I've read that it is around 12%, but I did a quick check and I came up with a different number. Could someone check my numbers, or explain what I did wrong?

Player starts with a pocket pair in texas hold em. The odds of him flopping a set are Y, odds of not flopping a set are X.

I calculated X using a binomial: Pr(n, k ) = n!/(k!(n-k)!)*pr(k)^k(1-pr(k))^k

I define X as the odds of not hitting either of the 2 cards to improve to a set (or quads, since first you have to have 3 of a kind before improving to 4):
X=Pr(50, 48)+Pr(49, 47)+Pr(48, 46)

First I found Pr(50, 48)=0.276232807
Then Pr(49, 47)=0.27634952
Then Pr(48, 46)=0.276471236

Then X=0.829053564, and Y=0.170946436

This would mean that a pocket pair improves to a set ~17% of the time on the flop, which goes against most of what I googled.

i dont really have any clue what you've done there. i don't think you understand what the binomial distribution does at all.

the proper way to calculate this would simply be:
(48 c 3) / (50 c 3) = 0.882, which gives you the probability that of the remaining 50 cards in the deck, the flop will contain three of the 48 cards which don't match your pair.

subtract this from 1 to get 0.118, the probability of flopping a set.

Oh. Yeah I did this wrong, I was finding the odds of something unrelated to the question by thinking that n=number of cards rather than the number of times something is done. Thanks for clearing that up

you start with a pair (2 cards). there is 50 cards left in the deck, only 2 remaining cards that can give you a set so your odds are 50-to-2 or 25-to-1 chance of flopping a set. 1 divided by 25 is 0.04 which comes out to 4%. so you have a 4% chance of flopping a set.

You forget that there are 3 cards on the flop

well the odds of you flopping a set on the turn is still 47-to-2 or little under 24-to-1. 1 divided by 23.5 is 0.04255... which is still 4% lol.

You can't flop a set on the turn. You flop a set on the flop. And you have 3 chances at it - the odds are about 12% as shown above.

The term 'flopping a set" is a little vague. Does it mean a set and only a set. What about a full house or quads? Anyway, DarkMagnus' answer of 11.8% is for at least a set.

You will get a PP 5.88%
Now, that you've got that pair you will flop a set or better 11.7%

In more more practical terms, what to expect in real live game:
You will flop 11.7 sets for each 100 PP that you hold. In order to get 100 PP you will have to "play" 1,700 hands.

As you can see, you will have a set on the flop once per (1,700*11.7)/100=199 hands. Now, how many hours takes you to play 199 hands in live poker? Well, you have 199/33=6 hours. On average you can say that you will have at least a set on the flop once every 6 hours of live play. If you game is much faster than 33 hands/hour then you tune 199/(# of hands per hour)

You can associate the set frequency with the frequency of getting a specific premium PP like AA. As we all know, we will get AA once in 220 hands.

So, I would say this: In 6-7 hours of live play we can expect on average to get pocket AA once and besides that to flop some kind of any set also. In a full day of play you can see that you’ll get on average the supreme PP plus some kind of set. We will have at least two big hands during one day of live play. But don't go crazy with your AA after the flop. You can shove all-in preflop if you have the opportunity but be cautious post flop with unimproved AA.

To flop a set (or better):
You know what 2 cards are because you have them as a pair in your hand. First card off the deck then you have a 2/50 chance to hit. (A)
If that fails you will have a 2/49 chance to hit on the second card (B)
If that fails you will have a 2/48 chance to hit the last flop card. (C)

As you aren't bothered (from the assumption at the start) whether you hit a set, quads or a boat, the maths then becomes [A OR B OR C] which is [2/50 + 2/49 + 2/48] which comes out at a little worse than 7/1 or the 12% previously quoted.

There are lots of situations (particularly live play where you don't have your calculator handy) where a rough and ready calculation will do - so [2/50+2/50+2/50] = 6/50 = 12%. You get a small rounding error, but moderate maths ability could get you there during a hand if you wanted to know if you were getting your price.

