Problem with outs....

[arabie]

I'm noticing far too often that i am incorrectly counting my outs. Once my opponents cards are revealed at the end, the outs/cards i was counting are quite often sitting in my opponents hands (add to this what i don't know from folding hands!). This must have a huge affect on the pot odds i've been relying on and can't seem to find a mathematically sound way of remidying the problem. Does anyone have a mathematical system for taking off outs (for example always consider a flush draw 7hearts vs 9 still drawing,(with a logical justification though))

I am not talking about finding tells or putting players on hands and then taking off the outs because this is already a given.

[arabie]

Comon guys and gals... this post needs a response! /images/graemlins/smile.gif thanks a lot in advance to those who take te time

[arabie]

Comon guys and gals... this post needs a response! /images/graemlins/smile.gif thanks a lot in advance to those who take te time

[SossMan]

I'm not sure what you're getting at here.

If it's the old, "my opponents may have folded my flush outs preflop" argument, well, that's simply false. You cannot discount your flush outs and not discount the non-flush outs.

[arabie]

So your answer is suck it up? Do you not think that if you are overcounting your outs, you can change you from a profitable to a non-profitable player?

[SossMan]

[ QUOTE ]
So your answer is suck it up? Do you not think that if you are overcounting your outs, you can change you from a profitable to a non-profitable player?

[/ QUOTE ]

there is no such thing as overcounting your outs if what you are talking about is fearing that some of your outs were folded preflop, or are in the burn pile, or are on the bottom of the deck....unknown cards are unknown cards.

give a concrete example for me, and I'll see if i can help...

[arabie]

okay, lets say Hero has Ah Jh on the button
UTG (Ad 5h) folds, UTG+1 (7h 2h) folds, UTG+2 (Jd Jc) calls, MP1 (Kh 4h) folds, MP2 (As 3h) calls, CO (10s 9h) calls, Hero raises, SB (7d 4d) calls, BB (Ac 3s) folds. Everyone else calls.

Flop (11 SB): 10h 2d 6s
UTG+2 bets, MP2 folds, CO raises, (14 SB) Hero?

now i've got:
1.5 outs for backdoor striaght
1.5 outs for backdoor flush
1.5 outs for each overpair
=6 outs needing 6.7:1 to see the turn.
the pot is offering 7:1 thereby justifying the call.
Now, 7 hearts are missing, 3 aces are missing, 1 king is missing for my backdoor straight, and 2 jacks are missing.

Now are my odds still correct to call? don't forget this includes the fact the overpair outs have been cut down because they will only make the best hand about half the time.

[arabie]

okay, lets say Hero has Ah Jh on the button
UTG (Ad 5h) folds, UTG+1 (7h 2h) folds, UTG+2 (Jd Jc) calls, MP1 (Kh 4h) folds, MP2 (As 3h) calls, CO (10s 9h) calls, Hero raises, SB (7d 4d) calls, BB (Ac 3s) folds. Everyone else calls.

Flop (11 SB): 10h 2d 6s
UTG+2 bets, MP2 folds, CO raises, (14 SB) Hero?

now i've got:
1.5 outs for backdoor striaght
1.5 outs for backdoor flush
1.5 outs for each overpair
=6 outs needing 6.7:1 to see the turn.
the pot is offering 7:1 thereby justifying the call.
Now, 7 hearts are missing, 3 aces are missing, 1 king is missing for my backdoor straight, and 2 jacks are missing.

Now are my odds still correct to call? don't forget this includes the fact the overpair outs have been cut down because they will only make the best hand about half the time (on top of sometimes putting me in a dominated situation).

[SossMan]

the problem is that you don't know what the other players folded, so how can that factor into your decision making for the hand?
You don't know if they folded your backdoor st8 outs, or your backdoor flush outs, or all the blanks.

You simply have to use the information presented to you. Unknown cards are unknown and therefore are treated with the same frequency as cards still left in the deck.

[MLG]

All right, I think I see your conceptual problem here. When you calculate the pot odds that you need, you are calculating the amount of outs that you have devided by the number of unknown cards. Those unknown cards include the cards that have been folded. For example after the turn there are 46 unknown cards (52 cards in a deck - 4 visible cards - two cards in your hand), so when calculating pot odds you divide your number of outs by 46. This calculation includes the chance that your outs have been folded.

