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View Full Version : a newcomer's query - HE folds effecting winning chances

05-24-2003, 07:44 PM
been reading the forums for a couple months and have really enjoyed them. am venturing on to make my first post. i've only been playing online for a couple of months (HE and O8) but at least I am a semi-consistent winner. thanks to the various ideas on this board which have really helped get me started analyzing the game properly.

so pardon the potential ignorance of this question please - (as well as the length...explaining things ideas succinctly is not my strong suit). i know only a little bit about the various mathematics involved in determining which HE hand is at an advantage pre-flop but i believe it has to do with running a computer sim of all the possible 5 card combinations that can come out of the deck.

On the WPT on Travel Channel today (which I saw for the first time and which wasn't as bad as i thought it would be) they had a 77 vs. AK at one point. I can't remember if the AK was suited or not (i think it wasn't). They mentioned that this was pretty much a coin-flip hand (they were all-in). i believe they graphically gave the 54-46 advantage to 77.

I started thinking about my hunch as to how the advantage is generated and wondered if 54-46 is an accurate stat if there were other players involved at the table.

for example, if there were 6 players at the table...but 4 of them folded and it just came to those two guys pre-flop then I suspect the AK is a little stronger then 46% based on the number of folds. My thinking is that each fold represents a hand that is less likely, albeit slightly, to hold an ace. And that each fold is more likely, albeit slightly, to hold a 7.
Obviously, it's difficult to quantify since all players play differently at different times.

But I look at it this way. There are 48 unknown cards and there are 3 Aces remaining. The 4 folds (8 players) represents a slightly higher chance that there were few, if any, aces among those hands. Consequently, there is a slightly higher chance that the 3 A's are in the remaining 40 cards to be dealt (same for the King as well I should think).
However, if any of those 4 players held one of the remaining two 7's they were more likely to have a foldable hand. Thus, the 40 remaining cards are slightly less likely to hold a valuable 7.

So instead of viewing it as 3 A's and/or two 7's remaining out of 48 cards....why not think of it as something like 2.8 A's out of 40 and 1.6 7's out of 40 since the 7 is the more fold-able card?

Obviously this doesn't hold true in heads-up games where one can do a sim of all the possible combinations and be certain that each combination has an equal chance of being dealt. But in a normal, non-psychotic game, where most players play somewhat normally then it seems each previous fold should represent something when it comes to gauging the calculations.

I'm sure this is not an original concept but I would be interested in any info about this that is out there.

please feel free to inform me if i have made no sense at all or if my concept is too simplistic.

lorinda
05-24-2003, 07:53 PM
I am posting this straight away because it is my initial reaction and I wanted to share it with a first time poster.

This seems like a very very valid point that I have never seen mentioned, at least not with regard to this situation, in 11 years of playing and 2 years of posting.

I suspect that the differences ARE slight, but certainly your theory carries a little weight.

I have always played AK a little more carefully if there are multiple callers in a pot for fear of one of them having an ace, and me getting into the "coin flip" with a coin that I don't like!

Your assumption as to how it is calculated is dead on, it is assumed that we have 0 knowledge of the other cards, when in fact we do have some.

I doubt it is enough to swing more than 1%, because some Aces and Kings will have been folded, but your observation is definately correct.

Lori

schroedinger
05-24-2003, 08:10 PM
I have heard it in reverse -- "I am not THAT crazy to call a four way cap before the flop with 87s because all the other hands obviously hold each others' high cards" -- OHHHHHHH-KAAAAAAAYYYYY.

I think this was even from a published poker writer.

Also, I don't know if it's just me, but a lot of cards get exposed at live tables -- whether by players or by dealers throwing them out too high.

I don't think his inference could ever be strong enough to affect the odds of this particular confrontation, however. Maybe in a 20 handed game or something. With all solid players.

Jimbo
05-24-2003, 10:46 PM
"This seems like a very very valid point that I have never seen mentioned, at least not with regard to this situation, in 11 years of playing and 2 years of posting."

Lorinda, you need to get out more often! /forums/images/icons/smile.gif Someone posts this very concept every month or so. Usually it is ignored but ocassionally someone explains why it makes no difference.

Nottom
05-25-2003, 01:44 AM
Think of it this way. Heads up 77 is a slight favorite against AK - true. But now add a 3rd player. Give him QT for example. Now 77 has to dodge an A,K,Q or T or hit his set to win. AK is still basically going to win if he hits an A or K most of the time so the added player doesn't hurt him as much.

