Odds of pairing AK by the river 62.55%

[glworden]

I just found this thread today and it's kind of funny. James posted the correct and clearly explained answer and then got three pages debating simple statistics.

Maybe it's kind of funny, but it's not ludicrous. One of the hardest things about learning is recognizing what your limits and deficits are. Sometimes that means revealing your ignorance to others and asking for comment and correction. If I don't know how to approach statistical problems like this and my ego prevents me from asking questions, I'll never learn, will I?

I've already learned a lot on CardsChat. Perhaps the most important thing is that there is so much I thought I knew that I really don't. Others give correction or a different perspective and if I'm open to it, I can improve my understanding and my game. Some posters are harsher, more critical, and even demeaning than others. I guess you just take that with a grain of salt. But the fact of the matter is, if I close myself off to criticism and correction, then I'm pretty well stuck where I am, encumbered by my baggage of false or limited concepts and knowledge.

I think the fact that there are three pages here with a lot of initial debate speaks to the complexity of poker math problems. Although James is lucid in his explanations, many of us still struggle with the concepts. At this point, I don't hear anybody saying they still feel like James is wrong. Do any of you feel that way?

Last night, I downloaded a poker calculator that arrives at results by running all possible combinations. I designated the hole cards AsKc and asked the program to run one opponent and all possible flop, turn and river combinations. The text just started spinning like crazy, and six hours and a couple million hands later, the thing was still going strong. That's a pretty decent illustration of just how complex all the possibilities are.

So those who took a while to get it, and even those who still don't get it - don't feel bad. Don't duck your head just because you fear somebody might think you're stupid. Ask your questions. Dare to be wrong.

But I still think the OP is a raving idiot.

Gary

 

[jdeliverer]

What does OP stand for? I've been hearing it a lot.

 

[dj11]

Original Post/Poster

 

[Dwilius]

Yeah I didn't realize glworden was the OP when he first called him an idiot I thought that was harsh.

The actual probability is 1 - (44/50)*(43/49)*(42/48)*(41/47)*(40/46) = 48.74%.

To get the probability of pairing on the flop...
1 - (44/50)*(43/49)*(42/48) = 32.43%,

Actually it was the 6th post not 21st that answered it all even doing it the faster way as zach later suggested. Its good for people to get to the answer glworden but this thread seems to have gone round and round.

The problem with a thread that goes on this long is it won't be read all the way through and all the different #'s are just going to confuse people. As Chuck T said, "this thread is now causing head asplotion."

 

[yoru72]

ok this thread is now causing head asplotion for me. Thought I knew what I was saying but now my mind is just going in circles.

Not saying you're right james, but I'm not saying you're wrong either. I'm just lost

Either way, what we're calculating here has little bearing or usefulness in poker...I mean aside from keeping our math skills in check. Equity is what we should be concerned with. For the beginners, don't go leaning on the (correct) math from this thread. Punch AK into an equity calculator like pokerstove and see how it does vs certain hands on all streets. That'll give you an idea of what you want to do with the hand.

Well, is a step in the right direction... because James is completly right about what he is saying.
You cannot simply add the probabilities.
Sorry about my english, but I will try to clarify this with a simple example:
Lets say that you want to calculate the probability to have at least one male child, and you are pretty sure that you will have exactly two childs.
Doing that by extension is pretty easy:
1) you have first a female, and later another female
2) you have first a female, and later one male
3) you have first a male, and later one female
4) you have first a male, and later another male
In the four possible scenarios you have at least one male child in three of them, so you can say that the probability is 75%.

Ok, lets do using a formula:
Using the additive (an incorrect) approach:
probability to have a male in first place 1/2 = .5
probability to have a male in second place 1/2 = .5
.5+.5 = 1 so the conclusion could be that anybody can have two female childs having exactly two childs (something smells wrong here, isn't it?)
Using the correct way (like James did), you can calculate the probability to have TWO females and substract this result from 1 to answer the original question.
So, the probability to have TWO females (because the births are independient events) can be calculated by the formula:
P(A and B)=P(A).P(B)=(1/2).(1/2)=0.5*0.5=0.25
So, you will have exactly TWO females with a probability of 25% and you will have at least one male with a probability of 75% (1-25%)

About the additive approach, you must use
The Addition Rule:

Consider events A and B. P(A

B)= P(A) + P(B) - P(A

B)

​

(cant read about it here: Addition Rule of Probability)

So, everyone using this approach are forgetting substract P(A

B)

I'm tired (is late here), and I'm pretty sure that with the example and the link everyone here can find their own answers.

