Archive for Tagpoker Forum A friendly place for winning poker players to share their strategy to making money playing poker.
 


       Tagpoker Forum Forum Index -> Heads Up Poker Strategy
N86xps

Mathematics behind hand reading

Hello everyone! I am a beginning poker player, and I am not a professional mathematician, though mathematics was my passion long before I even knew he rules of poker. So, I feel, that I MUST write on this topic. Maybe you all know it, in which case I will look funny and stupid. But I feel, that many poker players lack mathematical background, so they might make false assumptions and stick to them.

Let's start with an example:

you have carefully analyzed your opponents actions preflop, on the flop and on the turn, and came to a conclusion that your opponent has a flopped set of 8's 90% of the time. Here comes the river which is also an 8... This changes a lot! there is only one "combo" of 8s left, so you call his all in, and he stacks your overpair with quad eights. Bad beat! Or is it?...

Let's try to formulate a generic mathematical problem, and then solve it. Even if you do not know much of probability theory I still recommend you read it, I'll try to make it as simple as possible.

First let's define two events

A - your opponent holds a particular hand.
B - some particular card appears on board.

When we make some assumptions about our opponents holdings we assign (at least implicitly) some probability to each hand. So, for example if we are 14% sure, that our opponent holds 72o we can write

         P(A) = P(he has 72o) = 0.14

When the next street comes, this probability usually changes. For that we need a POSTERIOR probability of A given that event B occured. It is a conditional probability and we write:

          P(A|B)

Note, that events A and B are NOT independant, therefore the probability of B is affected by whether even A occured or not.

Now we will need the Bayes' theorem:

          P(A|B) = P(B|A) P(A) / P(B)

here P(B|A) is the conditional probability of B given A (the probability that 8 comes on the river IF our opponent holds a pair of eights). As can be seen, we need to multiply our initial estimation by the ratio

          R = P(B|A) / P(B)

This ratio indicates how much does our probability change after the next street comes. Now our formula looks simpler


          P(A|B) = R * P(A)

Let's find out how much does the probability of pocket eights change after another 8 comes on the river.

P(B|A) = P(8 on the river IF he holds 8's) = 1/44

P(B) = P(river is an eight) = 0.9*P(B|A) + 0.1*P(B|A') = 0.9*1/44 + 0.1*3/44 = 1.2/44

R = P(B|A) / P(B) = (1/44) / (1.2/44) = 5/6

P(A|B) = 5/6 * 0.9 = 0.75

so, if you were 90% sure, that your opponent has a set of eights before the river you SHOULD still be 75% sure, that he now has quad 8s.

Let's consider another simplified example.

Our opponents range is JJ+ and flop comes A22. For simplicity we do not have aces or deueces in our hand. Before the flop the probability of him having aces was 0.25. How did it change with that ace on the flop? Maybe some of you think that it is now 12.5%, because there are only 3 combos left as opposed to 6 combos before the flop. Well, it is WRONG. Let's calculate it:

P(B|A) = P(flop is A22 IF he holds AA) = 12 / 17926 (two aces left times 6 combos of 22s divided by the total number of possible flops)

P(B) = P(flop is A22) = 0.25 * P(B|A) + 0.75 * P(B|'A) = 0.25 * 12 / 17926 + 0.75 * 24 / 17926 = 21 / 17926

R = P(B|A) / P(B) = 12/21 = 4/7

P(A|B) = R * P(A) = 4/7 * 0.25 = 1/7 = 0.14...

so it is 14% and not 12.5%.

I hope that it makes sense. I am sure, that in order to be a good poker player it is not necessary to know all this maths. But at least it is usefull to know, that this math is not just counting "combos". All players know the combinatorics, yet some players misapply it. It is usufull to note, that usually when the number of combos goes down the probability also goes down, but not as fast. In the second example it is irrelevant. But in the first example this knowledge is a difference between a clear +EV call and a clear -EV call.

I appologize to those who know all this (and possibly much more).
Kim Cardassian

This could be interesting, but I think there's a flaw in your thinking applying Bayes' Theorem. It states that actual statistical proof needs to be observed, not?

