# The probability of flopping a set in holdem

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• Apr 11th 2008, 01:53 PM
Flippy
The other people playing might have the card(s) you need.

If you are alone, the 2 cards you need will always be in the remaining deck.
If there are other people, you don't know if the 2 cards are still in the deck. There is a pretty good chance they are both out.
• Apr 15th 2008, 01:46 AM
squarerootof2
i am aware that the conditional probability that you flop a set given you are dealt a pocket pair is about 1/8 (this is the probability a pro poker player has in his head, also the probability that they consider when they make their decisions). as for the exact calculations i don't remember my combinatorics well enough to do, but 1/8 should be fairly accurate.
• Apr 15th 2008, 02:05 AM
angel.white
Quote:

Originally Posted by ThePerfectHacker
The probability of a Royal Flush is 1 in 50,000.
But for some reason when I deal the cards it is much much much more common, hmm, did I do something wrong in my calculation :rolleyes: .

Interesting, I used to play every day, and only ever saw a Royal Flush two or three times. And this is among all tables that were playing (I don't think I've ever seen one at my own table). In fact, the best hand I've ever had (not counting internet poker) is 4 of a kind. And I played poker for probably about a year.
Quote:

Originally Posted by Quick
First, I assume no one else is playing.

That shouldn't matter, as you can't see their cards.
Quote:

Originally Posted by Flippy
The other people playing might have the card(s) you need.

The cards you need may be in the bottom of the deck as well, we don't rule out the bottom half of the deck, though. As far as the player is concerned, there should be 50 cards that the player could potentially have show up in the community cards.

For example, if there are three cups of kool-aid, grape, strawberry, and cherry, and you must choose one cup, you have a 1/3 chance of getting your favourite flavour. If you add another person to the mix, and that person chooses blindly, and doesn't tell you what they got, then you still have a 1/3 chance of getting your favourite flavour even though one of the cups has been removed. You can even take that one step further, add two people to the mix, each chooses blindly, and you get the remaining cup with no ability to choose, you still have a 1/3 chance of getting your favourite flavour.
• May 28th 2008, 05:19 AM
kraemer
Quote:

Originally Posted by angel.white
For example, if there are three cups of kool-aid, grape, strawberry, and cherry, and you must choose one cup, you have a 1/3 chance of getting your favourite flavour. If you add another person to the mix, and that person chooses blindly, and doesn't tell you what they got, then you still have a 1/3 chance of getting your favourite flavour even though one of the cups has been removed. You can even take that one step further, add two people to the mix, each chooses blindly, and you get the remaining cup with no ability to choose, you still have a 1/3 chance of getting your favourite flavour.

Absolutletly NOT. If the first person chooses my favorite flavor then my chance of getting it is 0, zero , Nada, Strawberry is OUT.
You are right if we repeat this situation a thousand times. Then I get my favorite about 1/3 of the times.

But in ONE given situation Your chance of catching Your flavor is 50% or ZERO with one person choosing before You and 100% or ZERO with 2 persons choosing before You.

Since You dont know their choice You cant calculate it, but that doesnt mean that Your chance of getting the cup that is already out are still 1/3.

I think this has become popular as "The Goat Problem".
Consider You make Your choice out of the 3 cups.
Now I remove one cup and tell You that it isnt Your favorite.

Now I allow YOu to choose again between the remaining cups.
Since Your first choice has a probability of 1/3 the remaining
cup's probability must be 2/3.

I know this sounds weird, but I once read a book (The goat problem or similiar) which prooves this to be right.
• May 28th 2008, 05:28 AM
CaptainBlack
Quote:

Originally Posted by kraemer
Absolutletly NOT. If the first person chooses my favorite flavor then my chance of getting it is 0, zero , Nada, Strawberry is OUT.
You are right if we repeat this situation a thousand times. Then I get my favorite about 1/3 of the times.

But in ONE given situation Your chance of catching Your flavor is 50% or ZERO with one person choosing before You and 100% or ZERO with 2 persons choosing before You.

Since You dont know their choice You cant calculate it, but that doesn`t mean that Your chance of getting the cup that is already out are still 1/3.

The frequentist interpretation of probability embeds this instance in an ensamble of similar instances where the other choosers choose at random.

Then you have exactly the situation of your second sentence, that is to a frequentist the probability in question is 1/3.

To a Bayesian also the probability is 1/3.

So under the two most common interpretation of applied probabilities (as opposed to mathematical, which is almost a branch of measure theory) the probability is 1/3.

RonL
• May 28th 2008, 06:55 AM
angel.white
Quote:

Originally Posted by kraemer
But in ONE given situation Your chance of catching Your flavor is 50% or ZERO with one person choosing before You and 100% or ZERO with 2 persons choosing before You.

Lets say 1 person chooses before you, they choose your favourite 1/3 of the time, so 1/3 of the time you have 0% to get your favourite. They do not choose your favourite 2/3 of the time, so 2/3 of the time you have 50% chance of getting your favourite.

$\frac 23 *\frac 12 + \frac 13 *0 = \frac 13$

Lets look at the other situation, two people choose before you. Then you are forced to take their leftovers. Their leftovers will be your favourite 1/3 of the time, and not your favourite 2/3 of the time.

$\frac 13 * 1 + \frac 23 * 0 = \frac 13$
• May 28th 2008, 08:27 AM
colby2152
Quote:

Originally Posted by fobster
Hi

I was just trying to work out the probability of flopping a set on the flop in texas holdem but don't know if its exactly. I'm assuming only a set and not four of a kind.

My workings:

P(Flopping a Set)=[(2/50)*(48/49)*(47/48)]+[(48/50)*(2/49)*(47/48)]+[(48/50)*(47/49)*(2/48)]

= 0.11510204 or about odds of 8.688:1

Does that seem correct?

Set on the flop or set between the flop and your two card hand?

You only have knowledge of the two cards in your hand, so if you include a pair in your hand, then it is what you say.
• Jun 17th 2008, 01:12 PM
esbo
I play poker quite a lot, assuming you have a pocket pair there are only 2 cards left in 50 which will make you a set, that's a 1 in 25 chance for each card flopped so as a rule of thumb I use 3 in 25 to make a set on the flop, I approximate that to a 1 in 8 chance, and 5 in 25 or 1 in 5 for a set by the river (the last card, so thats 5 cards in all). OK so the maths is approximate but it is a good enough rule of thumb. 1 in 8 on the flop. 1 in 5 by the river. If you do make a set you usually win a big pot, or occasionally lose a bit pot.
• Dec 24th 2008, 02:39 PM
Jameson
Just read through this thread and there is a lot of confusion.

Just to clarify: a set is when a player is dealt a pair (22-AA) and the community board brings a third card of that value. This is different than 3 of a kind, which could be a board of 22xxx and you hold a 2. When doing these calculations, it doesn't matter how many players are in the game since all cards besides yours and the board are unknown. Whether they are in the remaining deck or in someone's hand doesn't matter. I believe the general rule that poker players go by is that on the flop you have a 1 in 8 chance of flopping a set, like the above poster said.
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