 

[santa fe slim]

I won't attempt the math, I'll just quote from Phil Gordon's Little Green Book: "If I am unpaired in the hole, I will flop at least one pair 32.40%" of the time. Unpaired certainly describes big slick.

 

[jdeliverer]

32.42% actually. I think we've pretty much cleared up this thread, people aren't reading anything and almost everything is just repeats. Everything yoru said was correct.

 

[glworden]

Yeah I didn't realize glworden was the OP when he first called him an idiot I thought that was harsh.

The problem with a thread that goes on this long is it won't be read all the way through and all the different #'s are just going to confuse people. As Chuck T said, "this thread is now causing head asplotion."

Is there any way I can go back and edit my original post in order to briefly explain that the question is resolved and that my initial assertions were wrong? I'm not seeing an editing function.

Thanks to James - and others - for all your smarts.

Gary

 

[RickH2005]

A-K paired odds?

Just ask Chris Ferguson--I bet HE knows! Buy put simply, 6 outs to pair either one would be sumpin' like--lessee--6 outs time 5 cards to be flopped=30x2=---YUP! 60%! (I think!):smile:

 

[Dwilius]

Is there any way I can go back and edit my original post in order to briefly explain that the question is resolved and that my initial assertions were wrong?

Only a mod can do that now, but I just looked over the thread and it was pretty lucid at the beginning, since people start at either the beginning or end maybe make a summary at the end

...and Chuck T is right about looking into hand equity, you can hold on to AK to make one pair by the river just to lose to aces up, etc
...knowing its 1/3 to hit flop is good, but other than that we should be concerned with equity, new thread .

 

[RickH2005]

OK--A-K

WOW!!! Lots of math theory here looks like!:deal: Thanx fer the note, by the way! Now I remember I flunked math TWICE in college----an' now I know why!

 

[jdeliverer]

RickH what are you doing in your math?? Why do you multiply 6x5? And how does multiplying by 2 do anything or make it a percent?

*confused*

 

[pimpinalovabkln]

i just have a question , you know how people say hitting pocket aces is like 1 in every 210 or 220 hands, but shouldn't that be the same for any pocket pair?

 

[glworden]

i just have a question , you know how people say hitting pocket aces is like 1 in every 210 or 220 hands, but shouldn't that be the same for any pocket pair?

YES. Any specific pocket pair. The odds against hitting AA are 220/1. The odds against hitting 33 are 220/1.

Please, lord, let me be right about this one. James?

Gary

 

[ChuckTs]

YES. Any specific pocket pair. The odds against hitting AA are 220/1. The odds against hitting 33 are 220/1.

Please, lord, let me be right about this one. James?

Gary

Yep, you're correct. Just to clarify though, odds are written in ratios of x:y. Writing it as a fraction can definitely confuse things.

So the odds of hitting any pocket pair are 220:1 against, or 1 in 221.

 

[gis1239]

i have'nt read the whole thread so don't know if this has come up but if you are not to sharp with maths you can calculate outs roughly by multiplying with 2 per out or so the chance of hitting a A or K is roughly 6*10=60% i know it's not 100 % accurate but it's a pretty good way to do it if you just started

 

[nateogreato]

What a great discussion

I really like Jamesda...'s response. When I was in school that's the way I learned statistics. This is also consistant with what I've learned in poker books.

Bottom line: AK is a coin flip against any pair lower than KK (QQ, JJ, ....). Don't go all in with it preflop unless you are in blind trouble or you are trying to do a quick double and are happy with >50% chance of ending your tourney.

Thanks for all your insites.

 

[gis1239]

I really like Jamesda...'s response. When I was in school that's the way I learned statistics. This is also consistant with what I've learned in poker books.