If he holds 10 possible combo's, each has 10% (unless we weigh a certain combo higher) but in poker we never have any evidence of it being true or false. If there's 2 combo's on the flop that's 20%, but if the turn reduces that to 1 combo, wouldn't that reduce it to 10%? If we thought 8d was in villains range, and 8d appears on board, there is no chance he has it. And this 8d is completely removed from his range, despite what we thought. He would have only 1 combo of 8's. Not 1.8, 1.5 or 1.3. Could you please elaborate on this for people without mathematical backgrounds?

Because the way I see it, the 8d being removed from the range (there is 0% chance it is in there, due to it being on board) means that whatever we thought, whatever we weighed previously, is most likely wrong. That 8d on board is proof to the contrary. Bayes' Theorem is (I think) designed with the idea in mind that we should trust actual statistical evidence over probability (IE in the weighing of known statistics vs probability, the statistics have a higher weight). Because poker is a game where no evidence is seen except the cards that are in your hand and appear on board.

Either way, great thread. I love these type of posts where you are forced to think about the game before being able to fire off a reply.
_red_dog

I like what you have posted here.  However, this isn't exactly "hand reading" per se.  What you have given us is how card removal effects the range our opponent has, which is still useful information.

Yes you have used simplified examples for the purposes of people being able to comprehend it, but it is very rare (in HUSNG's) to find an opponent whose range is JJ+.  

What you also neglected was how an opponents actions can change how we perceive his range, does he check, bet, check/call, check/raise.  All these things can change what we view our opponents range to be, yes future streets are going to alter this and that is where your post comes in.  But once the turn card comes down his actions are going to, for the most part against bad-mediocre players, narrow his range.

All in all, I like your post, and tbh, this will be interesting for a lot of people on the forum.  But they just need to keep in mind that just because they have an idea of an opponents range preflop, their actions are going to be more relevant and more revealing about their hand.

Like I said, nice post.  Please post more around here Smile
N86xps

_red_dog wrote:
I like what you have posted here.  However, this isn't exactly "hand reading" per se.  What you have given us is how card removal effects the range our opponent has, which is still useful information.

Yes you have used simplified examples for the purposes of people being able to comprehend it, but it is very rare (in HUSNG's) to find an opponent whose range is JJ+.  

What you also neglected was how an opponents actions can change how we perceive his range, does he check, bet, check/call, check/raise.  All these things can change what we view our opponents range to be, yes future streets are going to alter this and that is where your post comes in.  But once the turn card comes down his actions are going to, for the most part against bad-mediocre players, narrow his range.

All in all, I like your post, and tbh, this will be interesting for a lot of people on the forum.  But they just need to keep in mind that just because they have an idea of an opponents range preflop, their actions are going to be more relevant and more revealing about their hand.

Like I said, nice post.  Please post more around here Smile


I was only writing about exactly what you said: the way card removal affects probabilities. Obvioulsy your opponents actions will influence your decision. If for example you had a very good read, that your oponent NEVER pushes quads on the river, then, if he pushes it, then he does not have it:). What math gives us is just a piece of information, that we must take into account. How we weigh it against other information that we get from our opponent is already a different subject.

Yeah, JJ+ is not realistic. Was too lazy to take on some more realistic example like TT+, AQs+, AK for example (I feel that some low stakes nit will 3bet with this range, but I may be wrong).

The purpose of that post was actually to show, that the way some people see card removal is flowed.

Futhermore, ANY evidence that we get on later street, not only the cards on the board, will "update" the probabilities through the Bayes' theorem. If I come up with a simple framework of how this could be applied on the fly I will post it.
N86xps

Kim Cardassian wrote:
This could be interesting, but I think there's a flaw in your thinking applying Bayes' Theorem. It states that actual statistical proof needs to be observed, not?

If he holds 10 possible combo's, each has 10% (unless we weigh a certain combo higher) but in poker we never have any evidence of it being true or false. If there's 2 combo's on the flop that's 20%, but if the turn reduces that to 1 combo, wouldn't that reduce it to 10%? If we thought 8d was in villains range, and 8d appears on board, there is no chance he has it. And this 8d is completely removed from his range, despite what we thought. He would have only 1 combo of 8's. Not 1.8, 1.5 or 1.3. Could you please elaborate on this for people without mathematical backgrounds?

