Two Plus Two Poker Forums Probability of set over set
 Register FAQ Search Today's Posts Mark Forums Read Video Directory TwoPlusTwo.com

 Notices

 Probability Discussions of probability theory

 01-05-2009, 01:17 PM #1 rich.mann23 newbie   Join Date: Dec 2008 Posts: 31 Probability of set over set Is this roughly the calculation for flopping set over set in a nine handed game of texas hold em assuming anyone with a pair sees the flop? Also, what is the probability of flopping an under set when set over set is flopped? ASSUME: you hold 10-10 and your opponent holds Q-Q. (We make no assumptions regarding the distribution of other cards to other players at the table.) This leave 48 cards unaccounted for. Therefore, based on our assumption, there are C(48,3)=17296 equally likely flops possible. (Note: C(n,k) = the number of possible combinations of k cards from n cards. The formula is n!/(k!(n-k)!).) In order for both players to flop sets, the flop must include one of the two remaining queens, one of the two remaining tens, and one of the other 44 cards in the deck. The number of ways such a flop can occur is 2*2*44=176. Therefore, the probability of set-over-set occurring, given our initial assumption, 176 in 17296. This is about 1.02%, or roughly one in 98. (Note: this rules out the possibility of the flop bringing quads vs a full house. This is extremely unlikely - the probability of a TTQ flop, given the above assumptions, is a very remote 2 in 17296, or about 0.0116%; the probability of a QQT flop is, of course, the same.)
01-05-2009, 01:29 PM   #2
Actually Shows Proof

Join Date: Aug 2008
Location: This looks interesting.
Posts: 7,903
Re: Probability of set over set

Quote:
 Originally Posted by rich.mann23 Is this roughly the calculation for flopping set over set in a nine handed game of texas hold em assuming anyone with a pair sees the flop? (snip) Therefore, the probability of set-over-set occurring, given our initial assumption, 176 in 17296. This is about 1.02%, or roughly one in 98.
That is the odds for two opponents flopping sets (but not quads) when they both start with pocket pairs and both see the flop, at any size table. That is not the same as the odds of set-over-set when you simply assume all players with pocket pairs will see the flop, as you have to multiply this probability by the chance of two pocket pairs being dealt at an X-handed table. That's not as important a question anyway, since what you really want to know is the chance of being beat by another set when you have one. 98:1 is the answer if you know opponent started with a pocket pair.

 01-06-2009, 11:18 AM #3 rich.mann23 newbie   Join Date: Dec 2008 Posts: 31 Re: Probability of set over set Okay. On average should you flop an over set 50% of the time?
01-06-2009, 02:04 PM   #4
Actually Shows Proof

Join Date: Aug 2008
Location: This looks interesting.
Posts: 7,903
Re: Probability of set over set

Quote:
 Originally Posted by rich.mann23 Okay. On average should you flop an over set 50% of the time?
So you are asking that when set-over-set occurs, will your own set will be higher 50% of the time?

If you have top set, it will be best 100% of the time. If you have bottom set, it will lose 100% of the time. If you have middle set, it will be best when opponent has bottom set and worst when opponent has top set. So I suppose you could deduce that in set-over-set flops you will win them 50% of the time in the long run. But you won't play enough poker in your entire life for that to matter.

As mentioned before, you have to multiply the 98:1 odds by the chance that two players at an X-handed table are dealt pocket pairs AND by the chance that they both see the flop. The chance that two players at a 10-handed table are dealt pocket pairs of different ranks is about 9.5% (if same rank there is no set possibility). Then, for reasons I analysed in another thread, we can suppose that 24% of the time they will both see the flop (use any number you like better here).

Thus 98:1 x 24% x 9.5% = 4377:1
or once in 4378 deals at a 10-handed table, you might see set-over-set at the flop between two opponents at that table. If we assume you will be one of those players 1/10th of the time, then you should encounter this scenario once in about 44,000 hands played.

