You've answered a different question than what was asked (and which I answered correctly). When two players see the flop both holding pocket pairs (of different ranks), at any size table, they will both flop sets once in 99 times, on average.

Oh yeah, you are correct about that, sorry if I misread your post.

can someone tell me the probability of getting oversetted by showdown?

i think its useful because there are many players who's range consist of only overpairs (preflop action) due to action/stats/etc and they are the type to stack off their overpairs regardless of flop. i wanted to account for the times they overset me on flop turn and river to see my real profit from setmining them.

You're all wrong.

It's about tree-fitty.

Case closed.

Okay, great thread. I am trying to apply this to some HM numbers and wanted to check out my school boy Math.

I flopped bottom set 63 times from 53339 hands this year. So I expect to be oversetted every (53339/63)*12 hands

(assuming I see the flop v a single player and that player always has a pocket pair)

I flopped middle set 61 times from 53339 hands this year. So I expect to be oversetted every (53339/61)*23 hands

(same assumptions apply)

So ~ every 6666 hands for both events

Is my Math way out?

nevermind

You joined in 2005, have only graced the forum with 50 posts and this is one of them. Its probably best to say nothing at all, the brain being capable of thoughts that do not need externalisation.

Pokerprotools/pql seems like the way forward for these questions,

select count(handtype(hero, flop) = trips and handtype(villain,flop) = trips)
from game="holdem", hero="55", villain="AA-22"
where equity(hero, flop) < equity(villain,flop)
= 6562/600000

spadebidder:

I'm glad you posted this. I experienced a very unusual situation not once (but twice) playing cash games this past weekend in Tunica. Both of these experiences involved being on the losing end of a set-over-set confrontation. In the first situation, my flopped set of 7's lost to a flopped set of Kings, but somehow my "ESP detector" must have been working flawlessly as I somehow managed to keep my loss down to somewhere around $100 chips in a $1/$2 no-limit game. I wasn't so lucky the second time ... My flopped set of 6's ran in to a flopped set of Aces and the other guy got my entire stack. (I probably could have played my hand better, but I thought my opponent had AK and had flopped TP with a good kicker and I was going to double my stack.)

After the second set-over-set disaster, I was wondering what the odds are that I would suffer this beat not once but twice in a period of two days? I talked with a dealer who had played professionally for ten years. He told me that losing to set over set is rare - only occurring about five percent of the time. (Presumably he meant five percent of the time when you flop a set and your opponent flops a better set.) So I thought, "OK, I lost to a 19:1 shot twice. I guess I was just unlucky." However, if your calculations are correct, then I was struck with a 4377:1 blow not once but twice in a period of two days.

The fact that this highly unlikely event occurred not once, but twice, back-to-back in two days makes me wonder if "something fishy" wasn't going on in that card room? However, before I start sounding too paranoid, I witnessed another hand where two players got all-in with sets. One of them had a set of 9s and the other one had a set of 2's. So, having played (maybe) 500 hands of poker over a period of four days, I saw three set-over-set confrontations.

Former DJ

If we accept my earlier assumptions to arrive at 1/4377 for 10-handed, then the chance to see this event at least 3 times in 500 hands is:

=1 - BINOMDIST(2,500,1/4377,TRUE) = 0.000226849
or 1 in 4408 times that we see 500 hands, 3 or more set-over-set flops will happen.

Going off topic, but I don't like this number much I would like to see your analysis, do you know the topic of that thread?

For the record, It's going to take two raises cold (or some silly 10BB open raise) to push me off 22. Quite honestly, I wish I was dealt 22-77 every hand.

Actually only play Omaha now. Bottom set doesn't look as good anymore ^_^

OD

Looking back, I don't like it much either. So let's say the typical player sees the flop 70% of the time when holding a pair. That would mean when 2 players have pocket pairs, 49% of the time they both will see the flop. I think that's a bit on the high side but closer than 24%. Somewhere in between is probably good for most games.

.

Those of you pulling your hair out due to a set-over-set disaster might be interested in the last paragraph of this:

http://forumserver.twoplustwo.com/sh...9&postcount=17

Former DJ

Yes, you read that right, I might be the only person in the world or universe who has ever lost with set under set on the flop on two consecutive hands.