Because the way I see it, the 8d being removed from the range (there is 0% chance it is in there, due to it being on board) means that whatever we thought, whatever we weighed previously, is most likely wrong. That 8d on board is proof to the contrary. Bayes' Theorem is (I think) designed with the idea in mind that we should trust actual statistical evidence over probability (IE in the weighing of known statistics vs probability, the statistics have a higher weight). Because poker is a game where no evidence is seen except the cards that are in your hand and appear on board.

Either way, great thread. I love these type of posts where you are forced to think about the game before being able to fire off a reply.


Well, what do you mean the actual statistical proof needs to be observed? Your experience and/or some other mathematical concepts give you a "best guess". After you acquire additional information, (in this case it is a card on the board, but it could be the guy scratching his head in live poker), and if this information does not tell you something with 100% certainty (which rarely the case unless you accidentally see your opponents' cards), then you need to weigh this additional information against the one you had before. Since opponents' range DOES affect the probability of some card appearing on the board, we need to calculate the POSTERIOR probability of that range. This is exactly what the Bayes theorem designed for.
N86xps

Just another thing to clarify all this.

Hand reading is making a hypotesis about your opponent's holding, and updating that hypotesis whenever you are presented with new information. This "updating" part is where Bayes' theorem comes to help. This is very general, and I can not even come up with an example where Bayes' theorem would be of no use. It is only if new evedence makes you 100% sure about something, which is rarely the case.
_red_dog

Yes totally agree that people really do not know how to appropriately apply card removal.  The general line of thought is "Well the flop was AJ2 and he called my cbet, oh look a jack on the turn, he "can't" have a jack now!!  My AQ MUST be good" which is retarded, but I have heard this expressed from a lot of, even winning, players.
N86xps

_red_dog wrote:
Yes totally agree that people really do not know how to appropriately apply card removal.  The general line of thought is "Well the flop was AJ2 and he called my cbet, oh look a jack on the turn, he "can't" have a jack now!!  My AQ MUST be good" which is retarded, but I have heard this expressed from a lot of, even winning, players.


The funny thing is that it is in the books:). Sometimes I feel, that those professional players just do not want their readers to know the math. I've seen so many times things like "poker math is not as difficult as might seem.. there are 47 unknown cards, we have 9 outs... etc"Smile. Bullshit poker math is easy! It is an unsolved game in fact, so it can not be easy.
U Cook Socks

_red_dog wrote:
Yes totally agree that people really do not know how to appropriately apply card removal.  The general line of thought is "Well the flop was AJ2 and he called my cbet, oh look a jack on the turn, he "can't" have a jack now!!  My AQ MUST be good" which is retarded, but I have heard this expressed from a lot of, even winning, players.


I'm not an expert at this, but in the example that you gave there.

You have to use other information combined with the combinations right?

So say you know for sure that he will only ever raise you for value and not as a bluff from past history, then all of a sudden a Jack seems far more likely, as he is unlikely to raise an ace for value, he is less likely to have AA than a Jack, he is even less likely to have JJ as he would have to have the last two in the deck. So a Jack or 22 seem the only hands he can have.

Obviously you can narrow ranges down in other ways, like he would have probably 3Bet AA etc, and it is only a quick example just to see if I am thinking about things in the right way ?
chesslw

Using Baye's Theorem is a mathematically nice way of looking at it. But imo the best way to estimate ranges is always to actually count them, given all the information you know, since what you are trying to work out is just that- card combinations.

Take the AQ on AJ2J. Basically on the turn you can look back to on the flop, and count all the different hands in villain's range, knowing that he can only have 2 jacks in his combination of hands (i.e. you KNOW that another J is coming on the turn- also this J is a specific J, which is important).

So if you know that he will call on the flop with (for e.g.) KK QQ J10, QJ, KJ and raise with AJ and sets (and he would have reraised AK preflop) and bluffs, then basically the part of villain's range that can have J can be properly counted, giving him a choice of 2 jacks instead of 3:

6 combos of KK, 3 combos of QQ, 8 combos of J10, 6 combos of QJ, and 8 combos of KJ.
chesslw

The ranges I gave are not realistic at all imo but I'm just using it as an example of estimating ranges.
N86xps

chesslw wrote:
Using Baye's Theorem is a mathematically nice way of looking at it. But imo the best way to estimate ranges is always to actually count them, given all the information you know, since what you are trying to work out is just that- card combinations.

