- Status
jdeliverer
Legend
Silver Level
I am paying attention to what they're saying. I am not calculating the probability of hitting exactly a single ace or king; rather, the others are double-counting the probability of hitting more than one.
And for this argument, you should 'stop calculating the probabilites if we hit an ace or king' because you've already counted it. And if you already counted it, it doesn't matter whether it pairs or not. I don't think there's any more I can say to convince you though, just try to look at it again.
Irexes
Legend
Silver Level
At least we agree you are answering the wrong question and the answer to the OP is nearer 60%.
jdeliverer
Legend
Silver Level
NO!! We don't agree at all.
If you look at my original post, you can see that I calculated the probability of NEVER pairing anything. Then I subtracted that from 1, leaving us with the probability that ANYTHING paired, one pair or two pair or trips, a FH or even quads. I am taking into account all cases.
If you go by the original answer, you are counting the first pair you get, in addition to the second pair, and anything else. Thus you are counting two pair or trips twice and a full house or quads three times.
G
glworden
Visionary
Silver Level
TO FURTHER CONFUSE THE ISSUE
James, there is no animosity at all. I appreciate the discussion, even if I'm still not getting my head around it totally. It's an odd thing, isn't it, that we have all these poker players who can't agree on what is seemingly basic and common statistic. I am the OP; my logic makes sense to me and many posters are backing me up. But heck - maybe we're all wrong!
To add further to the confusion, here's an online calculator which came up with a COMPLETELY different answer. I think I entered the info correctly. We're drawing from 50 cards, drawing 5 cards with 6 outs, and only require one hit to fulfill the parameters. Now this thing is saying our probability of hitting is 38.4%.
Can you explain that one?
James, there is no animosity at all. I appreciate the discussion, even if I'm still not getting my head around it totally. It's an odd thing, isn't it, that we have all these poker players who can't agree on what is seemingly basic and common statistic. I am the OP; my logic makes sense to me and many posters are backing me up. But heck - maybe we're all wrong!
To add further to the confusion, here's an online calculator which came up with a COMPLETELY different answer. I think I entered the info correctly. We're drawing from 50 cards, drawing 5 cards with 6 outs, and only require one hit to fulfill the parameters. Now this thing is saying our probability of hitting is 38.4%.
Can you explain that one?
Attachments
Irexes
Legend
Silver Level
Ok, I'm going to do an about turn here I think.
There is a 54.5% chance of not drawing an A or a K. And so therefore a 45.5% chance of drawing at least one.
The logic of your other posts makes sense too. So unless someone comes along with further evidence to the contrary, apologies James.
Zach, where are you??
There is a 54.5% chance of not drawing an A or a K. And so therefore a 45.5% chance of drawing at least one.
The logic of your other posts makes sense too. So unless someone comes along with further evidence to the contrary, apologies James.
Zach, where are you??
jdeliverer
Legend
Silver Level
glworden, I have no idea what the software is doing here, but I'll try to figure it out. anything under 40% is ridiculously low, as AK has a higher chance than that of beating pocket 2's. I'm assuming AK doesn't have a 10% chance of hitting a straight or flush 
jdeliverer
Legend
Silver Level
Ok I figured it out.
That calculator is doing what Irex thought I was doing. It calculates getting exactly one out.
Replacing the Required box with 2 we get 0.0938 and with 3 we get 0.009.
0.3844+0.0938+0.009 = 0.4872
Which is our answer
That calculator is doing what Irex thought I was doing. It calculates getting exactly one out.
Replacing the Required box with 2 we get 0.0938 and with 3 we get 0.009.
0.3844+0.0938+0.009 = 0.4872
Which is our answer
G
glworden
Visionary
Silver Level
Isn't this fun?
I'm just amazed that this seemingly simple question is so hard to grasp. I think James is winning converts by the force of his conviction; he really sounds like he knows what he's talking about.
I'm not trying to convince anybody. I'm just trying to figure out how to approach it and why I'm wrong.
How many of you (us) thought you knew the answer to this? How many thought 62%? How many still think that?
Gary
I'm just amazed that this seemingly simple question is so hard to grasp. I think James is winning converts by the force of his conviction; he really sounds like he knows what he's talking about.