With a situation that happens that rarely, the long run odds don't matter to your winnings at all. Bottom line is if you flop a set, be willing to put all your money in. Even if you have bottom or middle set and the board is not straightening or flushing, still put all your money in. Don't get hung up on worrying about getting beat by set-over-set. It happens so rarely that it's practically irrelevant, and if you just shove your stack every time you have any set, you'll come out way ahead.

 01-20-2009, 11:32 PM #5 2SHAE banned   Join Date: Nov 2008 Location: Swagger back Posts: 3,576 Re: Probability of set over set Happened twice to me in 1300 hands of six max. Ouch.
 02-27-2009, 10:46 AM #6 shorn7 old hand   Join Date: Aug 2007 Posts: 1,938 Re: Probability of set over set I have just experienced a run of under 100 hands where I flopped and underset 3x to another player at full ring. Talk about "tail" of a distribution.
 06-30-2009, 02:14 AM #7 googleit123 adept   Join Date: Mar 2007 Location: play me a good country tune Posts: 917 Re: Probability of set over set OK, so we know the formula for calculating set over set...but what about set over set over set. It happened to me today on Stars and have been searching for the probability of that happening. Anybody know what the probability is or how to calculate it? Here's the hand fwiw, What were the poker Gods thinkin' when this went down? PokerStars 25c/50c FR Limit Pre Flop: (1.5sb) Hero is in BB with AA UTG raises, 2 folds , MP 3!, 3 folds, SB calls, Hero CAPS, UTG calls, MP calls, SB calls Flop: (16sb) JAK SB checks, Hero bets, UTG calls, MP calls, SB folds Turn: (19sb) 8 Hero bets, UTG raises, MP 3!, Hero CAPS, UTG calls, MP calls River: (21.5bb) 6 Hero bets, UTG calls, MP calls FINAL POT: 24.5bb UTG mucks KK MP mucks JJ Hero shows AA (Three of a kind, Aces) Hero wins pot. Set over set over set! Never seen it before. Today turned out to be +75bb. Man, has this month turned around. June graphs thread?
 11-12-2009, 12:39 AM #8 UziStuNNa banned   Join Date: Oct 2009 Posts: 181 Re: Probability of set over set Odds of flopping set over set over set, assuming you know that all 3 players have pocket pairs preflop and that each player does not know each others hands would be 1 to (50^3)/(2^3). Equaling, 1:15625.
11-12-2009, 01:24 AM   #9
BruceZ
Carpal \'Tunnel

Join Date: Sep 2002
Posts: 11,876
Re: Probability of set over set

Quote:
 Originally Posted by UziStuNNa Odds of flopping set over set over set, assuming you know that all 3 players have pocket pairs preflop and that each player does not know each others hands would be 1 to (50^3)/(2^3). Equaling, 1:15625.
If 3 players have pocket pairs, then there are 46 cards unaccounted for, and there are C(46,3) = 46*45*44/(3*2*1) = 15,180 possible flops. There are 2 cards of each of the 3 ranks that make sets for a total of 2*2*2 = 8 flops that make set over set over set, so the probability of these 3 players all flopping a set is 8/15,180 or odds of 1896.5-to-1.

11-12-2009, 02:03 AM   #10
BruceZ
Carpal \'Tunnel

Join Date: Sep 2002
Posts: 11,876
Re: Probability of set over set

Quote:
 Originally Posted by rich.mann23 Is this roughly the calculation for flopping set over set in a nine handed game of texas hold em assuming anyone with a pair sees the flop?

The probability of set over set over set at a 9-handed table, with all pairs seeing the flop, is exactly

C(13,3)*43 / C(52,3) *
C(9,3) * 9/C(49,2) * 6/C(47,2) * 3/C(45,2)

=~ 1 in 111,666

The top line is the probability of 3 different ranks on the flop. The second line is the number of ways to choose the 3 players C(9,3), times the probability that each player has a different one of these 3 ranks. This is exact since only 3 players can have these 3 sets at one time.