Take the AQ on AJ2J. Basically on the turn you can look back to on the flop, and count all the different hands in villain's range, knowing that he can only have 2 jacks in his combination of hands (i.e. you KNOW that another J is coming on the turn- also this J is a specific J, which is important).

So if you know that he will call on the flop with (for e.g.) KK QQ J10, QJ, KJ and raise with AJ and sets (and he would have reraised AK preflop) and bluffs, then basically the part of villain's range that can have J can be properly counted, giving him a choice of 2 jacks instead of 3:

6 combos of KK, 3 combos of QQ, 8 combos of J10, 6 combos of QJ, and 8 combos of KJ.


Unfortunately this way you will have a wrong result. What you are szying is that the probability of him having a Jack is 2/3 less after the turn. Let's calculate haw much less it really is (I will use this range of hands for the flop: [KK, QQ, KJ, QJ, JT, 22] = 45 hands, I excluded JJ for simplicity, so that the calculations are easier to follow):

P(B|A) = P(jack on the turn IF villain holds a jack) = 2/45

P(B) = P(jack comes on the turn) = P(A)P(B|A) + P(A')P(B|A') = 33/45 * 2/45 + 12/45 * 3/45 = (66+36)/45^2 = 102 / 45^2

R = (2/45) / (102/45^2) = 90 / 102 ~ 90%

As you see again, the removal of a jack does decrease the probability of villain holding a jack, but only by 10% and NOT by 33%, though there are really 33% less combos.
chesslw

N86xps wrote:
chesslw wrote:
Using Baye's Theorem is a mathematically nice way of looking at it. But imo the best way to estimate ranges is always to actually count them, given all the information you know, since what you are trying to work out is just that- card combinations.

Take the AQ on AJ2J. Basically on the turn you can look back to on the flop, and count all the different hands in villain's range, knowing that he can only have 2 jacks in his combination of hands (i.e. you KNOW that another J is coming on the turn- also this J is a specific J, which is important).

So if you know that he will call on the flop with (for e.g.) KK QQ J10, QJ, KJ and raise with AJ and sets (and he would have reraised AK preflop) and bluffs, then basically the part of villain's range that can have J can be properly counted, giving him a choice of 2 jacks instead of 3:

6 combos of KK, 3 combos of QQ, 8 combos of J10, 6 combos of QJ, and 8 combos of KJ.


Unfortunately this way you will have a wrong result. What you are szying is that the probability of him having a Jack is 2/3 less after the turn. Let's calculate haw much less it really is (I will use this range of hands for the flop: [KK, QQ, KJ, QJ, JT, 22] = 45 hands, I excluded JJ for simplicity, so that the calculations are easier to follow):

P(B|A) = P(jack on the turn IF villain holds a jack) = 2/45

P(B) = P(jack comes on the turn) = P(A)P(B|A) + P(A')P(B|A') = 33/45 * 2/45 + 12/45 * 3/45 = (66+36)/45^2 = 102 / 45^2

R = (2/45) / (102/45^2) = 90 / 102 ~ 90%

As you see again, the removal of a jack does decrease the probability of villain holding a jack, but only by 10% and NOT by 33%, though there are really 33% less combos.


? Sorry I haven't really bothered to read the above- but I assume the calculations to be correct... What I'm saying is that you can actually count the no. of hands villain can have on the turn, having made a call on the flop.

So with AJ2J board if I count all the possible combinations how can the number be wrong? Did I count it wrong? With Baye's theorem surely you are doing the exact same thing I'm doing, but you are calculating the prob on the flop, then updating it by then event that the turn is a particular J.

What I'm doing is actually counting the total no. of hand combos villain can have, given that he can't have the cards shown on the board (and the cards you hold).

Surely the two have to arrive at the same answer?
chesslw

I "think" what you are calculating is:

prob(villain holds J on AJ2 flop given turn is J)

And what I'm calculating is:

prob(villain holds J given board is AJ2J)

and the two are the same?

