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Brian Alspach
Poker Digest Vol. 4, No. 26, December 14 - 27, 2001
The definition of the term ``set'' in hold'em is three-of-a-kind for which two of the cards are the player's hole cards and the third card of the same rank is on board. Sometimes you will hear players say someone has a set when there is a pair on board and someone has a third card of that rank in her hand. This is an incorrect use of the term. When I use the term ``set'', I am referring to the correct definition above.
Some of the most dramatic hands in hold'em arise when two players have sets. The main reason for the drama is that a set is about as well hidden as a strong hold'em hand can be. If you see a board with ranks A-2-6-8-Q and no flush possible, then two players holding, say, a set of 6s and a set of 8s are going to be confident of being ahead. If both players flopped sets, there likely is going to be some action. I've seen players eliminated early in no-limit tournaments without rebuys because of two players having flopped sets.
There was a message posted on rec.gambling.poker in November in which the poster asked about the chance of two players flopping sets, given that each player has been dealt a pocket pair. I'm going to answer this question, along with some related questions, in this column.
Let's suppose one player has been dealt a pair of rank x and another player, a pair of rank y, different from x. We make the assumption we know nothing about anyone else's hand. Thus, we are assuming the flop is coming from 48 unseen cards. The total number of possible flops is C(48,3) = 17,296. The number of flops of the form x-y-z, where z is different from both x and y, is 176 because there are two choices for each of the cards of ranks x and y, and there are 44 choices for the remaining card. Dividing by the total number of flops yields a probability of .0102, which is approximately 1/98.
Below is a table of some other flops of interest for two players holding pocket pairs of ranks x and y, respectively.
flop probability flop probability
x-x-y .00012 y-y-x .00012
x-x-z .0025 y-y-z .0025
x-z-z .0076 y-z-z .0076
x-y-z .0102 x-z-w .1018
y-z-w .1018 z-z-z .0025
z-z-w .1526 z-w-u .6105
Next, let's consider a few board probabilities. The total number of boards is C(48,5) = 1,712,304. Exactly 44 of those boards result in both players making quads. This gives a probability of .000026, or about 1 in 38,916.
There are 10,560 boards of the form x-y-z-z-w. Both players have full houses in this case. The probability of this outcome is .00617 or about 1 in 162.
There are 42,240 boards of the form x-y-z-w-u. Both players have trips (we are not going to worry about straights and flushes on board) in this case. The probability of this outcome is .02467, or about 1 in 41.
All of the above probabilities were derived under the assumption two players have been dealt pocket pairs. Most people probably are more interested in how often deals occur in which two players flop set over set. In order to calculate this, we must take into account how often exactly two players are dealt pocket pairs.
In my article entitled ``I'm In ... No, I'm Out: Part 5'' (Poker Digest, Vol. 2, No. 18), I gave a table of probabilities for various numbers of pocket pairs. In a hold'em game with nine players, the probability for precisely two players being dealt pocket pairs is .082. For 10 players it is .096, and for 11 players it is .11.
Let's concentrate on the 10-handed game. There is a probability of .096 that precisely two players are dealt pocket pairs, but some of the time the two players have pairs of the same rank. Clearly, in the latter situation, no one can make a set. What we really need is the probability two players are dealt pocket pairs of distinct ranks. We can use the .096 figure and proportionality to get a very good approximation.
I like proportionality arguments because of their simplicity and elegance. If you go back and look at my articles deriving the .096 figure, you will discover the method is complicated. In order to get the exact probability of two players being dealt pocket pairs in a 10-handed game, we would have to go through a modified version of the complicated process.
There are 13 ranks and three ways of choosing two pairs of a given rank. Thus, there are 39 ways of dealing two pocket pairs of the same rank. There are C(13,2) = 78 ways of choosing two distinct ranks, and there are six pairs of any given rank. Thus, there are 2,808 ways of dealing two pocket pairs of distinct ranks. This means there are 2,847 ways of dealing two pocket pairs of which 2,808 have distinct ranks. So the proportion of two pocket pairs of distinct ranks to two pocket pairs is 72/73. Multiplying .096 by 72/73 gives the approximate probability precisely two players have been dealt pocket pairs of distinct ranks. The reason this proportion is not exact is fairly subtle and I leave it up to readers to figure out the reason.
The product of 72/73, .096 and .0102 gives us the probability of a deal with a set over set flop in a 10-handed game, where precisely two players have been dealt pocket pairs. The product is .0009658 which is approximately 1/1,035.
The preceding number may strike you as being too small, but remember, there are other ways for a set over set flop to occur. Going back to the same table referred to above, we find a probability of .016 that precisely three players have been dealt pocket pairs. It is obvious that the chances of a set over set flop occurring when three players hold pocket pairs of distinct ranks is going to be greater. In fact, following the same type of arguments used above, we find the probability of a set over set flop deal with precisely three players holding pocket pairs to be .00492, or about 1/203. So we see, there is a much bigger contribution to the probability of a set over set flop coming from three players being dealt pocket pairs than from two players being dealt pocket pairs.
In the same way, we determine the probability of a set over set flop deal, where exactly four players hold pocket pairs, to be .00011886.