The trick here is to consider the flop first. If you consider the players first, then the more pocket pairs that are out, the more likely it is that 3 players flop sets, so this becomes complicated. The above calculation avoids that.

 02-05-2010, 11:40 PM #11 ncceche stranger   Join Date: May 2007 Posts: 8 Re: Probability of set over set Playing on Full Tilt Rush poker 3 tables at a time. I got it 11 times this week. And 3 times today. I won 2/11. I just started saving hand history the other day so i will dig through them to find the hands. I see about 600 hands per hour at 3 tables. And I've played about 18 hours this week. 600 X 18 = 10,800 hands. so I shouldve only saw it once. Yet i'm some how still up for the week. Does that make me lucky in a way? for beating the odds?
02-20-2010, 07:10 AM   #12
NoCaution
stranger

Join Date: Nov 2008
Posts: 7
Re: Probability of set over set

Quote:
 Originally Posted by ncceche Playing on Full Tilt Rush poker 3 tables at a time. I got it 11 times this week. And 3 times today. I won 2/11. I just started saving hand history the other day so i will dig through them to find the hands. I see about 600 hands per hour at 3 tables. And I've played about 18 hours this week. 600 X 18 = 10,800 hands. so I shouldve only saw it once. Yet i'm some how still up for the week. Does that make me lucky in a way? for beating the odds?
Well I have no doubt that at least one was against me. I have lost to set over set at least 6 times in the last week playing Rush (maybe more - I don't feel like digging through the HH). I've been playing about 2500 hands per day and found this thread while searching for just how bad my luck has been lately. I'm on a sick downswing (10 buy ins or so), and have experienced some of the ugliest coolers of my life.

As soon as I make Ironman, I'm taking a break. I was destroying the Rush tables for the first 2-3 weeks, but I've given it all back now, so I guess it's time to reevaluate.

Cheers-

03-30-2010, 04:28 PM   #13
aise5668

Join Date: Dec 2009
Posts: 850
Re: Probability of set over set

Quote:
 Originally Posted by spadebidder The chance that two players at a 10-handed table are dealt pocket pairs of different ranks is about 9.5%
How do you get this 9.5%?

03-30-2010, 11:30 PM   #14
Actually Shows Proof

Join Date: Aug 2008
Location: This looks interesting.
Posts: 7,903
Re: Probability of set over set

Quote:
 Originally Posted by spadebidder The chance that two players at a 10-handed table are dealt pocket pairs of different ranks is about 9.5% (if same rank there is no set possibility).
Quote:
 Originally Posted by aise5668 How do you get this 9.5%?
The first pocket pair at a 10-handed table is

1 - (1-(78/1326))^10 = 45.46%

The second pocket pair of a different rank, given that the first one happened, is

1 - (1 - (72/1225))^9 = 42.03%

The product of those is 19.1%, which is the correct answer. So I have no idea where I got 9.5% in the earlier post. It was wrong.

These are mutual-exclusive approximations but are very accurate. You could use inclusion exclusion to get the exact answer and it would be within about a tenth of a percent or so.

03-31-2010, 06:17 AM   #15
BruceZ
Carpal \'Tunnel

Join Date: Sep 2002
Posts: 11,876
Re: Probability of set over set

Quote:
 Originally Posted by spadebidder The first pocket pair at a 10-handed table is 1 - (1-(78/1326))^10 = 45.46% The second pocket pair of a different rank, given that the first one happened, is 1 - (1 - (72/1225))^9 = 42.03% The product of those is 19.1%, which is the correct answer. So I have no idea where I got 9.5% in the earlier post. It was wrong. These are mutual-exclusive approximations but are very accurate. You could use inclusion exclusion to get the exact answer and it would be within about a tenth of a percent or so.

First of all, they are independent approximations, not mutually exclusive. Secondly, if you had used a mutually exclusive approximation, you would have quickly seen that the answer can be no bigger than 15.6%:

C(10,2)*78/C(52,2)*72/C(50,2) =~ 15.6%

Thirdly, if you were going to use an independence approximation, the following one would have given an even tighter upper bound of about 14.4%, though it takes some thought to realize that this is still an UPPER bound:

1 - [1 - 78/C(52,2)*72/C(50,2)]C(10,2) =~ 14.4%

Fourthly, this turns out to be one case where the independence approximation is not very accurate as the exact answer is about 11.3%.