Additionally, if you are playing 77 heads up against someone who raised preflop (with AK but you don't know that) and the flop comes 8 T J and he bets you can't really profitably call here even though you you are beating Mr. AK.

All the worring about there being a better chance of there being aces in the deck becasue people folded is pretty insignificant compared to the fact that you can't confidently bet 77 unless you spike your set.

** noticed the first part doesn't really apply to your post, but I think the second part is still semi-applicable **

Cyrus
05-25-2003, 02:03 AM
"Think of it this way. Heads up 77 is a slight favorite against AK - true. But now add a 3rd player. Give him QT for example. Now 77 has to dodge an A,K,Q or T or hit his set to win. AK is still basically going to win if he hits an A or K most of the time so the added player doesn't hurt him as much."

The assumption is that 77 (or lower) will most probably have to improve against a large field. Against isolated opponents it might not have to. The large field of opponents however might also give the 77 holder the proper odds to call and see the flop. (The odds against making a set on the flop are 8.3 : 1 , while against making at least a set, i.e. incl. boat, quads, the odds are 7.5 : 1)

So, as you say, the 77 has to "dodge" more hostile flop cards when against more than one opponent, but, generally, the middle-to-small pair is also closer to getting proper pot odds when the field gets larger. And having said that, I believe that the usual concluding caveat is that the rest is up to the specific conditions of the game, the deal and the round.

05-25-2003, 04:40 PM
Thanks for your responses and ideas on this topic.

Here's another thought on the whole deal....
It's a virtual guarantee that you could not have 5 or 6 of the 6 remaining A's or K's among the 4 hands that were folded since that means someone would be folding KK, AA or AK. Having 4 of the remaining 6 A's or K's among the folded hands seems highly unlikely also. Doesn't this mean that you can look at hitting an A or K really as a 6 out of 46 proposition as opposed to 6 out of 48?? Actually, I think that you can conservatively take it down further to 6 out of 45.5 or something (just because the odds of the 4 a's or K's being distributed into 4 seperate hands PLUS the odds that those 4 would all be fold-able hands is so great).

No, it doesn't make that big of a difference...but I suspect it might be greater than 1%. But probably still pretty trivial.

Obviously in my view I would see the chances of hitting a 7 as being 2 in 48.1 or something.

Addressing another valid point - yes, the idea that the 77 is weaker against 2 hands staying in...and the idea that you can't bet it as strongly in later rounds are both accurate and interesting. I was specifically looking at the scenario in which 4 other players folded pre-flop leaving 2 players head to head having gone all-in. there is no real strategy involved here....I am simply disputing the validity of putting the 7's at a 54-46 advantage when in fact they possibly aren't.

Interestingly though...it brings up another idea which may have more relevance -
5 other players at the table table and you have 77.
Nobody folds. Perhaps, using my theory, your pot-odds are slightly increased. You now have 40 cards left in the pack...plus the 10 unknown cards in the other hands. I would not view this as 2 out of 50 that a specific remaining card in the pack is a 7. i would look at this as more of a 2 out of 45 scenario (or something)....due to the unlikelihood that 1 or both of the 7's are in the other hands. There's the chance that someone else has both 7's of course and they could have stayed in. But other hands involving 7's are more likely to be folded.

When looked at in this way i wonder if there is any validity in using these ideas in deciding calls where the pot-odds seem to be right on the borderline.

Thanks again for your input and for sparking more ideas in my teeny-little noggin that hopefully get me thinking like the kind of quality poker player i strive to become.

Cyrus
05-25-2003, 05:48 PM
"It's a virtual guarantee that you could not have 5 or 6 of the 6 remaining A's or K's among the 4 hands that were folded since that means someone would be folding KK, AA or AK. Having 4 of the remaining 6 A's or K's among the folded hands seems highly unlikely also."

OK, all this presumes that we know nothing else about the opponents, eg we don't know with what hands they're likely to call early, what kind of players they are, etc.

Under this premise, we cannot have any information about the cards in the folded hands from the number of folded hands! Simply put (and this is very similar to the debunking of card clumping myths in BJ), you can have all 4 Aces and all 4 Kings, for example, being folded: Say you're the SB, in a 10-handed game, and they fold before you, starting from UTG, with ATo/A4s/K9s/A2o/A5o/KTo/K3o/K2o. Now you can dispute the correctness of all those folds, especially the late-position ones, but the fact is that those eight pairs are as likely to be dealt as every other group of eight pairs.