Adding the three probabilities gives us .00600466 or about 1/167. There will be a small contribution for deals with five or more players being dealt pocket pairs, but it will not change the number by much. So if you are playing in a game where players holding pocket pairs stay to see the flop, you should be seeing flops with two or more players making sets on the flop about once every 167 hands.
Statistics: Posted by JSBC — Tue May 09, 2006 9:01 am
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The 1 in 2900 number means this: One time in every 2900, you have a set AND someone else to have a set on the same hand.
If two people start with a PP (not the same pair! -- if I have QQ and you have QQ then our odds of hitting our sets are obviously 0), then their odds of both hitting a set go like this.
We know 4 cards now, not two. I have a pair (say aces) and you have a pair (say kings). We both have two outs.
For player 1, any of the 3 cards can hit his set. (46/48) * (45/47) * (44/46) = 0.877659
So the odds of hitting his set is 1-0.877659, or 0.12234 -- about 12%. [This is slightly higher than the odds of hitting a set on the flop knowing no other cards -- the difference is that we know that the two cards in your hand do NOT match the two cards in my hand.]
For player 2, assuming player 1 hit his set, there are now only 2 cards that can hit player 2. Under this assumption, we know 5 cards, so we start counting down from 47: (45/47) * (44/46) = 0.91582. So the odds of hitting this set as well are 1-0.91582 = 0.08418 -- about 8%.
0.12234 * 0.08418 = 0.010298. Almost exactly 1%.
So the odds of a set over set situation, any time exactly two players see the flop and both have pocket pairs, is 1%.
In a ring game where EVERYBODY sees the flop with a pocket pair (and no two pairs are the same), I would imagine the odds of set over set would be around 80%-90%, just because almost any card that comes is going to give somebody a set. But that ain't gonna happen too often (duh).
I'll try to answer Rhound's question about percentages when multiple people have hands.
Some other time, though.
Statistics: Posted by k3nt — Mon Jan 31, 2005 10:16 am
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I actaully disagree with k3nt as to what those percentages we both calcualted mean. These calcuations don't mean for every hands played you have a .113 % chance of set over set, the calulation that you did is assuming that both players have to have PP which doesnt happen every hand. That percentage is the odds of both players flopping sets with their PP. If you could figure out how often two people have PP in the same hand you could then multiply that to that percentage you calculated and get the odds of set over set per hand played.
I'm not exactly clear where the calcualation with 9 players comes in. I actaully dont agree with the 5% just based on my intution, that seems really high to me. Can you explain your calculation.
Statistics: Posted by Rhound50 — Sat Jan 29, 2005 1:03 pm
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I agree with Rhound. I think that my initial number was about 33% too high, because I did not take into account the fact that once one person has a set on the flop, the odds of a second person flopping a set also are cut down because the only person really only has 2 cards to hit his set rather than 3.
So taking that into account, it's about a 0.130% chance that two people will FLOP a set. That's 13 times in 10,000, or 1 time in about 770. So now we're down to 1/3 as often as seeing AA as your starting hand, give or take.
Rhound's answer of 0.113% is pretty darned close to my revised answer. So there's some agreement there, I think.
However, I disagree with Rhound about what the numbers mean. The 0.113% or 0.130% chance is NOT the chance that it will happen only after people have pocket pairs. Rather, that chance is starting from the beginning of the hand. Run 10,000 deals of 10 hands each, then throw in a flop each time, and 2 of those 10 hands will show sets on the flop on the same hand a total of about 13 times (or about 11 times if Rhound is correct).
As for Ice's second question, I think that should be pretty straightforward to answer. Here goes, complete with stream of consciousness commentary.
You have a set. Just to keep it simple you have QQ, and the flop comes AKQ. What are the odds that any given opponent started with AA or KK?
We know 5 cards (your QQ, plus an AKQ). 47 remain.
3/47 * 2/46 he has AA
3/47 * 2/46 he has KK
Either one of those is a 0.277 percent chance.
Combined it's a 0.5543 percent chance for one opponent to have a set.
You have 9 opponents. Only one of them has to have a set to beat you, so we run everybody's favorite combination equation:
9! / [(8!)(1!)] * (0.005543) * [(1-0.00543)^8] = about 4.77%
I'm pretty happy with that answer. It makes intuitive sense.
Conclusion: a little less than 5% of the time that you have a set on the flop, somebody else will also have a set. This is assuming that nobody folded their pair preflop and that you're at a full 10-handed table.
So when you hit your set, it's going to be the best hand on the flop about 19 times out of 20. (Exept for possible flushes or straights, of course.)
Remember, the odds of you hitting your set in the first place are only about 1 in 145. So at the beginning of the hand, the odds that you yourself will be in a set over set situation is 1 in 145 times 1 in 20, or about 1 in 2,900. Wow, that's low. You'll typically get AA as your starting hand about 13 times for every time you'll find yourself in a set over set situation.
If the board is paired and you hit your set, what are the odds that somebody flopped quads? Damn small. Really, really small. Don't worry about it.
Statistics: Posted by k3nt — Fri Jan 28, 2005 11:27 pm
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