The exact calculation uses a variation of inclusion-exclusion where the adjustment factors are not just +/- 1, since we are starting with 2 players. Each line of inclusion-exclusion has multiple terms which each require their own adjustment factor which are shown in bold below. Getting these adjustment factors right takes some thought. The rest isn't difficult. Here are the first 4 terms:

C(10,2)*78*72 /C(52,2)/C(50,2) -

C(10,3)*(2*78*72*66 + 1*78*72*1*3) /C(52,2)/C(50,2)/C(48,2) +

C(10,4)*(3*78*72*66*60 + 2*78*72*66*1*C(4,2) + 1*78*1*72*1*3) /C(52,2)/C(50,2)/C(48,2)/C(46,2) -

C(10,5)*(4*78*72*66*60*54 + 3*78*72*66*60*1*C(5,2) + 2*78*72*1*66*1*15) /C(52,2)/C(50,2)/C(48,2)/C(46,2)/C(44,2)

=~ 11.3%

The first line is for 2 players, the second line is for 3 players, etc. The first summed term on the second line is for 3 different pairs. These are counted C(3,2) = 3 times by the first line, hence the adjustment factor subtracts 2. The second term is for 3 players holding 2 different pairs. These are counted twice by the first line, hence the adjustment factor subtracts 1.

The first term on each line is for N players having N different pairs, starting with N=2 on the first line, N=3 on the second line, etc. The second term on each line is for N players holding N-1 different pairs starting with N=3 on the second line. Starting with N=4 on the third line, the third term is for N players holding N-2 different pairs.

Finally, the question asked by the OP concerns set over set. For that problem, if you consider the flop first, then the above probability is not needed. It becomes an easy matter to make an exact calculation involving 2 or 3 players holding 2 or 3 specific pairs. See my earlier post in this thread with an example for 3 sets.

Last edited by BruceZ; 03-31-2010 at 07:29 AM.

 03-31-2010, 08:19 AM #16 spadebidder Actually Shows Proof     Join Date: Aug 2008 Location: This looks interesting. Posts: 7,903 Re: Probability of set over set Bruce - I believe this is a difference between EXACTLY two players have a pair, and AT LEAST two players have a pair, which satisfies the reason for calculating it, and so that is what I gave although I should have been more clear. You gave the answer for EXACTLY two players, which doesn't really help us estimate how often two players will see the flop holding pairs, which was my original intent. There's no doubt this is a close correct approximation for some player getting a PP at 10-handed: 1 - (1-(78/1326))^10 = 45.46% For someone else having a pair (excluding same rank) I used the same approximation method and got: 1 - (1 - (72/1225))^9 = 42.03% where Alspach calculated this the (very) long way as 42.06%. So that's about where this approximation should fall relatively and I believe it is correct for another pair at the table. http://www.math.sfu.ca/~alspach/comp35/ So this gives us the chance that AT LEAST one other player has a pair, not that EXACTLY one other player has a pair, and I'm thinking that difference is the entire difference between my answer and yours. All that leaves is the combined probability of the two events. It seems we should just multiply to get that answer, which gives 19.1% for at LEAST two players holding a pair. Agree? There's no reason the independent approximation should be much off at all in this case, and I don't think it is as long as we define what we are calculating. Last edited by spadebidder; 03-31-2010 at 08:39 AM.
03-31-2010, 08:45 AM   #17
BruceZ
Carpal \'Tunnel

