The probability of getting a pair is 16:1, because if my second card is dealt, there are 51 cards remaining, and 3 of those make a pair to my first card:
Wills = 3
Willnots = Total - Wills = 51 - 3
Probability = Wills : WillNots = 48:3 or 16:1

In the same way: pocket Aces, pocket kings, QQ, JJ, ... are all 220:1
Getting an AK suited, AQ suited, AJ suited, ... are all about 330:1

This is fairly easy to come by, but the question I'm looking an answer to is the following:
Let's say there are 9 people at the table. Knowing that getting a pair is 16:1, what are the odds that:
-> I alone have a pair?
-> 2 people have a pair?
-> if 4 people in front of me let me know they don't have a pair, what are the odds that one of the 4 remaining people behind me have a pair?

In the same way: Knowing that getting AK suited, AQ suited and AJ suited is about 330:1
Then with 9 people at the table, what are the odds that:
-> If I have AQ suited and there are 3 people behind me to act, what are the odds that one of those 3 people is holding AK suited?
-> If I am UTG with AQ suited and there are 8 people behind me to act, what are the odds that one of those 8 people holds AK suited?
-> If I have AJ suited and there are 5 people behind me to act, what are the odds that one of those 5 people is holding either AQ suited or AK suited?

You can roughly assume that the fact that one player has a pair has no effect on the probability that another one has a pair. So, each player independently has a pair 1/17 of the time. Of n given players, noone has a pair with probability (16/17)^n. For 8 opponents this works out to roughly 62%. So if you have a pair somebody else at the table has a pair 38% of the time.

Thanks for the reply, I appreciate it.

Let's see if I understand this: let's say I have pocket tens: The chance of someone having JJ or better is about 54.25:1

So each player independently has JJ or better 1/54.25 of the time. Of n given players this would mean noone has a pair with probability (53.25/54.25)^n

So let's say I'm with 9 guys at the table, first to act and I have TT, there is about an 86% chance ((53.25/54.25)^8) no one has a higher pocket pair?

If I were in mid position with my TT and 4 guys behind me, there is about a 92% probability ((53.25/54.25)^4) no one has a higher pocket pair.
Am I understanding you correctly?

Yep, although I didn't check your arithmetic.

There's no need to do the arithmetic in your head, simply get Tony Guerrero's book, Killer Poker By the Numbers.

Sorry, if I sound like an informercial....

On the contrary, Thanks for the reference man... I've been searching for a book that talks about the above stuff

Was just thinking about this stuff myself lately.

I use this:

The Blood Theorem

1-((1-h)^p) = L

where h = % of hands that beat yours, p = players remaining yet to act, and L = the likelihood that you are beat.



I put this in an excel program and get instant answers.

And your math is spot on. This equation is another way to look at it.

Good luck!

Hey, thanks for the suggestion man, I appreciate it. Always nice to have some options.

"h = % of hands that beat yours"... what exactly do you mean by this? How do you get to this percentage? Could you give an example of how you use this?

By the way "The Blood Theorem" is that something you coined or something that's used widely used?

I know exactly one person who uses it. For others it's just simple probability theory.

In Pokerstove, I manually select the hands that beat mine in my opponent's range and it will give a percentage at the bottom of the menu - that's what I use. Without Pokerstove arriving at this number is still simple: Remembering that there are 12 ways to make an offsuit unpaired hand, 4 ways to make a suited unpaired hand, and 6 ways to make a paired hand - and the total amount of hand combinations is 1326. Pokerstove also makes it easy to deselect cards from your opponents range they cannot have. E.g. if you are holding the A of diamonds, you can remove that from your opponent's range. To do this manually, know that removing one card from your opponent's range leaves them 9 ways to make an offsuit paired hand, 3 ways to make a suited unpaired hand, and 3 ways to make a paired hand.

So if you have AJs on the HJ (4 people left to act) the hands you absolutely don't want to see are AK, AQ, AA, KK, QQ, and JJ. There are 9 ways to make AKo and AQo each (18 total), 3 ways to make AKs and AQs each (6 total), 3 ways to make AA, 6 ways to make KK and QQ each (12 total) and 3 ways to make JJ - totaling 42 hands. 42/1326 = 3.17% of 1 opponent holding any of these hands.

1-((1-h)^p) = L

1-((1-.0317)^4) = 12.09% of the time one of your 4 opponents has one of the 42 hands out of 1326 that beat you.


Yeah, I coined it. It was something I wrote and thought useful and decided to share it. If you find it useful, use it. It's free. If you don't find it useful, ignore it. It's helped me on many occasions, especially jamming or folding in tough tournament situations.

My friend trumpet (who is a math genius) had this to say on it:
And I say:

"Blood Theorem just sounds so coooool!"



Good luck to you.

I'm jealous of those who had a solid education in math. One man's simplicity is another man's struggle.

I own 13 poker books and have read poker websites for the past 5 years and have never seen this simple probability theory anywhere. Maybe I just missed it, but I'm sure there are others out there like that as well. Even if this is as simple as 2+2 to you, I hope it will help others simply to see it in print.

(nb of players behind you)*(nb of pocket pairs higher than yours)/2 gives a rough estimate.

As an example, if you got Queens on the button, there is a 2% chance that SB or BB picked up Aces or Kings.

Since you called me a math genius I will allow you to call your statement a thorem. Seems fair, doesn't it?

I am going to stop focusing on the theory of poker and focus on using Quantum Mechanics to change my hole cards.

Quanum mechanics makes me scared.

While this is a quite reasonable assumption, just to set the record straight, the fact that one player has a pair actually INCREASES the chance that another player will have one. We had an interesting thread on this a while back titled Probability is Tricky.

Shortcut: http://forumserver.twoplustwo.com/15...ciples-433273/

lol