Someone who is good at math, please calculate odds for me on the parlay of losing with flopped set under set on two consecutive hands.

This game was 10 handed and assume every player stayed in every pot with every pair, as that's pretty close to reality in this game (action game).

Thanks in advance.

Not by a long shot.

This must be set over set time on the site. There are at least three other postings recently with similar questions.

I developed a simple formula for being outsetted on the flop. It is:

Pr(your set of rank R is out-setted on the flop if N opponents see the flop) = N*(14-R)*0.0415%.

Using this and combining it with the chance N opponents see the flop (if a flop, there is at least 1 opponent), I came up with odds of 110,000 to 1 for you losing to oversets two consecutive hands given you had a set each time on the flop. If you want to include the chance of you getting a set for each of the hands, the odds zoom to several billion to 1 against.

So, yeah, you were a bit unlucky, to put it mildly (very mildly). But with billions of hands played each year, it’s not unusual that this will happen, but why to you?

It's only 44.32 million to 1, not several billion. I calculated it exactly 2 different ways and also by simulation. Even using your formula I only get 49.6 million, but I found yesterday that the factor of 0.0415% in your formula needs to be 0.0458% in order to match my exact calculation for the flop since we are requiring the flop to be unpaired. Your 0.0415% corresponds to what you would get if you allowed the flop to be paired with fh or quads. Using 0.0458% gives 1 in 40.7 million for just the flop which doesn't consider the possibility of getting quads by the river. If I take the quads into account, I get 44.47 million, and if I use my exact value instead of 0.0458%, I get the 44.32 million, basically dead nuts in agreement with the other calculation and the sim. The 0.0458% gives a combined probability over all pairs 2-K of 0.02679 and 1/0.02679^2 = 1 in 1393. Using 0.0415% gives 1 in 1697, so I don't know where your 110,000 to 1 came from.

P(underset on flop and lose) =

P(XYZ flop) *
P(dealt XX, YY, or ZZ) *
[ P(exactly 1 other set out) * P(lose to higher set) *
(1-P(quads and no quads for opponent)) +
P(2 other sets out) * P(lose to higher set) *
(1-P(quads and no quads for opponents) ]

Note that the probability on each line is conditional on the previous lines.

52/52 *48/51 * 44/50 *
9/C(49,2) *
[ (9*6/C(47,2) - 2*C(9,2)*6*3/C(47,2)/C(45,2)) * 1/2 *
(1-1*43/C(45,2)) +
C(9,2)*6*3/C(47,2)/C(45,2) * 2/3 *
(1-1*41/C(43,2)) ]

=~ 0.00015 or 1 in 6657

Squaring for 2 in a row gives about 1 in 44.32 million


Simulation:

Output:

> p
[1] 0.000152
> p-error
[1] 0.000124
> p+error
[1] 0.00018


Exact table P(oversetted) for 222-KKK on flop (compare to your formula - this is what you want to match):

Average over all pairs:


Code for flop exact table and exact_probs:


I also have a very involved exact calculation for yesterday's problem that I spent a lot of time on. It matches the sim perfectly with particular assumptions, but I bagged it when I found that the other OP had filter issues and probably didn't even care whether we exclude paired flops.

I’m not sure we’re addressing exactly the same problem, and it may be related to how many opponents see the flop.

I assumed hero has a set with unspecified rank with 9 opponents pre-flop. I then used the binomial as an approximation that n of 9 opponents get a pair (and see the flop), given that at least 1 has a pair. I got for n = 1, 2, 3 -- 77.5%, 19.4%, 2.8%. I then used the simplified formula to estimate the chance for an overset on the flop for each n. It turned out that for n opponents seeing the flop, the average overset probability is n*0.0025 using my 0.0415% value. This compares to your value of about 0.00278, fairly close and that has to be due to the difference in the constants 0.0415% vs. 0.0458%. (I thought I excluded flop pairs, but maybe not.)