Join Date: Sep 2002
Posts: 11,876
Re: Probability of set over set

Quote:
 Originally Posted by spadebidder Bruce - I believe this is a difference between EXACTLY two players have a pair, and AT LEAST two players have a pair, which satisfies the reason for calculating it, and so that is what I gave although I should have been more clear. You gave the answer for EXACTLY two players, which doesn't really help us estimate how often two players will see the flop holding pairs, which was my original intent. There's no doubt this is a close correct approximation for some player getting a PP at 10-handed: 1 - (1-(78/1326))^10 = 45.46% For someone else having a pair (excluding same rank) I used the same approximation method and got: 1 - (1 - (72/1225))^9 = 42.03% where Alspach calculated this the (very) long way as 42.06%. So that's about where this approximation should fall relatively and I believe it is correct for another pair at the table. http://www.math.sfu.ca/~alspach/comp35/ So this gives us the chance that AT LEAST one other player has a pair, not that EXACTLY one other player has a pair, and I'm thinking that difference is the entire difference between my answer and yours. All that leaves is the combined probability of the two events. It seems we should just multiply to get that answer, which gives 19.1% for at LEAST two players holding a pair. Agree? There's no reason the independent approximation should be much off at all in this case, and I don't think it is as long as we define what we are calculating.

No, I computed the probability of AT LEAST 1 pair.

I showed you trivially that this probability can be no greater than 15.4%, and this is also for AT LEAST 1 pair.

I see no reason why your independence approximation should be accurate at all.

EDIT: Removed calculation of exactly 1 pair.

Last edited by BruceZ; 03-31-2010 at 09:08 AM.

03-31-2010, 08:56 AM   #18
Actually Shows Proof

Join Date: Aug 2008
Location: This looks interesting.
Posts: 7,903
Re: Probability of set over set

Quote:
 Originally Posted by BruceZ No, I computed the probabilty of AT LEAST 1 pair. It is easy to convert my calculation to exactly 1 pair by changing the coefficients. I just did this quickly, and it gives about 8.5%. I showed you trivially that this probability can be no greater than 15.4%, and this is also for AT LEAST 1 pair. I see no reason why your independence approximation should be accurate at all.
But yet it is accurate for another player having a pair. Something is wrong with my combined probability then.

Is there some reason we would not simply multiply the two probabilities together in this case? I'm confused. Apparently multiplying the chance one player has a pair, times the chance another player at the table has a pair, is wrong in this case but I'm not sure why. I know that given I have a pair, the chance another player has one is the number I show.

So once I have the chance some player has a pair, and the chance for another player to have one given the first, what is the correct way to combine those here? Is that just a wrong approach? Is there a shortcut?

PS - I know your ~11.3% is the correct answer. I just can't figure out why mine is so far off.

Last edited by spadebidder; 03-31-2010 at 09:05 AM.

03-31-2010, 12:29 PM   #19
BruceZ
Carpal \'Tunnel

Join Date: Sep 2002
Posts: 11,876
Re: Probability of set over set

Quote:
 Originally Posted by spadebidder But yet it is accurate for another player having a pair. Something is wrong with my combined probability then. Is there some reason we would not simply multiply the two probabilities together in this case? I'm confused. Apparently multiplying the chance one player has a pair, times the chance another player at the table has a pair, is wrong in this case but I'm not sure why. I know that given I have a pair, the chance another player has one is the number I show. So once I have the chance some player has a pair, and the chance for another player to have one given the first, what is the correct way to combine those here? Is that just a wrong approach? Is there a shortcut? PS - I know your ~11.3% is the correct answer. I just can't figure out why mine is so far off.

You want:

P(1 or more pairs)*P(2 or more pairs | 1 or more pairs)

but you are computing:

P(1 or more pairs)*P(2 or more pairs | player 1 has a pair)

These are not the same because the second probabilities are very different, even though they may sound similar. In the first case, the second probability is

(# hands with 2 or more pairs) / (# hands with 1 or more pairs)

In the second case, the second probability is

(# hands with 2 or more pairs AND player 1 has pair) / (# hands where player 1 has pair)

Notice that the numerator in both cases is dominated by hands with exactly 2 pairs, and the numerator in the second case contains exactly 1/5 of these. Now what about the denominators? Most of these are hands with only 1 or 2 pairs, and there will be more hands with 1 pair than with 2 pairs. The denominator in the second case contains exactly 1/10 of the hands with only 1 pair, and exactly 1/5 of the hands with exactly 2 pairs. So the denominator in the second case is not 1/5 of the denominator in the first case; it is between 1/10 and 1/5 and closer to 1/10. So comparing the 2 fractions, the denominator scales from something between 1/10 and 3/20, and the numerator scales by about 1/5, so the second fraction is larger by a factor of between 1.33 and 2.