[SossMan]

[ QUOTE ]
All right, I think I see your conceptual problem here. When you calculate the pot odds that you need, you are calculating the amount of outs that you have devided by the number of unknown cards. Those unknown cards include the cards that have been folded. For example after the turn there are 46 unknown cards (52 cards in a deck - 4 visible cards - two cards in your hand), so when calculating pot odds you divide your number of outs by 46. This calculation includes the chance that your outs have been folded.

[/ QUOTE ]

reminds me of the old bitter rocks at the card room who say "you know what the odds are of hitting a flush draw w/ 4 on the flop? 50-50, it either gets there, or it doesn't."

[arabie]

Okay, i don't think you guys are understanding my problem. I know how to calculate pot odds, i more than clearly understand that the folded cards are ignored in these calculations and because of my knowledge of this, i am writing these posts. My point is that ignoring the folded cards could be detrimental to one's game because one will often end up drawing to more outs than are available. My previous example shows that the odds were correct in drawing if you count your outs, however, when you see the folded cards the hand becomes incorrect to call. Therefore, if you are consistantly calling on incorrect odds, due to the fact that the correct amount of outs can only be known through your opponents hands, you are in mathematical jeopardy and easily have the potential to lose in the long run. I figured this problem would've been addressed many times and I am simply looking to see how others deal with it. Are you guys trying to tell me im hullicinating this concept?

-Adam-

[AceKQJT]

I know where you're coming from:

A-2 and A-5 are folded pre-flop. You hold A-A. In some weird twist of fate, the pot is laying you 30:1 odds on the Turn when you *KNOW* you need trips to win (because you saw your opponents hole cards when he peeked, and you know he has a smaller set), so you call. Unfortunately, you are drawing completely dead, and therefore were not actually getting the odds to call in this situation.

Well, you have to do the best you can with the information you have available. If you credit yourself with 21 outs, and someone has folded 2 of them, it doesn't really affect your decision much.

I suppose that if you based your decision on the theoretical outs that you have, that will help. For instance, if you calculate 21 outs, and a call comes to a slighlty +EV, then you make the call. If you have 2 calculated outs, you need EV to be very much in the plus to make the call. This should help compensate for those outs that you have counted that are not available.

--Casey

[BruceZ]

[ QUOTE ]
Okay, i don't think you guys are understanding my problem. I know how to calculate pot odds, i more than clearly understand that the folded cards are ignored in these calculations and because of my knowledge of this, i am writing these posts. My point is that ignoring the folded cards could be detrimental to one's game because one will often end up drawing to more outs than are available. My previous example shows that the odds were correct in drawing if you count your outs, however, when you see the folded cards the hand becomes incorrect to call. Therefore, if you are consistantly calling on incorrect odds, due to the fact that the correct amount of outs can only be known through your opponents hands, you are in mathematical jeopardy and easily have the potential to lose in the long run. I figured this problem would've been addressed many times and I am simply looking to see how others deal with it. Are you guys trying to tell me im hullicinating this concept?

-Adam-

[/ QUOTE ]

This post (http://forumserver.twoplustwo.com/showthreaded.php?Cat=&Number=566041&page=&view=&sb =5&o=&vc=1) should provide a detailed answer to essentially the same question you are asking. If not, come back.

[SossMan]

Okay, i don't think you guys are understanding my problem.

I understand your problem perfectly.

Maybe this will help:

On average you will have 9 outs (for a nut flush draw).
Sometimes the opponents would have folded an out preflop. Sometimes they will have had none of your suit preflop. If they have none of your outs preflop, according to you, you still have 9 outs right? 9 out of what? 47 unseen cards, right?
No. If 8 people folded preflop and nobody had any of your suit, then you have to subtract those 16 cards from the calculation and now you have 9 out of 31 (29% twice). All of a sudden you are 50-50 to hit your flush!!
If they folded 3 of your outs preflop, now there are 6 outs out of 31 and you are 35% to hit your flush by the river.
Maybe they folded 6 of your suit, and now you only have 3 left out of 31 and you would be about 18% to hit by the river.

Average all the possibilities out and you come to the correct (and same) answer as simply taking 9/47 twice (35%).

get it now?

[Cerril]

The only time you're miscounting your outs is if you have a reason to think they're in someone else's hand. If not, then the probability that they're holding them (that the hand that's being played contains those cards) is too difficult to bother with calculating on the fly.

You're figuring your outs based on the number of unseen cards, which includes the ones in each of your opponents' hands. So on the flop you can only discount 5 cards from the deck (your two and the flop).

