heh... What I mean is that he's assuming he is in a hand, and he has already flopped a set. That being the case, what are the odds that any individual person in the hand also flopped a set?

His contention is that because he's already flopped a set, you just need to calculate the odds of other people having a PP, and also flopping a set. In other words, the odds would not be 7.5*7.5:1

I'll just do the calculation for 1 opponent. Maybe I'll do the case of more opponents later.

Without loss of generality, lets say the flop is K92 and you hold p9. The only holecard-combinations giving your opponent a set are pK and p2. Since you know 5 cards, there are 47 cards unkwnown. So there are 47*46 combinations for your opponents holecards (i regard, for example, 3h4s and 4s3h as different combinations. It does not affect the result if you dont do this). Of this 47*46 combinations, there are 6 pK- and 6 p2-combinations. So the odds of your opponent holding one of these combinations is

(2*6) / (47*46) = 0.005 = 0.5 %

I would argue that this is the lowest possible possibility (if the guy checked the big blind) otherwise for example if villians range for his action preflop is 50%, then the percentage is doubled because half of your hole card possibilities can be eliminated.

Without doing the math, the real answer to this question is that for a scenario where you are heads up and you both have pocket pairs, the odds of both of you flopping a set are 98:1. This is greater than 7.5*7.5:1 because when you flop a set, you have taken away one of your opponents board card chances for his set so now he needs to hit his two outs in two of the flop cards, not three.

But calculating in general the probability of how often the bad scenario of when you have a pp and get outflopped set over set is pretty well impossible since players can throw away their pp pf (to a raise, low pp UTG, playing tight, etc.).

And really if you are heads up and flop a set, you need to know the probability of your opponent also holding a pp which again is pretty well impossible to predict accurately.

From a poker perspective, it's low enough not to factor it to much into your play much like not worrying that your K high flush is beaten by an A high flush (on a three flush board). You should probably worry about it a bit more in Omaha though.

His reply:

No, I still maintain you are wrong. If you ask the question what are the odds of two people flopping a set when both have pairs the answer is 1% (12% x 8%). If you ask the question what are the odds my opponent flopped a set (assuming he started with a pair) and you have already flopped a set the answer is 8%.

Well, I would say his reply is correct.

Yeah, except that's the answer if your opponent has a PP -- what would the percentage be of him both having a pocket pair, and also flopping a set? As was mentioned previously, you have to remove some cards from the possibilities because you already flopped yours.

If he plays 100% of his hands, then we can calculate these odds. If he doesn't, then we can't since we would need to know the percentage of hands he plays that aren't PPs and also which PPs he may fold. And then we need to figure out which PPs he folds given pf action, position, stacks, etc. It can't be done.

This has moved beyond pointless.

As we all know, we will get AA once in 220 hands. Uh, I thought the chances of getting AA were once every 221 hands. 4/52*3/51=1/221. The odds of getting a pair of aces are 220 to 1 against. As far as flopping a set, wouldn't the easiest way be to calculate the chances of not flopping a set and then subtract that from 1?
1-(48/50*47/49*46/48)=0.11755102=~11.75%

fyp.

Ummm while I didn't include the fact that it was the odds of flopping a set or better... you are a bit late on that.
See previous post on second page...

Originally Posted by LargeLouster and corrected by jay_shark (the part in red).
As we all know, we will get AA once in 220 hands. Uh, I thought the chances of getting AA were once every 221 hands. 4/52*3/51=1/221. The odds of getting a pair of aces are 220 to 1 against. As far as flopping a set, wouldn't the easiest way be to calculate the chances of not flopping a set or better and then subtract that from 1?
1-(48/50*47/49*46/48)=0.11755102=~11.75%

fyp. (written by jay_shark)

I see your point, Poo-Bah. Here's my first question. If I have AA, which of the following 3 flops, ignoring order and suit, would you include in your correction or better?

1. AAK
2. AKK
3. KKK

Here's my second question. Once you make the correction, would the difference show up before the sixth decimal place? Unless it did, and if it showed up on the fifth decimal place and was 5 or more, you would still end up rounding back to ~11.75%, wouldn't you?

Best wishes,
MasonLouster

I think next time someone has an obvious question they should try googling for the answer.

The probability that you flop a set or better using your hole cards is ~ 11.75%.

If you have pocket aces, the flops that you're interested in is

A-x-x, A-x-y, A-A-x,

we have

P(A-x-y or A-x-x) = 2*C(48,2)/C(50,3) ~ 11.510204%

P(A-A-x) = 48/C(50,3) ~ 0.2448979592%

Add the two and we get ~ 11.755102%

Thanks, now I see why you're a Poo-Bah. And I see how your phrase using your hole cards eliminates what I described as KKK or as you would put it, xxx. Add the two and we get ~ 11.755102% I believe, however, using either the odd-even rule or the IRS rules of rounding off, this would actually round off to ~11.76%. In my first answer I rounded off to ~11.75%, just as you did.

Also, I hate to be a pest (actually, I kind of enjoy it) but how could my calculation of 1-(48/50*47/49*46/48)=0.11755102=11.755102% come up with the exact same answer as yours did? Did mine implicitly take into account your more complete and mathematically accurate solution?

Best Wishes
MasonLouster

Because natural logarithm(e) raised to the power of (irrational number (pi)*imaginary number(i)) times -4=2+2

And as a matter of fact, I now propose >>>>natural logarithm(e) raised to the power of (irrational number (pi)*imaginary number(i)) times -4<<<< as the new name to this website.

I couldn't agree more (even though I don't understand anything you said, wayyyyyyyyyyyyyy beyond me). However, you gotta admit, I did get you on the rounding off thing.

Best Wishes,
MasonLouster

Based on my empirical experience, 4% sounds about right.

Originally Posted by TheWuzi
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.

Originally Posted by Ulkis
Based on my empirical experience, 4% sounds about right.

You overlooked that you get 3 chances to make the set, or, by your calculation, 12%. As I learned, actual chances of flopping a set, or better, are ~11.75% (~11,76% if you're a stickler for rounding off the fifth decimal place).

For the math, see this part of post #37, As far as flopping a set, wouldn't the easiest way be to calculate the chances of not flopping a set and then subtract that from 1?
1-(48/50*47/49*46/48)=0.11755102=~11.75% and also post #41.

Best wishes,
MasonLouster

This

Do what I'm going do right now. Thread>Unsubscribe from This Thread.

You are missing a term (which in this example equals zero, but it is worth mention if you want to apply this correctly to similar problems).

It should read

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 and B and C)

Clearly, it is impossible for the first, second, and third card to be of the same rank as your pocket pair (and thus the final term plays no role in the calculation).


Cheers,

docStats

I purposely omitted the last term because there are only two outs and this event has probability zero.

I've quoted this formula in the more general case (also known as the sieve principle) on many occasions specifically when dealing with the probability that you're facing an over-pair pre-flop with n players behind you.

http://en.wikipedia.org/wiki/Inclusi...sion_principle

Thanks for this. Nice, elegant and practical approach. The 36% chance of flopping A or K translates into odds of 1.72:1 against hitting it. Is that right? I am asking because my poker book put the odds at 2.1:1. Don't think that is right.