If you are playing a hand and there is a pot of $585 and it is $150 to call, for all but the very gifted the decimal places are immaterial as you can't work it out in your head and it shouldn't make a difference anyway if the pot is laying you 12% OR 12.2% Even if you are capable of working it out, it is so close that the difference shouldn't be the difference for you between one play and another. Knowing that you are 'a bit worse than 4/1' from the pot should be all the maths you need to play the hand and all the other factors can come into play.

Relevance being, most weak players can see flushes or straights on a flop but a set catches them off guard. How big of a raise will you call with a pocket pair under 9s in hopes of flopping a set? The implied odds in low limits are huge.

With a big pair like AA, KK, QQ I will raise if I’m the first one in or 3bet if opponent raises upfront. Now if I play deep stack, and have PP under 99, I will call based on stack odds. If the raiser is deep too I will call 5% to 7% of my stack. If for example I have $1,000 and pocket 77, the guy up front open raise for $50 to $75, I will call if I figure I’m not going to be re-raised in the back. I call up to 7% of my stack and see the flop even if I suspect the upfront raiser has AA.

In fact, I want him to have AA and I want him to believe that he should play that pair real fast and hard all the way to the river. That is the ideal scenario for your 77. If the raiser bets about 1/15 of his stack and you cover than I would say CALL. If on the other hand the raiser has $100 stack and he open-raise for $25 and you suspect him for having a big pair or AK I will not call. Why? Because you will be a big dog against any overpair and about even money against AK and the reward when you hit your set will not cover the losses you accumulate when you miss. He doesn’t have enough dough for you to attack after the flop.

When you hit your set on the flop and there is a flush or a straight draw possible, don’t rush to raise and re-raise to knock your opponent out of his draw. Just bet enough for him to call as a mistake but do not raise him too much so he’s making the correct play of folding. Bet about halve of the pot. You want him to call you! You have more outs to make a FH or Quads against him. When you have a set on the flop and his drawing to flush he’s got only 7 clean outs while you’ve got the rest of the deck working for you. Actually, from 47 cards left he’s got 7 clean outs to flush while you don’t have to hit anything to beat him unless he hits his flush, after that point you still have 10 outs to kill his hand.

Never slow play a flopped set! Nobody knows you have a monster anyway, therefore, it is stupid to conceal the strength of your hand. Backing off to a raise and then check-raising on the turn is a valid strategy (although not necessarily best). I do not back off when there is a third suited card on board. I feel that I have enough outs to disregard the possibility of a made flush against me.

Note:
1. If you lose with a set, you'll lose a lot of money. If you don't, you are not playing your sets correctly

2. The reverse it’s also true: If you win with a set, you'll win a lot of money. If you don't, you are not playing your sets correctly

Your approach has some errors.

To compute the probability that you will hit a set or better, then using a Venn diagram, you can set the problem up like this:

Deal the cards one at a time and label the first card A, the second card B, and the third card C.

P(A or B or C)

= P(A) + P(B) + P(C) - P(A and B) - P(A and C) - P(B and C)

P(A or B or C)

= 2/50 + 2/50 + 2/50 - 2/50*1/49 -2/50*1/49 - 2/50*1/49

~ 11.755%

The easy way of doing this is by starting with the approximation that the chances of a particular card coming up is about 2%.

So if there are two cards that can help you and there are three cards flipped up, the chance is 2(good cards)x2(%)x3(flop)=12%.

Let's delve into this even deeper. Say you have AKoff. You now have 6(good cards)x2(%)x3(flop)=36%.

Thank you, thank you, I'll be here all week.

To me all it looks like you have done here is worked out a set or better, and then subtracted the probabilities of hitting a better hand. So your answer looks correct for Probability of hitting a set and only a set. In poker, whilst you are working out your price of flopping a set, you presumably won't mind hitting quads either, so the assumption of "set or better" seems fair.

No matter either way, most of my point is that whether you use a quick and relatively accurate method such as the 2% rule - it is something you can do real time and is therefore helpful to live poker players. Real time, Venn diagrams aren't any use, the maths is interesting from a 'settling a prop bet' angle to 4 decimal places, but for poker, quick and approximate ftw.

If you took the time to read it, you'll realize that I actually computed the probability of flopping a set or better. I specifically mention this.