Weighting these overset probabilities by the chance of n opponents seeing the flop gave me an overall probability of an overset of 0.00315. Given an overset, I assumed hero had 1 out to win by the river for a loss probability of 43*42/(44*43)=0.96, assuming villain doesn’t hit quads. This resulted in final loss probability of .00301, which for two consecutive hands gave me the 110,000 to 1.

If I use your average overset values over all ranks for n=1,2, etc.,e.g., 0.00278 0.00554, etc. this becomes 91,000 to 1, fairly close and much larger than the 1393 to 1 you got. So, I’m not sure why the big difference unless you are making a different assumption on the number of players seeing the flop. I should be able to figure that out from your equations so I appreciate the effort in doing this along with the sim programs.

Those numbers are not probabilities given that n opponents see the flop with a pair. They are probabilities given that n opponents were dealt cards. The probability that they get a pair and see the flop is already built in. It's also built in to the 0.0458%*R*n where n is players dealt cards, not players who saw the flop with a pair. This agrees closely with my exact calculation which calculates the probability that 1 or 2 players were dealt pairs that made an overset as you can see:


This considers all C(47,2) hands that they could have been dealt. Using 0.0415%*R*n will get within 0.45% of those values worst case, and it will be nearly exact when you don't exclude paired flops. It sounds like you misinterpreted these numbers and effectively applied the probability that they have a pair twice, which is why you got such a small number. You don't average over those numbers. The only one you use is the last one 0.02478 for 9 opponents. Then it's just 1/0.02478 = 1 in 40.355 for 1 hand, and for 2 hands square that to get 1 in 1629. The 1391 came from taking quads into account which added a factor that effectively changed the 0.02478 to 0.02679.

To get the overall probability before any cards are dealt that the hero gets an underset on the flop and loses the hand, all you have to do with that 0.02478 is multiply it by the probability that the hero gets dealt a pair and makes a set on an unpaired flop, and then doesn't get quads when another overset doesn't have quads. Then square that for 2 consecutive hands which gives 1 in 44.31 million in agreement with the sim, and also in agreement with the other exact calculation I posted. The probability that the hero gets dealt a pair and makes a set on an unpaired flop is

P(set) = 1/17 * 2/50 * 48/49 * 44/48 * 3 = 0.006338535

Multiplying that by the 0.02478 (actually 0.024775) and squaring gives 1 in 40.6 million. The probability that he makes quads when another overset doesn't have quads is

P(no quads) = 1 - 1*43/C(45,2) =~ 0.9565657

which you seem to have gotten. The only thing that doesn't consider is quads when you are up against 2 oversets which is mice nuts but my other calc considered that too. So we have

1/[P(set)*0.024775*P(no quads)]^2 =~ 1 in 44.31 million

In agreement with the other exact calculation and the sim.

Okay, I see the problem now.

Set Over Set Expert.....

Just finished a 2c 5c nl session due to tilting, played about 900 hands and lost 3 hands by set over set...effective stacks every time...super tilted...have to shut it down for a bit....

Wonder what the odds of losing set over set 3 times in 900 hands is???

Depends on:
- how many players are at the table
- whether you're talking about flopped sets or any time in the hand
- how likely you both are to see the flop (and the turn/river if relevant)

For the 3rd point, maybe you'll see the flop almost every time but the same can't be said of the turn or river. If you have 55 and the flop is 8AK there's a good chance you'll fold before the turn and river. (For an upper bound estimate on the probability, we can pretend you're both incapable of folding at any point in the hand.)

I experienced 211 Set over Set on the flop, when i start with a pair and opponent has a pair. This is with 4598 flopped Sets of mine. Makes 4,5%!

How likely is that? Not trying to prove anything, but lately I see Sets all over the place. Escpecially with my monster hands (AA/KK). Kind of strange.

.

You are saying that you have 4598 flopped sets, and of those, 211 times a villain holding a pair also flopped a set, or 4.59%. That's low. I think we would expect it to happen about:

1 - (C(45,2) / C(47,2)) = 8.4%

That's the chance one of the other two flop cards matches his rank given you have a set and he has a different rank pair than you. And the times he has the same rank are so rare we can disregard.