I love math! (subject: poker/hold'em)
I love math! (subject: poker/hold'em)
In preflop the probability of being dealt connected suited cards is:
~3.92%
and for a specific hand, and by that i do not mean specific suit, that is):
~1.96%
The probability og being dealt Pocket pairs is:
~5.88%
And for a specific hand, again not caring about the suits, it is:
~2.66%
If this is correct, and im assuming that it is, allthough i might have made a mistake, you are ~0.7% more likely to be dealt AKs than AA.
The problem is that i have been unable to determine wether AKs is infact better than AA. It seems than theres some confusion about it as i have seen charts both saying one and the other, however, most of those have shown AA to be the better hand. (or maybe i just remember wrong)
Anyhow, i would like to be able to determine it myself, but i havent been able to.
So, is AKs ~0.7% better than AA?
~3.92%
and for a specific hand, and by that i do not mean specific suit, that is):
~1.96%
The probability og being dealt Pocket pairs is:
~5.88%
And for a specific hand, again not caring about the suits, it is:
~2.66%
If this is correct, and im assuming that it is, allthough i might have made a mistake, you are ~0.7% more likely to be dealt AKs than AA.
The problem is that i have been unable to determine wether AKs is infact better than AA. It seems than theres some confusion about it as i have seen charts both saying one and the other, however, most of those have shown AA to be the better hand. (or maybe i just remember wrong)
Anyhow, i would like to be able to determine it myself, but i havent been able to.
So, is AKs ~0.7% better than AA?
Last edited by Zacariaz on Wed Sep 05, 2007 7:31 pm, edited 1 time in total.
- Combuster
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Well, chance computations on this scale are ugly. To answer the question wether pair aces are better than AK of a color, you build an expectation of the money you are going to win on the average case. The problem with that is that you can never put a number on how a player will behave. Consider the problem of how high you can raise the stakes without people dropping out or calling your bluff. Those things are hard to predict, and require thorough research.
For that reason, I won't even bother with trying to answer it.
If you want a satisfactory answer, consider playing poker with evolutionary algorithms and take large scale statistics, but pulling that off is far beyond the standard maths homework.
For that reason, I won't even bother with trying to answer it.
If you want a satisfactory answer, consider playing poker with evolutionary algorithms and take large scale statistics, but pulling that off is far beyond the standard maths homework.
This has nothing to do with playing poker or how different players react. This is pure math, probability, odds, stats whatever, and im doing it just for the kicks of it. (duh)
I know it is complex math, and maybe it is too much to ask of you to do it for me, further more, i am actually more interested in the math than the answer, so if you could point me in the right direction instead maybe?
The thing is that those sites, documents, book, etc. i have found, focuses on calculating the probability of you getting good preflop, good flop, turn and river, but little on the actual probability of winning over an opponent.
eg.
I have this hand and i calculate the probability of me getting full house, straight, 3 of a kind, etc.
OR
I calculate the chancce of winning, not taking betting and player reactions into account. (that may be a job for nupic though: www.numenta.com)
This i believe can be done allthough i'm allso aware that it will be very hard and complex math.
Sometimes you stumble upon charts which is ranking the various starting hands. What i wanna know if is those are mathematicly correct and, if that is the case, how it is calculated, because calculating the probability of getting a certain hand, which is easy, doesn't tell you how good the hand is.
Anyway, this is my little project, stupid as it may sound, and i'll take any help you can/will offer.
Regards
I know it is complex math, and maybe it is too much to ask of you to do it for me, further more, i am actually more interested in the math than the answer, so if you could point me in the right direction instead maybe?
The thing is that those sites, documents, book, etc. i have found, focuses on calculating the probability of you getting good preflop, good flop, turn and river, but little on the actual probability of winning over an opponent.
eg.
I have this hand and i calculate the probability of me getting full house, straight, 3 of a kind, etc.
OR
I calculate the chancce of winning, not taking betting and player reactions into account. (that may be a job for nupic though: www.numenta.com)
This i believe can be done allthough i'm allso aware that it will be very hard and complex math.
Sometimes you stumble upon charts which is ranking the various starting hands. What i wanna know if is those are mathematicly correct and, if that is the case, how it is calculated, because calculating the probability of getting a certain hand, which is easy, doesn't tell you how good the hand is.
Anyway, this is my little project, stupid as it may sound, and i'll take any help you can/will offer.
Regards
Instead of having Combuster tell you how the math works, why not just take some time and read? If you're not too experienced in the maths (I'm not say that I am, mind you. ) reading a few good books can help you a lot. Probability is this area apparently, so why not find something on that?
If you can stand hundreds of pages of dry theorems, Dover prints some excellent books that are very cheap.
It's just a suggestion...
If you can stand hundreds of pages of dry theorems, Dover prints some excellent books that are very cheap.
It's just a suggestion...
C8H10N4O2 | #446691 | Trust the nodes.
so the question hang with: which books? thats what i ment by pointing me in the right direction. Allso the internet is a great source, but there allso alot of crap.
And i am somewhat experienced in various areas of math, however theres allso some great holes in my education.
Edit: Juest saw the "Dover" comment. I have never hard it/them/him whatever and have no idea where to start looking.
And i am somewhat experienced in various areas of math, however theres allso some great holes in my education.