If you had solely mentioned that the approximate probability of flopping a set or better is 12%, then I would have let it go. You did actually make an attempt at a more rigorous approach which is theoretically wrong and I had to correct you.

It's only fair.

And why not do it the simplest possible way ie you have 50 cards left and 2 of which 1 at least you want to come at flop . That of course = 1 - chance the flop doesnt have any of the 2 . Since there are (50,3) possible flops and only (48,3) that dont have the card you want the chance of no (set or better) is (48,3)/(50,3) so

p(set+)=1-48*47*46/50/49/48=11.7551...%

Also people will recognize this as 1-(48/50)*(47/49)*(46/48) which is 1 - products of probabilities to miss the critical cards on each one of the three flop cards, since you want all 3 to miss it its a product!

(already used above by DarkMagus but in the form above people can relate to it with probabilities instead of combinatorics. However combinatorics brutal force method is the best way to handle such problems as they become more complex in other special questions and conditional probabilities can lead to confusions sometimes.)


Obviously this is close to 1-96%^3 (96% ~48/50 approx 47/49 approx 46/48 of cards dont get get you to set or better per trial) since 1-96%^3~12% you get a rough answer fully understanding the spirit of all above but using a fast approximation like (1-x)^n~1-nx for small x , here n=3 and x=4%



ps: (n,k)=n!/k!/(n-k)!

I have a related question that I believe I know the answer to, but I'll ask it anyway because there's some discussion amongst our poker players about it.

Let's assume you've flopped a set. What are the odds that any other unique player in the hand has also flopped a set?

My thought was that these are two discrete events, and thus it doesn't matter if you start from the assumed fact that you've flopped a set or not: you're still about 7.5:1 to flop your set (the event that already happened), and since an opponent is 7:5:1 to flop a set, the odds of this happening simultaneously are just the product, or, ~56:1

The argument was that because you started with the assumption that you'd flopped a set already, the calculation would be different. ie, you just need to run the 2/47 cards that help the guy flop his set.

That would result in other people with pocket pairs flopping a set against your already flopped set way too often, though, it doesn't make any sense... because you have to calculate the nexus of these two events, not just how often they happen in a vacuum... right?

If (A) happens, how often does (B) happen as well? You can't just calculate how often (B) happens saying "Well, (A) has already happened", because (A) has to have happened already in order for you to care that (B) happened.

These are dependent events, so you must calculate the odds of (A) happening, multiplied by the odds of (B) happening to come up with how often they happen concurrently.

chances of floping a set are like 1 in 7.5 or close to it.

Yes, I know this -- that wasn't my question.

I can't answer your question exactly, but I will give it a shot. Math guru's butcher my post after this. If you are calculating the odds of your flopping a set, you are roughly 1:7.5. However, when you look for your opponent also flopping a set, now there are only 2 cards on the flop that he could have hit a set with compared to the general "1:7.5" odds, so just for the odds that your opponent also has a set, it might be lower. Then for you to flop a set and him to flop a set, I'd say you'd have to multiply these two numbers and that would give your final statistic of both of you flopping a set.

Ok, here is what I think…

Odds of your not flopping a set = (48/50)*(47/49)*(46/48)=88.2%
Odds of your opponent not flopping a set = (1)*(47/49)*(46/48)=91.9%
(the (1) here being a sure thing because you have to have a set card)
Odds of your flopping a set= 100%-88.2%=11.8%
Odds of your opponent flopping a set= 100%-91.9%=8.1%

Odds of both events happening= 11.8%*8.1% = 0.96%

you'd also have to take into account the odds of another player having a pocket pair.

Thanks for your reply -- I guess that last line is the crux of the argument, though.

One of the guys at our game is saying that if we assume you have *already* flopped a set, and with that given, what are the odds that someone else also has a set in the hand.

My contention is that the odds would be as you describe them... because you need to know when two events happen at the same time. His contention is that the fact that you've already flopped a set means the calculation is different.

Well tell him that if you don't flop a set, the calculations are very easy. Then walk away.

But your posts always contain a flopping set, how am I supposed to just walk away?