You're giving your opponents credit for -possibly- having some of your outs (one of the reasons you generally reduce your overcard outs is the possibility of being reverse dominated), but in general you don't give them a higher probability that they have those card than any two other unseen cards. If you decided to calculate independenly what the chance of you drawing out on the river if they don't hold your outs and if they did hold your outs, weight both options by their respective probabilities (multiple probabilities if you're wondering about multiple outs) then you're going to get the same answer as if you just figured you were drawing from a full 46 card deck (rather than 46 - opponents' cards).

If you have a read that says 'pocket aces' when you're trying to complete an OESD (TJ)QK, then you can consider reducing your outs (and the cards left) by those aces. Generally though you can't be so sure that it's worth changing calculations around.

[ThinkQuick]

there is a recent post about this somewhere else, can't find it tho.

If your're looking for a mathematical remedy, then consider that the PROPORTION of outs in the deck is consistent with your calculations.
With 10 people in, there is a good chance that they had the outs you need, but since you can't know then you do the calculations we all know:
i.e. you have 10 outs after the turn with 46 unknown cards remaining

if you tried to account for the 18 out of play cards held by players, then you'd get more complex math with the same result.
ie. with 10 outs, and 6 cards already known, on average your opponents will hold 10*18/46 = 3.91 of your outs.
That means that (10-3.91)=6.09 outs remain in (46-18)=28 cards left.

46cards:10outs EQUALS 28cards:6.09 outs
I hope that's the mathematical reconciliation you were looking for,
Jonathan

[pudley4]

Take a very simple card game. There are 4 cards: 1, 2, 3, 4. Someone offers to bet you $5 vs your $1 that you can't pick card #1. The catch: he will remove one card (randomly, without looking) before you get a chance to pick. Do you take the bet?

OF COURSE!

But what about the times where he removed card #1 - you're drawing dead!

It doesn't matter.

If you look at it before he removes a card, you have a 1 in 4 chance of winning. If you look at it after he removes a card, you need to look at 2 scenarios: one where he removes the winning card, and one where he doesn't remove it.

He will remove the card 1 out of 4 times. These times you win 0%. 0% * 1/4 = 0.
He will remove a different card 3 out of 4 times. These times you will win 33%. 33% * 3/4 = 1/4

Add these two scenarios together (0 + 1/4) and you get.......

1/4 (the exact same as if you disregard which card he removes)


This is the exact same situation as when you are counting your outs - it doesn't matter whether the outs are in your opponents hands or not, because you don't know where they are. They could be in their hands, in the muck, in the burn cards, in the bottom of the deck, or they could be coming on the turn/river.

If you don't know, it doesn't matter.

[elwoodblues]

[ QUOTE ]
My point is that ignoring the folded cards could be detrimental to one's game because one will often end up drawing to more outs than are available

[/ QUOTE ]

Only if you know the value of the folded cards. You don't. There might be times when somebody who's out of the hand will say "I folded the nut flush." Okay, maybe then you can discount the Ace. Or if your opponent ONLY (100% of the time) open raises with AA -- you may want to discount 2 aces if he raised pre-flop.

[Mike Haven]

not quite what you are asking, but along the same lines:

i thought Ed Miller came up with a good concept that personally i can't remember noticing in print before, which is that if you are, say, drawing to overcards, instead of counting your outs as six, count them as three (four? i don't have his book in front of me) to allow for the times that you hit one of your cards but you lose anyway

obviously if you are drawing to an Ace four-flush you believe you will win if you hit, so you would count nine outs, but if you are drawing to an Eight-high four-flush you might consider reducing your outs-count by one or two before calculating whether or not you have the pot odds to call?

[js13_tps]

just looking at this from the flush perspective, does the fact hero has to catch runner-runner hearts to make the flush change the math any as opposed to needing either/or on the next two cards?

[Mike Haven]

as you know, runner runner anythings are difficult to hit

my own belief is that if you need runner runner flush cards to an Ace three-card flush you can probably call the flush draw the equivalent of two outs when you are checking pot odds on the flop

whether or not you hit the first flush card on the turn obviously changes things to your having a nine-outer or a zero-outer as far as the flush draw on the turn is concerned

if you have say an Eight-high three-card flush on the flop, i would think you should count it only as the equivalent of a 1-outer in respect of pot odds calculations

there is no reason not to give half-outs to draws - you might decide that a Queen-high three-card flush draw is worth the equivalent of one-and-a-half outs on the flop