Yes, you think this is unlikely, just as it is unlikely that the winning lotto numbers can be 1,2,3,4,5,6. Well, those six lotto numbers are as likely to come up as any other group of six numbers.

"I was specifically looking at the scenario in which 4 other players folded pre-flop leaving 2 players head to head having gone all-in. There is no real strategy involved here... I am simply disputing the validity of putting the 7's at a 54-46 advantage when in fact they possibly aren't."

If you accept the notion that, having no other info abt our opponents, we do not know what cards our opponents may be folding, then the 77 pair is a 54:46 favorite if the math says so.

"When looked at in this way I wonder if there is any validity in using these ideas in deciding calls where the pot-odds seem to be right on the borderline."

Well, I know there are situations where gambling with the worst of it is absolutely correct/necessary. There are also situations where pot odds are completely irrelevant! But this is getting too general so I must allow others, more experienced and/or knowledgeable about this, to speak up.

Jimbo
05-25-2003, 05:53 PM
"When looked at in this way i wonder if there is any validity in using these ideas in deciding calls where the pot-odds seem to be right on the borderline."

Nope!

schroedinger
05-25-2003, 10:19 PM
I think the far greater danger is that you will use this kind of "thinking" to justify calling when you should clearly be folding.

If you knew your opponent had 77, you would fold your AK (unless you were getting short stacked and in danger of being blinded out) every time.

But see my original post in this thread on the dangers of the line of thinking that you are pursuing . . . . Even a good player uses it to justify calling with what is clearly a fold.

Or in other words, listen to Jimbo.

clay1m
05-26-2003, 04:05 AM
I have a question along the same lines as the one posed about AK vs 77, but I think it would come up more often. For example say that I take a flop with two other players in a ten handed game of limit Holdem. My hand is 2 cards of the same suit. The flop gives me a flush draw. The mathematics you can find in any book on poker theory now tells me that since I have nine outs with 2 cards to come I have approx 35% chance of drawing to my flush (assuming my opponents have none of my suit) It seems to me that this is very unlikely because 7 players were dealt cards that were folded preflop. Surely some of those cards that are now in the muck are my suit. In fact it would be a remote possibility that all of the cards either in the muck or in the hands of my opponents are not of my suit. So if I know that I must make my flush to win and I continue playing based on the fact that the pot is offering me correct odds for a 35% chance of winning could this be an error?

lorinda
05-26-2003, 08:49 AM
If you knew your opponent had 77, you would fold your AK (unless you were getting short stacked and in danger of being blinded out) every time

Unless it was a ring game and the blinds tipped the EV in your favour at 46%

Lori

Cyrus
05-26-2003, 10:14 AM
"Say that I take a flop with two other players in a ten handed game of limit Holdem. My hand is 2 cards of the same suit. The flop gives me a flush draw. The mathematics you can find in any book on poker theory now tells me that since I have nine outs with 2 cards to come I have approx 35% chance of drawing to my flush (assuming my opponents have none of my suit)."

I haven't seen any serious book addressing this kind of flop with the assumption that the opponents have none of the player's suit!

"It seems to me that this is very unlikely because 7 players were dealt cards that were folded preflop."

It's not simply unlikely, it is quite wrong to assume such a thing. There is simply no basis for that kind of assumption.

This is similar to the erroneous thinking of BJ players who assume things about the cards in the shoe, thinking they see clumps, sequences, or biases.

"Surely some of those cards that are now in the muck are my suit. In fact it would be a remote possibility that all of the cards either in the muck or in the hands of my opponents are not of my suit."

This is a fallacy. When you have no other (reasonably reliable) information about the cards in your opponents' hands, then the correct thinking is assuming that all the cards are out there and all the combinations are possible.
In other words, the cards of your suit could all be in the muck, they could all be in the deck waiting to be dealt or they could be in both places.

--Cyrus

____________________________________________

The mathematics tell us that when you hold 2 suited cards and the flop gives you a 4-Flush draw, you will make that Flush 35% of the time. And this is not on the basis of any assumption about what "the opponents are holding". It is based on the solid premise that there are still (13 -2 -2)= 9 cards of your suit that are out there somewhere, anywhere.

C(9,2)= 36 combinations that can be made, with 2 cards in each combination made out of the 9 remaining of your suit. This means that there are 36 combinations whereby you hit your suit in both Turn and River.