Edit: Juest saw the "Dover" comment. I have never hard it/them/him whatever and have no idea where to start looking.
- Combuster
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That is manageable. The only thing you have to do in that case is to compute the amount of different dealings that give the desired result, and divide it by the total amount of different dealings.Zacariaz wrote:eg.
I have this hand and i calculate the probability of me getting full house, straight, 3 of a kind, etc.
(i don't know exactly what variant of poker you play)
consider having a pair aces. the chance that the next card will be an ace is 2 / 50 (two aces left of fifty cards)
the chance that you get at three aces over all the 5 dealt cards is as follows
2 / 50
48/50 * 2/49
48/50 * 47/49 * 2/48
__________________+
...
the first one is obvious - the chance that the next card is an ace. The second one reads that the first one is not an ace, but the second one is. then you get the option that the next two cards are not aces, but the third one.
for exactly one ace in the next three cards:
2/50 * 48/49 * 47/48
48/50 * 2/49 * 47/48
48/50 * 47/49 * 2/48
__________________ +
....
and then you have yet to make sure it isn't a full house instead.
In a similar fashion you can compute the probability of other hands.
If it doesn't have to be totally a priori and can involve computation, why not just do an exhaustive search of all possible card combinations for a given set of known cards?
For example, have 2 arrays: Cards held/available for use, cards unknown.
Initially, cards held holds only 2 cards, the ones you're holding. Unknown holds all the rest.
Then, algorithmically, try to make combinations of cards from the unknown array and see what the probability of those occurring is. For example, to check the probability of a pair on the flop given 2 known cards:
There are 2 cases. (a) a pair appears in it's own right on the flop (both cards being in the flop). (b) one card is in your hand, one card is in the flop.
(the third case is that you have a pocket pair, in which case of course the probability of a pair is 1.0!)
Because these are mutually-exclusive, you can use statistical theory to find the probability of either then add.
CASE A:
CASE B:
this is just 3/50 (as there are 3 other cards in the pack that could make you a pair for the first of your cards) + 3/50 (for the other card)
I'll give this a shot a coding this up later to make sure I don't get >1 probability or something
JamesM
For example, have 2 arrays: Cards held/available for use, cards unknown.
Initially, cards held holds only 2 cards, the ones you're holding. Unknown holds all the rest.
Then, algorithmically, try to make combinations of cards from the unknown array and see what the probability of those occurring is. For example, to check the probability of a pair on the flop given 2 known cards:
There are 2 cases. (a) a pair appears in it's own right on the flop (both cards being in the flop). (b) one card is in your hand, one card is in the flop.
(the third case is that you have a pocket pair, in which case of course the probability of a pair is 1.0!)
Because these are mutually-exclusive, you can use statistical theory to find the probability of either then add.
CASE A:
Code: Select all
find all possible combinations of two-card groupings in @cards_unknown, that make a valid pair
ensure that these are unique, i.e. if you have AS, AD, remove AD, AS.
$probability = 0.0;
for each unique pair,
$probability += 1.0 / 50 + 1.0 / 49;
this is just 3/50 (as there are 3 other cards in the pack that could make you a pair for the first of your cards) + 3/50 (for the other card)
I'll give this a shot a coding this up later to make sure I don't get >1 probability or something
JamesM
texas holdem was what i was think of using for an example, but come to think of it, it might be wiser to use 5 card stud or simular...Combuster wrote:(i don't know exactly what variant of poker you play)
Im not quite sure what you are trying to show me here?JamesN wrote:If it doesn't have to be totally a priori and can involve computation, why not just do an exhaustive search of all possible card combinations for a given set of known cards?
For example, have 2 arrays: Cards held/available for use, cards unknown.
Initially, cards held holds only 2 cards, the ones you're holding. Unknown holds all the rest.
Then, algorithmically, try to make combinations of cards from the unknown array and see what the probability of those occurring is. For example, to check the probability of a pair on the flop given 2 known cards:
There are 2 cases. (a) a pair appears in it's own right on the flop (both cards being in the flop). (b) one card is in your hand, one card is in the flop.
(the third case is that you have a pocket pair, in which case of course the probability of a pair is 1.0!)
Because these are mutually-exclusive, you can use statistical theory to find the probability of either then add.
CASE A:
Code:
find all possible combinations of two-card groupings in @cards_unknown, that make a valid pair
ensure that these are unique, i.e. if you have AS, AD, remove AD, AS.
$probability = 0.0;
for each unique pair,
$probability += 1.0 / 50 + 1.0 / 49;
CASE B:
this is just 3/50 (as there are 3 other cards in the pack that could make you a pair for the first of your cards) + 3/50 (for the other card)
I'll give this a shot a coding this up later to make sure I don't get >1 probability or something
JamesM
theres 52*51 / 2 possible handcombiations.
(3/51) * 100 % of those are pairs.
OR
(2*3*13 / (52*51 / 2)) * 100% of those are pairs.
Edit: i thik i might understand what you are trying to do, but im quite sure it is neither correct or wise.
I was showing an algorithmic method for determining the probability of a pair on the flop. While it may not be what you want to do (maybe you want to compute everything a-priori, using a formula as opposed to an algorithm), I would like to hear some explanation as to why it is incorrect. Also why it is not wise, but that is less of an issue!
JamesM
JamesM