N86xps wrote:
What you are szying is that the probability of him having a Jack is 2/3 less after the turn


I never actually worked out any numbers/probs in my calculations- just saying that I think it would be much easier to do a back of envelope calculation in your head if you just physically count the possible combos given all of the imformation available (i.e. the board, and your hand).
N86xps

chesslw wrote:
N86xps wrote:
chesslw wrote:
Using Baye's Theorem is a mathematically nice way of looking at it. But imo the best way to estimate ranges is always to actually count them, given all the information you know, since what you are trying to work out is just that- card combinations.

Take the AQ on AJ2J. Basically on the turn you can look back to on the flop, and count all the different hands in villain's range, knowing that he can only have 2 jacks in his combination of hands (i.e. you KNOW that another J is coming on the turn- also this J is a specific J, which is important).

So if you know that he will call on the flop with (for e.g.) KK QQ J10, QJ, KJ and raise with AJ and sets (and he would have reraised AK preflop) and bluffs, then basically the part of villain's range that can have J can be properly counted, giving him a choice of 2 jacks instead of 3:

6 combos of KK, 3 combos of QQ, 8 combos of J10, 6 combos of QJ, and 8 combos of KJ.


Unfortunately this way you will have a wrong result. What you are szying is that the probability of him having a Jack is 2/3 less after the turn. Let's calculate haw much less it really is (I will use this range of hands for the flop: [KK, QQ, KJ, QJ, JT, 22] = 45 hands, I excluded JJ for simplicity, so that the calculations are easier to follow):

P(B|A) = P(jack on the turn IF villain holds a jack) = 2/45

P(B) = P(jack comes on the turn) = P(A)P(B|A) + P(A')P(B|A') = 33/45 * 2/45 + 12/45 * 3/45 = (66+36)/45^2 = 102 / 45^2

R = (2/45) / (102/45^2) = 90 / 102 ~ 90%

As you see again, the removal of a jack does decrease the probability of villain holding a jack, but only by 10% and NOT by 33%, though there are really 33% less combos.


? Sorry I haven't really bothered to read the above- but I assume the calculations to be correct... What I'm saying is that you can actually count the no. of hands villain can have on the turn, having made a call on the flop.

So with AJ2J board if I count all the possible combinations how can the number be wrong? Did I count it wrong? With Baye's theorem surely you are doing the exact same thing I'm doing, but you are calculating the prob on the flop, then updating it by then event that the turn is a particular J.

What I'm doing is actually counting the total no. of hand combos villain can have, given that he can't have the cards shown on the board (and the cards you hold).

Surely the two have to arrive at the same answer?


Well:).... If you count the combos, you will see, that there are 33% less jacks in villains range. But it does NOT mean, that it is 33% less likely, that our villain has a jack. As I showed in my calculations it is only obout 10% less. So, when you see that jack on the turn it really should not make you that happy:).
chesslw

OK I think I understand now. I think we misunderstood each other.

I was never meant to calculate "how much more/less likely that villain holds J on turn compared to on the flop"- which isn't that helpful imo as a tool for handreading...

So I think you were trying to debunk a myth about cardremoval?
chesslw

Sorry- I should appologise- since I didn't bother to really read your first post. i thought you were trying to estimate ranges with Baye's theorem (which is horribly overcomplicating things).

Yeah- I agree with you- that reducing the total combos of Js by a third is not the same as making villain having a J 1/3 less likely- you can easily go to extremes to show this (mathematically)- e.g. if villain always/never has a J, then a J on the turn doesn't change things at all.
N86xps

chesslw wrote:
Yeah- I agree with you- that reducing the total combos of Js by a third is not the same as making villain having a J 1/3 less likely- you can easily go to extremes to show this (mathematically)- e.g. if villain always/never has a J, then a J on the turn doesn't change things at all.


Very nice and concise proof!

Yet Bayes' theorem is still helpful. Everytime you get new information, that changes our opponent's range (even a timing tell) it is useful to know how much does this new information actually change. I somehow feel, that it is possible to incorporate it in one's thinking even at the table. Say we have a hypotesis about our opponent's holding H, and we estimate the probability of that hypotesis being true P(H), then after some event E occurs the new probability of H is given by

                  P(H|E) = R * P(H)

where R = P(E|H) / P(E)

If we can estimate the R, then it is easy to find the new probability of H. I am sure, that with some practice it can be done.
chesslw

Yep- that is fine way of estimating villain's range... But to be fair- if you just count the total no. of combos- it should never be that hard to work out an updated range when the turn comes (you've had to do this in the first place on the flop anyway). Since you don't care about how much villain's range has improved/gotten worse (unless you have physical tells on villain when he sees turn etc but the value in this idea is very thin)- I believe just counting is the best way.