I'm not trying to convince anybody. I'm just trying to figure out how to approach it and why I'm wrong.
How many of you (us) thought you knew the answer to this? How many thought 62%? How many still think that?
Gary
G
glworden
Visionary
Silver Level
Do the maths for not hitting gl as per James' first post, using Excel or similar.
It's pretty convincing
OK, here's a thought. We're trying to hit a pair or better? You could use Jame's method of never pairing a card, yet still end up with a straight or flush. How do you figure those chances in?
G
glworden
Visionary
Silver Level
Does poker stove give yo an answer to the basic question: pairing an A or K by the river?
T
Toad
Rock Star
Silver Level
James,
Tell me if I'm on the right track here. When you say we can't simply add the probabilities I had to take a simpler example:
We are going to flip a coin 4 times and want to find out the probability that heads will come up at least once.
.5+.5+.5+.5=2 (obvioulsly not possible since prob has to be between 0 and 1)
By your math we take the odds that it won't happen, mutiply and subtract from one.
1-(.5*.5*.5*.5)= .875 or 87.5% that we will hit heads at least once.
Right?
My brain hurts...
Tell me if I'm on the right track here. When you say we can't simply add the probabilities I had to take a simpler example:
We are going to flip a coin 4 times and want to find out the probability that heads will come up at least once.
.5+.5+.5+.5=2
(obvioulsly not possible since prob has to be between 0 and 1)By your math we take the odds that it won't happen, mutiply and subtract from one.
1-(.5*.5*.5*.5)= .875 or 87.5% that we will hit heads at least once.
Right?
My brain hurts...
jdeliverer
Legend
Silver Level
D'wilius - exactly right
Toad - Yup (except you did something wrong since 1-(.5*.5*.5*.5) = 0.9375.
It's the same as 0.5 (chances we get heads 1st time) + 0.5 (chances we will get heads 2nd time)*0.5(chance we have not gotten heads yet) + 0.5(chances we get heads 3rd time)*0.25(chance we have not gotten heads yet + 0.5(chances we get heads 4th time)*0.125(chance we have not gotten heads yet) = 0.5 + 0.5*0.5 + 0.5*0.25 + 0.5*0.125 = 0.9375.
Note that for all calculations past the first flip, we disregard any situations in which we got heads in the first flip because we already counted those situations.
Toad - Yup (except you did something wrong since 1-(.5*.5*.5*.5) = 0.9375.
It's the same as 0.5 (chances we get heads 1st time) + 0.5 (chances we will get heads 2nd time)*0.5(chance we have not gotten heads yet) + 0.5(chances we get heads 3rd time)*0.25(chance we have not gotten heads yet + 0.5(chances we get heads 4th time)*0.125(chance we have not gotten heads yet) = 0.5 + 0.5*0.5 + 0.5*0.25 + 0.5*0.125 = 0.9375.
Note that for all calculations past the first flip, we disregard any situations in which we got heads in the first flip because we already counted those situations.
dj11
Legend
Silver Level
I was the original post who quoted CardPlayer mag in one of those little blurbs that fill space saying that AK will pair by the river 60% of the time.
Its a number I generally use, but won't go broke over too often.
Edit add, several minutes later...
I found my original post of this in this thread;
https://www.cardschat.com/forum/gen...ourite-hold-em-article-79328/?highlight=blurb
My post was;
blah blah blah,
Most interesting snipet from the issue currently in my reading room is a Poker Odds blurb which states;
'The chances of either an Ace or a King appearing on the board by the river are 60 percent.'
This does not say that those are the odds of pairing AK by the river. It is entirely possible tho that I extrapolated in another post (hopefully burried extra deep) that those odds were valid for pairing.
Its a number I generally use, but won't go broke over too often.
Edit add, several minutes later...
I found my original post of this in this thread;
https://www.cardschat.com/forum/gen...ourite-hold-em-article-79328/?highlight=blurb
My post was;
blah blah blah,
Most interesting snipet from the issue currently in my reading room is a Poker Odds blurb which states;
'The chances of either an Ace or a King appearing on the board by the river are 60 percent.'
This does not say that those are the odds of pairing AK by the river. It is entirely possible tho that I extrapolated in another post (hopefully burried extra deep) that those odds were valid for pairing.
Last edited:
- Status