Must have misundertood something then. Thought it's 1.02% Set vs Set on the flop.

That's correct, but isn't what you asked. You said you had 4598 flopped sets (bolded below). The expectation there for villain with a pair to also flop a set is 8.4%.

If you meant to say that you had 4598 total hands where you and an opponent both held a pair and saw the flop, counting all instances whether you flopped a set or not, then seeing 211 set-over-set there is essentially impossible and your information given is wrong. Yes, 1.022% is expected, and 4.5% with this sample size didn't happen. The chance is basically zero.

I better check again then...

I will look for all hands where I hold a pair and flop a set (trips excluded, since I am holding a pocket pair). I will then match those hands with all the players that hold a pair preflop as well and flop a Set as well.

Set vs Set on the Flop

Hero Holding Pocket Pair & Flopped a Set: 4204
Villain Holding Pocket Pair & Flopped a Set with Hero: 127
= 3,02%

Hero Holding Pocket Pair & Flopped a Set or Full House or Quads: 4703
Villain Holding Pocket Pair & Flopped a Set or Full House or Quads with Hero: 153
= 3,25%

So again this is way below the expected number, which is 8.4%. I suspect your denominator of 4703 is counting some hands where villain didn't have a pocket pair.

And you seem to be still confusing two different probabilities.

1. When you and an opponent both hold a pair and see the flop, the chance that both flop a set is 1.022%.

2. When you and an opponent both hold a pair and see the flop AND you actually flop a set, the chance that he does too is about 8.4%.

In #2, your denominator is much smaller as it only counts hands where you hit the set.

Thank you for making this clear. Yes, I was kind of confused. Is this correct?

1) count hero and villain hold a pair, saw flop / count hero and villain hold a pair, flop a set = 1,022%

2) count hero and villain hold a pair, saw flop, hero flops a set / count hero and villain hold a pair, saw flop, flop a set = 8,4%

Almost. Your numerators and denominators are backwards.

dang...you are right happens when you write it out...

Is it even possible to get a correct number with 1 + 2? Since villain might folds his pair, we will never see his hand in our hand histories and therefore can't account the total numbers correct. Or should we assume that he hadn't had a pocket?

What about probability No3? NewOldGuy was correct with my calculation. My denominator was counting hands where villain hadn't had a pocket. Is it necessary to make the assumption that villain is holding a pocket perflop, too - for our calculation? In other words: what are the odds of running into another Set (not trips) on the flop without assuming anything with villain.

1) count hero + villain holding a pair & flop a set (full house or quads) / count hero + villain holding a pair, saw flop = 1,0220%

2) count hero + villain holding a pair, saw flop, flop a set (full house or quads) / count hero + villain holding a pair, saw flop, hero flops a set (full house or quads) = 8,4000%

3) count hero holding a pair & flop a set (full house or quads) / count hero villain holding a pair & flop a set (full house or quads) = ?

The only way you can know for sure and do an accurate calculation, is to limit the set to only hands that were preflop all in + both players had pairs.

Just focusing on preflop allin sounds reasonable. No other way of calculating odds of Set vs Set (on the flop)? A way where we don't consider if villain holds a pocket or not.

Villain can't hit a set if he doesn't hold a pocket pair (matching pair on board is not a set, it's trips). But the problem is the times he does have a pair, and doesn't hit the set and then folds, we won't know if he had a pair or not. You can't know this which is why you can only accurately do the stat for all-in hands.

I understand. I already do distinguish between Set and Trips. So let's sum it up! We have...

1) Flop Set Test A (Hero + Villain Allin Preflop)
count hero + villain holding a pair & hero + villain flop set (full house or quads) / count hero + villain holding a pair = 1,0220%

2) Flop Set Test B (Hero + Villain Allin Preflop)
count hero + villain holding a pair & hero + villain flop set (full house or quads) / count hero + villain holding a pair & hero flops set (full house or quads) = 8,4000%

3) Any Street Set Test A (Hero + Villain Allin Preflop)
count hero + villain holding a pair & hero + villain flop turn river set (full house or quads) / count hero + villain holding a pair = ??%

4) Any Street Set Test B (Hero + Villain Allin Preflop)
count hero + villain holding a pair & hero + villain flop turn river set (full house or quads) / count hero + villain holding a pair & hero flops turn river set (full house or quads) = ??%

Would this be correct with or do I have to calculate it like: (Set_vs_SetFlop + Set_vs_SetTurn + Set_vs_SetRiver)/3 = x%

3) Any Street Set Test A (Hero + Villain Allin Preflop)
Possible Board Combos Flop,Turn,River: C(46,5) = 1370754
Possible Set vs Set Combos: 2*2*44*43*42 = 317856

Probability: 0,2319

I am not sure if I should collect searching Set, Full & Quads since the initial OP ruled this out!? I guess in this case I should only look for Sets.

Could someone confirm above calculation with No3? Full and Quads should be excluded, I believe.

So you just want the chance of set>set on any street given that Hero & Villain already have pocket pairs? Personally I would include boats and quads, but as you wish. I'm grunching so I don't know what these are needed for.
44*43*42 counts order of the other cards, which shouldn't be counted. Also, it allows for the board pairing which you don't want.

