Odds of set over set?

[iceman5]

Does anyone know the odds of 2 people hitting a set in the same hand? Lets assume that any player being dealt a pair does not fold.

[Rhound50]

I followed this one over from the NL forum. I am pretty good with odds but I really cant think of a way of putting that into numbers. I think the easiest way to get a number from it would be to anylize a bunch of Pt numbers and see how often it occured. I was trying to figure out a way to do an exact caluction and I cant think of one. My guess is its 2%-5%.
I still dont fold a set fearing another set almost ever.

[phatlad]

I once saw THREE flopped sets. Rainbow flop and no straight draw.

That was a fun hand to watch.

[k3nt]

I ran the math myself in an excel spreadsheet. This answer is not precise but it should be very close. I'll spare you the details of how I came up with it.

If anybody else wants to take a shot at doing the math and compare answers, I'd be grateful.

Anyway, my answer is (drum roll)....

0.20%, or about one time in 500, two players will FLOP a set. (Turn/River calculations not performed.)

Given that the odds of being dealt AA are 1 in 221, you should see a set-over-set flop situation occur at your table (not necessarily to you!) a little less than one time for every two times you personally are dealt AA.

This assumes a full ring game (10-handed).

Note: this is the odds at the beginning of a hand that two people will flop sets (assuming nobody folds a pair). It does not represent the odds that someone else will have a set given that you already have a set. The odds of that are much higher.

[Rhound50]

[k3nt]

Huh?

Oh. I figured out what you meant.

No. That's not what I was saying. I guess I wasn't clear.

All that my "AA is 1/221" comment was trying to do was to give a sense of what 1 chance in 500 means -- to give it some context that people can relate to easily. 1 chance in 500 is about half of 1 chance in 221 (the chance of AA = 1/221). So since you have a feeling in your head already for about how often you get AA, you can now cut that in half to get a feeling in your head for about how often you ought to see set over set happening at your table.

I did not use 1 in 221 anywhere in my calculation of the odds of set over set. That would be wrong. I did it right.

[Rhound50]

[iceman5]

Are you saying that 1 time in every 500 hands that I play...I will see a set over set at my table? That sounds way too often, but you could be right.

Now how about this. Of the times that I flop a set myself, how often will someone have a set also?

[Rhound50]

Those numbers are the odds of when two people start with pocket pairs both of them hitting a set. So that definatly doesnt mean that set over set happens 1 out of every 500 hands since you only get a pp every 1/17 hands, without actually doing the numbers the number is probably more like 5000 hands. This number also doesnt neccisarilly mean that if you make your set that the chance of your opponant having a set is .2% but it does show just how rare set over set is. There are way to many variables to know exactly what the odds of your opponant having a set when you do, i think the only way to know for sure is to do a statistical anylsis of hands played, but i do think it is safe to say that if you have a set the chances are very small that your opponant does. Of course in saying this I am probably going to get a rash of set over sets

[iceman5]

[k3nt]

Hold on for some stream of consciousness in a pretty long post. Sorry in advance. But you were warned!!

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.

[Kalle]

[iceman5]

[Rhound50]

I think we have gotten some really good numbers out of this but I a couple of small questions first off Iceman you fold a set with a P a lot more often than 1 every 2900 hands. By my calcuation more like 1 every 128 hands this one is pretty simple to calculate 1/17*1/7.5.

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.

[k3nt]

You will flop a set every 144.5 hands (1/17 * 1/8.5) [not 1/7.5, it's actually 7:5 : 1 which is one time in 8.5]. Here is how to calculate this. You have two outs. The odds that NONE of the 3 flop cards will match either of your two outs are: (48/50) * (47/49) * (46/48) = 0.88244. The chance that at least one will match are therefore 1-0.88244, or 0.11755, or 11.755%, which is shockingly close to 1 in 8.5 (it's actually 0.999 in 8.5).

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.