I still think using Baye's Theorem is overcomplicating the calculations particularly in Hold'em- when the first calculation on the flop is not that hard with only 2 hole cards. If you are fairly familiar with combinatorics it is a very simple excercise to count (taking account of pairs, none pairs, suits and also card removal), and even if not, it should be very easy for anyone to become proficient at.

However, I think your idea is useful for plo (or some other game where counting isn't easy)- when working out ranges is very complicated, and working out the conditional probabilities might be easier.

All in all, this is what almost all of the top high stakes players do when they play- i.e. working out all combos of ranges, so it is a very useful way of looking at things:

http://www.youtube.com/watch?v=r6-0iJOnH5I

[i.e. when Jungleman says the K is a good card for his hand combinatorically- it just means that it weaken's villain's range against his hand (since it reduces the no. of hands that was beating him, but keeping the no. of hands that he beat the same).]
N86xps

chesslw wrote:
Yep- that is fine way of estimating villain's range... But to be fair- if you just count the total no. of combos- it should never be that hard to work out an updated range when the turn comes (you've had to do this in the first place on the flop anyway). Since you don't care about how much villain's range has improved/gotten worse (unless you have physical tells on villain when he sees turn etc but the value in this idea is very thin)- I believe just counting is the best way.

I still think using Baye's Theorem is overcomplicating the calculations particularly in Hold'em- when the first calculation on the flop is not that hard with only 2 hole cards. If you are fairly familiar with combinatorics it is a very simple excercise to count (taking account of pairs, none pairs, suits and also card removal), and even if not, it should be very easy for anyone to become proficient at.

However, I think your idea is useful for plo (or some other game where counting isn't easy)- when working out ranges is very complicated, and working out the conditional probabilities might be easier.

All in all, this is what almost all of the top high stakes players do when they play- i.e. working out all combos of ranges, so it is a very useful way of looking at things:

http://www.youtube.com/watch?v=r6-0iJOnH5I

[i.e. when Jungleman says the K is a good card for his hand combinatorically- it just means that it weaken's villain's range against his hand (since it reduces the no. of hands that was beating him, but keeping the no. of hands that he beat the same).]


You are right. If all the hands in villain's range are equally likely, then you can just count the combos. I did not realize it before:). Say in the JJ+, flop A22 example when flop comes, there are 3 out of total 21 combos left which is 1/7 = 14%, just do not forget to update the total number of hands (21 as opposed to 24 preflop). In the quad 8s example we could also just use combos. We were 90% sure he had a set and there were 3 combos of eights. We can say, that he had 3 out of 3.3333... total number of hands, and then, after eliminating 2 combos we are left with 1 out of 1.33..... total number of hands = 75% which is the correct result.

Though there is a lot of other information that changes our perception of opponent's range, it can be some betting pattern, or some kind of tell.. The correct and the most general way of incorporating this new knowledge is by using the Bayes' theorem.
chesslw

Probability in the rigorous sense is all about the state space and the weight function you assign to that space.

Baye's theorem is a nice way of working things out when you don't have all of the info.

In poker- the state space is finite, and the number of hands villain can hold is very easy to count, so you have all the info effectively. Each hand may be assigned the same probability (or more technically measure), or if you like, you may say that villain decides to play 78s 50% of the time- in which case you can assign that hand with half the weight/measure you assign other possible villain hands preflop and take that into account.

So if you want to work out his range, all you need to do is just count the hands in the state space (which changes by the flop/turn/river when villain's range can no longer contain some hands).
chesslw

So yeah- Baye's Theorem is obviously always fine to use.

But the most general way to work out probabilityin a general sense is just to "count" (and take into account of different weights if you insist the different hands to have different probs) all of the combinations- since fundamentally in probability whatever method you use you always end up doing just that.
N86xps

chesslw wrote:
So yeah- Baye's Theorem is obviously always fine to use.