You can also hit 1 out only, in either the Turn or the River, and still make your Flush of course. This case will combine any one of the 9 remaining cards of your suit with one card from the rest of the cards, none of which is of your suit. The rest of the cards out there are (52 - 2 cards in your hand - 3 cards on the flop - 9 remaining cards of your suit) = 38, so you can have 9*38 = 342 combinations with 1 card of your suit and 1 blank.

36+342= 378 is the number of all the combinations for Turn and River that hit you with 1 or 2 of your outs for a Flush. If this is compared with the total number of all possible Turn+River combinations, which is C(47,2)= 1081, we get the probability of hitting either 1 or 2 outs and making your Flush. It's 378/1081 = 0.349676225716928769657724329324699 or 35%.

Nottom
05-26-2003, 02:56 PM
All the worring about there being a better chance of there being aces in the deck becasue people folded is pretty insignificant compared to the fact that you can't confidently bet 77 unless you spike your set.

[/ QUOTE ]

I think this is the most relevant portion of my post. Sure maybe based on a player betting habits you can adjust the odds slightly if you reads are accurate, but in a hypotheticl NL game if an opponent goes all in in front of you and shows you his 77 (assume the blinds are small enuf to not be a factor) short of another opponent flashing the other 2 7's, I think my AK is going in the muck.

Cyrus
05-26-2003, 03:33 PM
"Say that I take a flop with two other players in a ten handed game of limit Holdem. My hand is 2 cards of the same suit. The flop gives me a flush draw. The mathematics you can find in any book on poker theory now tells me that since I have nine outs with 2 cards to come I have approx 35% chance of drawing to my flush (assuming my opponents have none of my suit).

It seems to me that this is very unlikely because 7 players were dealt cards that were folded preflop."

This is false but I guess one must elaborate a bit more than my blank dismissal.

Since we do not know (because if we do, it's a different ball game!) where the remaining cards of our suit are, we must assume that they can be anywhere. They could all be in the pack of cards waiting to be dealt, they could all be in the muck, some could be in our opponents' hands -- or they could be spread among the three places.

We compute mathematically the probability of hitting the Flush by assuming that the suited cards are somewhere among the 47 cards that are unknown to us : 52 - 2 cards in our hand - 3 cards on the flop = 47. Those 47 cards include the muck, the undealt cards, the cards still at play among our opponents, everything.

This may seem intuitively to be incorrect -- as you said, How can NO ONE of our opponents hold/mucked a single card of my suit??

It is nevertheless correct for the simple reason that the calculation takes into account ALL the possible cases, from the suited cards being out of play upto them being all in the deck still. The calculation essentially has "weighed" all the possible cases and since they are all equally probable (equaprobable), it correctly accepts that the cards can be anywhere among the 47 unknown cards.

If the other cards of your suit are all in the muck, your chance of hitting the Flush is 0%. (Same thing if the cards are all in your opponents' hands.) At the other extreme, if the suited cards are all in the deck and waiting to be dealt, the probability of hitting the Flush, if you go to the River, is no longer 35% but higher, actually 45% *. It should be obvious that the correct probability should lie somewhere between 0% and 45%. And since the cards waiting to be dealt (35 cards) are more than the cards in our opponents' hands (12 cards), it should "make sense" that the correct probability, i.e. 35%, lies much closer to 45% rather than 0%.

--Cyrus

_______________________________________________

* We take the original post's premise, with 7 players at the table. We assume that all the other cards from our suit are still in the deck. The unknown cards are now (52 - 2 cards in our hand - 3 cards on the flop - 12 cards in our opponents' hands) = 35. This makes for C(35,2)=595 possible combinations of Turn+River.

We still have C(9,2)=36 combinations whereby we hit a suited card in both Turn and River.

Then we have (35 - 9 cards of our suit) = 26 cards that can be combined with each of our 9 suited cards to produce a Turn+River whereby we hit only 1 of our outs -- and still make a Flush of course. That gives 9*26= 234 such combinations.

36+234= 270 total combinations whereby we make a Flush by hitting 1 or 2 outs, and 270/595 = 0.45378151260504201680672268907563 or 45%.

Cyrus
05-27-2003, 12:21 AM
DRAWING TO A FLUSH WHEN YOU HAVE FLOPPED A 4-FLUSH DRAW AND ALL THE OTHER CARDS FROM YOUR SUIT ARE IN THE DECK :

<pre><font class="small">code:</font><hr>

Number of opponents Prob of completing
1 36.4%
2 37.9%
3 39.5%
4 41.3%
5 43.2%
6 45.4%
7 47.7%
8 50.3%
9 53.2%

</pre><hr>