By the inclusion-exclusion calculation, the second probability should be about 11.3% / 45.46% =~24.9%. Compare this to the 42.03% which you used. This is different by a factor of about 1.7, in agreement with the above analysis. Offhand I know of no shortcuts to getting the 11.3% other than the inclusion-exclusion calculation that I've shown.

I recomputed the probability of EXACTLY 2 pairs by changing the adjustment factors in the inclusion-exclusion method, and this gave about 9.5%, which coincidentally is the number you had posted originally.

03-31-2010, 12:46 PM   #20
Actually Shows Proof

Join Date: Aug 2008
Location: This looks interesting.
Posts: 7,903
Re: Probability of set over set

Quote:
 Originally Posted by BruceZ I recomputed the probability of EXACTLY 2 pairs by changing the adjustment factors in the inclusion-exclusion method, and this gave about 9.5%, which coincidentally is the number you had posted originally.
I don't think it was coincidence, I just don't remember how I got it. I probably looked up someone else's calc on it.

Thanks for the explanation, makes sense.

 03-31-2010, 01:02 PM #21 Alizona temp-banned   Join Date: May 2008 Location: In my own little world Posts: 3,022 Re: Probability of set over set How about the odds of set over set (I had the higher one) with us both beaten by the flush draw that gets there on the goddamn Barry Greenstein Ace On The River©? I'd say the odds are slim, but... this kinda crap happens to me daily. Yeah yeah, so he had correct odds to chase his draw, but the friggin board coulda paired (prob giving my other opponent quads LOL). Poker is a stupid friggin game.
 03-31-2010, 01:24 PM #22 BruceZ Carpal \'Tunnel     Join Date: Sep 2002 Posts: 11,876 Re: Probability of set over set Here's a somewhat decent approximation to the 11.3% figure using independence only to compute the probability that I get a pair and someone else gets a different pair (accurate), and multiplying this by 5 since I take part in 1/5 of the 2 pair cases. This will be slightly high since the cases of more than 2 pairs mean I take part in more than 1/5 of the 2+ pair cases. P(I get pair) * P(someone else has different pair) * 5 1/17 * [1 - (1 - 72/1225)^9] * 5 =~ 12.4% Last edited by BruceZ; 03-31-2010 at 01:32 PM.
 04-02-2010, 11:46 PM #23 MrBubbleBoy adept   Join Date: Mar 2010 Posts: 1,099 Re: Probability of set over set this just happened to me in my final table of a 6 handed tornament 4 people. gaaaaay
 04-03-2010, 01:25 AM #24 bbarker703 grinder   Join Date: Jul 2004 Location: Arlington, VA Posts: 483 Re: Probability of set over set Was playing donkey limit poker with friends at the AC Showboat around Pres Day '09 and saw set over set over set on the flop...it was like a spoon in a microwave. I'd have laid bottom set down by the turn, but all three went capped to the river. Think I figured the flop at like 2200-1.
04-29-2010, 04:08 PM   #25
aise5668

Join Date: Dec 2009
Posts: 850
Re: Probability of set over set

Quote:
 Originally Posted by spadebidder 98:1 is the answer if you know opponent started with a pocket pair.
If you know that your opponent started with a pocket pair then these odds are way off. Assuming that I flopped a set, then the opponent, who we know has a pocket pair, will have two cards to also flop a set

