This is a discussion on High pocket pair frequency within the online poker forums, in the Learning Poker section; I don't know if this is just the conspiracy theorist in me or what, but it feels like anytime I have pocket Kings someone else 


#1




High pocket pair frequency
I don't know if this is just the conspiracy theorist in me or what, but it feels like anytime I have pocket Kings someone else at the table has Aces, or if I have aces someone else will have kings, or if I have pocket queens, they'll have kings.
Is there some explanation for this phenomenon? It also seems like pocket kings almost always flops a set and wins over pocket aces. I hate to think like this. I'd like to believe Pokerstars/FTP are on the up and up, but it feels like it happens too frequently to just be a coincidence. 
#2




I often get the same feeling especially on FTP, followed by scolding myself for my usual sin of greed. By this I mean, at times, when I feel I'm very strong I will over bet looking to score big and end up pissing off the poker gods.
Below are some useful probabilities though I recommend everybody commit to memory if they haven't already. 
#3




Yeah it just really has to do alot on the random number generator which you can google to find out how it works exactly...you can't really worry about those things but just try to make sure you get a set which isn't up to you but yeah

#4




Quote:
I suspect that if you used poker tracker and went through all the ocasions that you were dealt KK and someone else had AA that it would be no higher than it should be. 1st column = # of Opponents 2nd column = Chance at least one opponet has AA when you have KK 1 0.49% or 1/204 2 1.00% or 1/100 3 1.52% or 1/66 4 2.07% or 1/48 5 2.64% or 1/38 6 3.23% or 1/31 7 3.84% or 1/26 8 4.48% or 1/22 9 5.15% or 1/19 
#5




re: Poker & High pocket pair frequency
Well, it is all up to chance...There is always the same amount of cards used in each game...the more players there are the higher the chance will be that they have a pair higher then yours...that where you have to take the risk...if you've got kings, you're pretty set but there's will always be that small chance

#6




When you have K/K, assuming a full table, you'll be up against A/A about one in 25 times. (Remember, you're against 8 other people who can cold deck you with A/A)
You will win with K/K vs A/A about 1 in five times. Some schmuck with 6h/7h hearts will crack your AC/Ad about 23% of the time. Suckouts are pretty common. That's the nature of the game. 
#7




I thought I read somewhere that if u have a pocket pair, and opponent is more likely to have a pocket pair than if you did not have a pocket pair. The pair vs pair paradox?
just googled it, since i forgot the numbers myslef but here it is: "The odds of more than one player on a tenhanded Hold 'Em table receiving a pocket pair is 11.4%. But when YOU see your pocket pair, the odds of another player holding one is 42.5%" another example is if u flip a coin for heads or tails..the chance to flip it 3 times Heads in a row is 50%*50%*50%=12.5%, BUT if you have flipped heads twice already than its 100%*100%*50%=50% chance to hit heads "The bottom line: When you have a pocket pair in Hold 'Em, the chances another player on your 10player table also has a pocket pair is over 42%. When you see those hands, don't think the site is rigging the deal to build pots. Think that this is how things work in a random system"  http://www.pokerdiy.com/pokerforums...afv/topic.aspx I googled it and found that info there ^^ is the link if u want to read for urself, or if dunno if i'm supposed to quote where I c/p, so I did. I remember reading about it in a book i thought, H on cash vol.1? Not sure, but before i read about this I was always saying to myself, how come every damn time I get a pp someone else has a bigger one/ lower one that hit set...Hope this helps. 
#8




Eep. It's true, but that's a bad example. Coin flips are statistically independent, so flipping 2 heads doesn't increase the odds of the next flip being heads. It's always 50/50. Dealing cards DOES affect the distribution because they're making pairs with the remaining 50 of 52. Basically, if you don't have a pair, and 2 randomly selected cards from the deck are 144/2450 (5.8876%) to be a pair. If you do have a pair, it rises to 146/2450 (5.9592%). The odds are better, but only by a very little bit. If you have no pair at a 10 person table, 42.0252% of the time, someone else does. If you do have a pair, 42.4762% of the time someone else does, too. It's only half a percent more.
Corrections to my math are welcome if I am in error. RF 
#10




re: Poker & High pocket pair frequency
The chance of getting dealt a Pocket Pair in the hole are about 5%. Unfortunately, those odd's apply for every other player at the table too. So at a 10 seat table, 10 x 5% chance....50% chance of a player having a PP.
If your dealt a pair, everyone elses individual chance remains (theoretically) the same. So when you get your Kings, there is around a 45% chance of being up against another pair. And with 13 card values, that leaves about a 3/4% chance of seeing Aces in your opponents hand. Already been said above, just slightly less long winded. 
#11




That's a really common mistake that happens to be pretty close to the right answer anyway in this case. 10 independent events at 5% each isn't actually 10 x 5%.
The best counter example I saw was a student from a well known pricey university who complained about the equity in ticket distribution. Each year, any given student had about a 1/3 chance of getting season tickets. Masters students, they reasoned, had a 2/3 chance while undergrads, at 4 years, had a 4/3 chance of getting season tickets! Uh, no. Probability ranges from zero to one. There's no such thing as a 4/3 chance. The easiest way to figure out the chance of something happening at least once is to figure out the chance of it not happening at all, and to get that, you multiply the chances (always with independent events). So, no pair in one hand is about 95% (see above for more exact numbers). 2? 95% x 95%. 3? 95%^3 10? 95%^10, which is 0.5987 (59.87%). P(at least one pair in 10 trials) = 1  P(no pairs in 10 trials) = 1  0.5987 = 0.40126 (40.126%) It's off a little due to rounding to 5% instead of using 1/17. RF 