But the most general way to work out probabilityin a general sense is just to "count" (and take into account of different weights if you insist the different hands to have different probs) all of the combinations- since fundamentally in probability whatever method you use you always end up doing just that.


I agree with you completely.

We can of course work everything out from scratch on every street. But what I am offering here is "updating" your result every time you get new evidence. Often this will be more complicated then working it out from scratch. But say on the turn we have a hypotesis H that our opponent holds a hand X. Here comes the river and we get new evidence E (not only the river card, it can be ANYTHING). I somehow believe, that in some cases it will be easier to estimate that

            R = P(E|H) / P(E)

then to remember EVERITHING that was on preflop, on the flop, and on the turn and analize it all together.

I'll try to come up with a good example a little later:).
chesslw

N86xps wrote:
chesslw wrote:
So yeah- Baye's Theorem is obviously always fine to use.

But the most general way to work out probabilityin a general sense is just to "count" (and take into account of different weights if you insist the different hands to have different probs) all of the combinations- since fundamentally in probability whatever method you use you always end up doing just that.


I agree with you completely.

We can of course work everything out from scratch on every street. But what I am offering here is "updating" your result every time you get new evidence. Often this will be more complicated then working it out from scratch. But say on the turn we have a hypotesis H that our opponent holds a hand X. Here comes the river and we get new evidence E (not only the river card, it can be ANYTHING). I somehow believe, that in some cases it will be easier to estimate that

            R = P(E|H) / P(E)

then to remember EVERITHING that was on preflop, on the flop, and on the turn and analize it all together.

I'll try to come up with a good example a little later:).


Ok. I'm very interested in this example Smile

Imo Baye's Theorem is a good thing to talk about in poker anyway, and is very helpful for people in this forum who haven't heard about it or thought about(not just for poker imo- there are sooo many cases of criminal courts misusing statistics/probability- check Sally Clark and the more US-famous OJ Simpson trials).

To be honest, in my time playing poker- I have rarely used counting combos in any decision (since I can rarely have a good enough estimate anyway). Though if you are a limit player you definitely need to use the concept a lot (since you see a lot of showdowns, and the decisions are very mathematical in nature anyway).
N86xps

chesslw wrote:
N86xps wrote:
chesslw wrote:
So yeah- Baye's Theorem is obviously always fine to use.

But the most general way to work out probabilityin a general sense is just to "count" (and take into account of different weights if you insist the different hands to have different probs) all of the combinations- since fundamentally in probability whatever method you use you always end up doing just that.


I agree with you completely.

We can of course work everything out from scratch on every street. But what I am offering here is "updating" your result every time you get new evidence. Often this will be more complicated then working it out from scratch. But say on the turn we have a hypotesis H that our opponent holds a hand X. Here comes the river and we get new evidence E (not only the river card, it can be ANYTHING). I somehow believe, that in some cases it will be easier to estimate that

            R = P(E|H) / P(E)

then to remember EVERITHING that was on preflop, on the flop, and on the turn and analize it all together.

I'll try to come up with a good example a little later:).


Ok. I'm very interested in this example Smile

Imo Baye's Theorem is a good thing to talk about in poker anyway, and is very helpful for people in this forum who haven't heard about it or thought about(not just for poker imo- there are sooo many cases of criminal courts misusing statistics/probability- check Sally Clark and the more US-famous OJ Simpson trials).

To be honest, in my time playing poker- I have rarely used counting combos in any decision (since I can rarely have a good enough estimate anyway). Though if you are a limit player you definitely need to use the concept a lot (since you see a lot of showdowns, and the decisions are very mathematical in nature anyway).


"A lot of players make handreading out to be far more difficult than it really is. They trouble themselves over extensive weighted range analysis, Bayes Theorem, and complex expected value calculations. At its most basic level, handreading is much simpler than that. I tell my students to focus on one simple question: Is he aggressive or passive?..."

This is a quote from Balugawhale's Easy Game. So, this guy knows the math:).

I am interested in that example as well, yet I don't have it yet:).
chesslw

yeah... unless you play limit hold'em or the very highest of stakes at deepish stacks, counting exact hand combos are not accurate enough to justify the useage.
       Tagpoker Forum Forum Index -> Heads Up Poker Strategy
Page 1 of 1