2/49+(47/49*2/48)= 8.08%

8.08% ~ 11.5:1

i.e. not nearly as uncommon as your fuzzy math suggests

 Thread Tools Display Modes Linear Mode

 Posting Rules You may not post new threads You may not post replies You may not post attachments You may not edit your posts BB code is On Smilies are On [IMG] code is On HTML code is Off Forum Rules
 Forum Jump User Control Panel Private Messages Subscriptions Who's Online Search Forums Forums Home Two Plus Two     Two Plus Two Magazine Forum     The Best of Two Plus Two     The Two Plus Two Bonus Program     Two Plus Two Pokercast     Two Plus Two Videos     Marketplace         Classified Listings         General Marketplace         Staking - Offering Stakes         Staking - Seeking Stakes         Staking - Selling Shares - Online         Staking - Selling Shares - Live         Staking Rails         Transaction Feedback & Disputes     Commercial Marketplace     Classified Listings     Staking - Offering Stakes     About the Forums Fantasy Sports     Fantasy Sports         DraftDay         Sporting Events General Poker Discussion     Beginners Questions     Live Casino Poker         Poker Venues         Regional Communities     Poker Goals & Challenges     Books and Publications     Poker Theory     Poker Tells/Behavior, hosted by: Zachary Elwood     News, Views, and Gossip, Sponsored by Online Poker Report     Twitch - Watch and Discuss Live Online Poker     Televised Poker     Home Poker     Poker Legislation & PPA Discussion hosted by Rich Muny     That's What She Said!     Poker Beats, Brags, and Variance Coaching/Training     Coaching Advice         SpinSng.com - Smart Spin Team     Cash Game Poker Coach Listings     Tournament/SNG Poker Coach Listings     SpinSng.com - Smart Spin Team International Forums     Deutsch         BBV [German]     Français     Two Plus Two en Español Limit Texas Hold'em     High Stakes Limit     Medium Stakes Limit     Small Stakes Limit     Micro Stakes Limit     Mid-High Stakes Shorthanded     Small Stakes Shorthanded     Limit-->NL     Heads Up Limit No Limit Hold'em     High Stakes PL/NL     Medium Stakes PL/NL     Small Stakes PL/NL     Micro Stakes PL/NL     Medium-High Stakes Full Ring     Small Stakes Full Ring     Micro Stakes Full Ring     Heads Up NL     Live Low-stakes NL Tournament Poker     STT Strategy     Heads Up SNG and Spin and Gos     High Stakes MTT     Midstakes MTT     Small Stakes MTT         MTTSNG     MTTSNG     MTT Community     MTTc - Live         WPT.com Other Poker     High Stakes PL Omaha     Small Stakes PL Omaha     Omaha/8     Stud     Draw and Other Poker General Gambling     Backgammon Forum hosted by Bill Robertie.     Probability     Psychology     Sports Betting         DraftDay     Other Gambling Games Internet Poker     Internet Poker         Winning Poker Network         nj.partypoker.com         Poker Bonuses, Rakeback, and Affiliates     Internet Bonuses     Affiliates, RakeBack, and Bonuses     Commercial Software     Software         Commercial Software         Free Software     nj.partypoker.com         WPT.com 2+2 Communities     Other Other Topics         OOTV         Game of Thrones     The Lounge: Discussion+Review     EDF     Las Vegas Lifestyle     BBV4Life         omg omg omg     House of Blogs Sports and Games     Sporting Events         Single-Team Season Threads         Fantasy Sports     Wrestling     Golf     Pool, Snooker, and Billiards     Chess and Other Board Games     Video Games         League of Legends         Hearthstone         Starcraft 2         World of Warcraft     Puzzles and Other Games Other Topics     Politics         Economics     Politics Unchained     Business, Finance, and Investing     Travel     Science, Math, and Philosophy     History     Religion, God, and Theology     Health and Fitness     Student Life     The Studio     Laughs or Links!     Computer Technical Help     Programming

All times are GMT -4. The time now is 12:59 PM.

 Contact Us - Two Plus Two Publishing LLC - Privacy Statement - Top