Bottom line: AK is a coin flip against any pair lower than KK (QQ, JJ, ....). Don't go all in with it preflop unless you are in blind trouble or you are trying to do a quick double and are happy with >50% chance of ending your tourney.

Thanks for all your insites.

you can go all-in with it sometime you just can't call an all-in it can be +ev to push if you think your opponent will fold like 5-10% of the time

 

[nateogreato]

Well poker is often about opinions

If you want to play TAG (tight-aggressive) like many succesful poker pros do, it's a good idea to limit the gambling you do and bet on high odds pots. In time, the donkeys will eliminate themselves and hand you their chips. If this was not true, then no one would win regularly at the tables. I know some very tallented people who do just that. I am not saying that I am that good - far from it.

I'm OK with going all in with AK off. I'd just say it's not to smart if you want to protect your stack and play low risk hands. Honestly, I think you have to gamble a little whern you get in blind trouble or when you have insight into you opponants weaknesses. First hand with unknown opps, it's a gamble.

 

[jdeliverer]

I really like Jamesda...'s response. When I was in school that's the way I learned statistics. This is also consistant with what I've learned in poker books.

Bottom line: AK is a coin flip against any pair lower than KK (QQ, JJ, ....). Don't go all in with it preflop unless you are in blind trouble or you are trying to do a quick double and are happy with >50% chance of ending your tourney.

Thanks for all your insites.

The real value of AK is its fold equity. You can win pots much of the time with it just from betting 4BB and everyone folding. And also, AK is one of the better hands heads up, which is where you will usually be if you don't steal the blinds. It is far better than 22-99.

Why? Let's look at some examples.

If you have AK you could be:

Way behind: AA or KK, which dominates any pp up to 99 as well.
Coin flip: Against any other pp
Way ahead: Any ace or king not AA, AK, or KK

If you have a low pocket pair you could be:

Way behind: Any pp higher than yours
Coin flip: Against two overcards

---

Very rarely will you be way ahead with 22-99, unless someone calls with A9 or 33 in which case they should be a good source of money for you On the other hand, AK will be dominating against plenty of people willing to go all in with AQ, KQ, etc.

Combined with the fold equity, this makes AK a very strong hand when played well pre and postflop.

 

[jdeliverer]

i have'nt read the whole thread so don't know if this has come up but if you are not to sharp with maths you can calculate outs roughly by multiplying with 2 per out or so the chance of hitting a A or K is roughly 6*10=60% i know it's not 100 % accurate but it's a pretty good way to do it if you just started

This is only accurate postflop, the mistake of being 10% off in your calculation is huge when calculating pot odds. This is good enough when calculating from the flop to the river or turn to the river, but for the whole board the margin of error is not within acceptable limits.

 

[SavagePenguin]

I take James' side on this one. I remember a teaser question during an episode of the WPT along these lines. The answer was that you'd pair your hand 50% of the time if you saw all five cards. I've always just assumed it was true, and James seems to support that.

 

[glworden]

THE QUESTION IS RESOLVED

Hey Folks,
THE QUESTION IS RESOLVED. James solved it on the first page, there were a couple pages of doubt and discussion, but now it's clear that the probability is just under 50%, as James calculated.

So there's no need to side with James, or to side with the Original Poster, which is ME. I posted the question to check whether my methodology is correct and sound. It's not. I learned something.

Thanks for all the input and discussion - but the question is RESOLVED!
glworden

 

[PurgatoryD]

Resolved? But I want to play, too.

Actually, I won't touch the resolved issue because, well, it's resolved. But you did post a question about the "poker odds Calculator" back on page 1 of this thread. I was playing with some numbers (before I figured out that this thread was 3 pages long) and I think I figured out what it's computing.

On the poker odds calculator, it appears that they calculated the odds of getting *exactly* one of your outs. The odds of getting exactly one of your outs in any of the 5 draws is 7.69%. Since there are a total of five draws, you have a total probability of 5 * 7.69% = 38.45% of getting exactly one of your outs. If you want more on the 7.69% number, I can go into it. But you may not. But if you do, just ask. I just had to post since I did the @$%^ math!

Great thread, BTW. Especially the bold red type at the end!
dave

 

[jdeliverer]

*whacks self in head*

I answered that on page 2. But good observation.