2*2*(44*40*36 / 3!) = 42240
or 2*2*C(11,3)*4³ = 42240
44240 / C(48,5) = 80/3243 =~ 2.46685%

Sorry, that should say: 2*2*(44!!!! / 32!!!!)/3!
or 2*2*(44 PPPP 3)/3!

Thank you heehaww. I would love to see both chances (with and without counting full and quads). But as you can see, I am no good in calculating those probabilities.

1) Flop Set Test A (Hero + Villain Allin Preflop)

count hero + villain holding a pair & hero + villain flop set / count hero + villain holding a pair =~ 1,0220%

count hero + villain holding a pair & hero + villain flop set (full house or quads) / count hero + villain holding a pair =~ ??%

2) Flop Set Test B (Hero + Villain Allin Preflop)

count hero + villain holding a pair & hero + villain flop set / count hero + villain holding a pair & hero flops set =~ 8,4000%

count hero + villain holding a pair & hero + villain flop set (full house or quads) / count hero + villain holding a pair & hero flops set (full house or quads) =~ 8,4000%

3) Any Street Set Test A (Hero + Villain Allin Preflop)

count hero + villain holding a pair & hero + villain flop turn river set / count hero + villain holding a pair =~ 2.4668%

count hero + villain holding a pair & hero + villain flop turn river set (full house or quads) / count hero + villain holding a pair =~ ??%

4) Any Street Set Test B (Hero + Villain Allin Preflop)

count hero + villain holding a pair & hero + villain flop turn river set / count hero + villain holding a pair & hero flop turn river set =~ ??%

count hero + villain holding a pair & hero + villain flop turn river set (full house or quads) / count hero + villain holding a pair & hero flop turn river set (full house or quads) =~ ??%

How could I filter my database to check what percentage of the time I am out-setted?

Did anyone ever come up with a solid answer of what percent of the time a set over set happens in 6 max? I'm talking if you played a million hands how many of those hands would be set over sets? What percent does that leave us?

Playing heads-up Texas hold 'em I think when Hero is dealt a pair that will end up as a set, his one opponent will also have been dealt a pair that will end up as a set about 1.2%. And I think that number is small enough to simply multiply by the number of players to get a reasonable approximation. Thus with five opponents, when you get dealt a pair that will end up as a set, at least one of your opponents will also be dealt a pair that will end up as a set about 6.0%. I think that's a pretty good approximation.

Thus if you are dealt a pair of deuces and end up with a set of deuces, one of six opponents will have been dealt a pair that would end up as a higher set about 6.0% of the time. And if you are dealt a pair of eights and end up with a set of eights, one of six opponents will have been dealt a pair that would end up as a higher set about 3.0% of the time.

Will every opponent who is dealt a pair that would end up as a higher set than Hero see the flop or continue past the flop or turn? I don't know. I think it's opponent dependent.

Alas, the game that interests me is Omaha-8, and the number I get for one opponent in that game is, I think, too large to simply multiply by the number of opponents.

I'm not a mathematician.

Buzz

Oh, I was going to say Bruce did it for 9max, but he actually did set > set > set. I'll do "at least set > set" for N-max (any 2 or 3 players at the table; Hero doesn't have to be included).

C(13,3) * 4^3 / C(52,3) *
[ C(N,2) * C(3,2)^3 / C(49,4) / 3!! -
2*C(N,3) * C(3,2)^3 / C(49,6) / 5!! ]

For 6max that gives an answer of about 1 in 1910 hands.
For 9max, about 1 in 801 hands.

Edit: above doesn't assume Hero / anyone is already dealt pocket pair. (Buzz' numbers do which is why they're much larger percentages. I haven't checked their accuracy but they don't sound unreasonable.)

Bad way to phrase that. I did set>set with the inclusion of set>set>set. Put another way, the chance of at least 2 sets on the flop (without assuming the flop is unpaired, hence my first line of work is the probability that it's unpaired).

This is wrong if I'm understanding what you are trying to calculate.

1.02% is the percent of the time that, when any two players see the flop both holding pairs, they both hit sets. It assumes we already have two players holding pairs and seeing the flop. One way to calculate this is shown in the very first post in this thread.

The chance your heads-up opponent has a pair given that you have one is 73/1225. So your scenario would multiple those together.

Thank you. I greatly appreciate your response.

As a mathematician, I might be interested in how often two players both end up with sets.

But although I'm interested in math, and am greatly appreciative of responses and problem solving techniques shown by mathematicians in this forum, I'm not a mathematician.

I'm a card player.

As a card player, I'm interested in how often an opponent has a set when I also have a set.

I hope that makes sense without offending anyone.

Buzz

Yes. In order for two players to end up with sets, they both must be holding a pair (and the board must not be paired).

Yes. Player A cannot make a set if he doesn't hold a pair. Same for Player B.

My 1.2% estimation above was for a five card board on the river.

For a three card flop, my estimation is one tenth of what I got for a five card final board. (I think when Hero, holding a pair, flops a set, heads-up, his single opponent, who sees every flop, also flops a set 0.12% of the time).

Yes. (I think so too).
6*12+1=73 and C(50,2)=1225.
But since both players could not make sets if they both had the same pair, I'd use 72 as the numerator (and still 1225 as the denominator).

At any rate, thank you very much. I greatly appreciate your response too.

Buzz