two pair, and that's one of your outs that's dead.
 
Stu_Ungar

Stu_Ungar

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I think you should begin discounting outs when there are multiple players continuing after the flop.

Let's say you have KhQh in a limped pot and the flop comes Jh9c4h.

If 5 people see the flop and there is a 3/4 PSB bet thrown out, let's say four people continue after the flop. You decided to call because you have a strong drawing hand: 9 outs to a king-high flush, 3 more outs to a nut straight. Plus you might count 3 Ks and 3 Qs as possible outs if someone is betting on TP.

However, in a multiway pot, chances are running extremely high that some of your outs are gone and many of them may be dirty. Eg., limped pots invite people who play ATSC. It would be a disaster if the Ahxh is out against you. Also, someone could be betting/calling with TP and a straight draw themselves, so you could be reverse-dominated by KJ/QJ. QT is a possibility here, so hitting your K or Q would be disastrous and could give someone else the nuts or two pair, and that's one of your outs that's dead.

In your example, with 4 to the second nut flush, Id play it as though I was holding the nut after the flop.

The Ah shouldnt be too much concern to you right now. Yes it can beat you but honestly, if you wait for situations where you have the absolute nuts, you wont be playing very often!!.

So the flush gives you 9 outs

The K may give you top pair, but then again it may not. It is certainly worth say 1.5 outs, (discounted from 3)

a Ten gives you the nut straight so id include 4 good outs there.

Infact one of those outs (Th) gives you the nut STRAIGHT FLUSH draw!!

So I make it 14.5 outs

THis gives you about a 30% chance of hitting buy the turn and 50% by the river.

Even if the Ah is out there, its far more likely that the guy dosnt have two hearts in his hand.. only one, the Ah.

Id say this is a good pot to build!!!

Get your money in.. and if you do suckout.. play the hand the same way next time.. in the long run this is a winning situation for you!!!
 
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it feels like everytime someone goes on me with a flush draw they win, but if i play it i never win.
 
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lmille4574

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you are going under the assumption that no one has or did have your outs. and the all your outs are live cards. Whether they are or not you will never know. but percentages speak for themselves. test on your own. I did
 
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sd great 1

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When calculating your percentages to win the hand you best first should put your opponent on a hand. Are you assuming they have the nuts at that point, or are they playing a mediocre hand, or are they on a draw as well. Once you put them on a hand, then take away any cards from your draw or add if you think you might have overs that could win as well.

Also if you are willing to call a large bet a on flush draw, why not put out a large bet yourself, if they fold to your bet your percentage of winning the hand just went up to 100% ; I like those odds !!
 
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dollabill315

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Its kinda easy at first to not see the obvious, but yes the "discounting" cancels each side out. Yes folded players hands might have contained ALL of ur flush draw outs(or whatever draw), but then again its a random hand so maybe EVERY card of that suit(except your hole cards and the 2 on board) are left. Sorry to ramble and probably repeat things already said... even though the topic is almost useless because, well nothing is effected... it was still a interesting read...

and Stu, if there is no need to discount odds from what we just discussed then what were Malmuth and them saying in the book you read?

Thanks.
 
Stu_Ungar

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and Stu, if there is no need to discount odds from what we just discussed then what were Malmuth and them saying in the book you read?

OK firstly that was nothing to do with discounting outs because other players MIGHT hold them. Just wanted to reiterate that for clarity.

When they discount outs they do so to give a truer picture of where you stand at that point and wether you are getting enough pot odds to continue.

Miller suggests discounting the odds of making say top pair by around a half.. so if you have 3 outs to top pair, you play as though you have only 1.5

Its done when there is a high chance that you make your hand you may still loose (you are drawing dead). You have no way of knowing this for sure and your hand cannot simply be folded because of the possibility of a better hand being out there.

So with a flush, you may have 9 outs, but how many of them are good? If you hold

7h 6h

the board is 5h 4h x

Then (ignoring the straight flush possibility!!)

If anyone holds 8h xh or above then you are drawing dead.

You cant easily fold just because this is possible ao assume you dont have a good enough read on someone to say for sure if the do or dont have a flush draw.

So here you discount the flush outs to say 6.

It means that you can continue, but not quite as agressively as if you felt all 9 outs were good and you need slightly better odds to continue if you feel you are currently behind. This helps make up for the times that you do hit your flush and loose to a higher flush.

If you instead held Jh xh then you would discount less.
 
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