calculating basic hand combinations NLHE - Gambling and Probability

weeblewoble · 07-27-2012, 11:44 AM

I am trying to figure out how the author calculated the hand combinations for NLHE. I figured out how they got some of the solutions I think, but I can't figure out how they got the following listed below. Could you provide the equation, and explain where the numbers came from? Thanks.

1) 9 combos of 2 pairs
2) 12 combos of a pair + kicker
3) 16 combos of 2 unpaired cards

Are these equations correct?
sets = c(3,2) // 3 cards b/c 1 card is a community card
any 2 suited cards = c(4,1) // 4 suits
any pocket pair = c(4,2) // 4 suits, 2 b/c each card has a different suit

NewOldGuy · 07-27-2012, 12:49 PM

You really need to provide what is being described by 1-3, I don't think anyone can give you an answer with the information provided.

Combo math is pretty simple but it isn't clear what you are asking.

C(3,2) = 3, but 3 sets in what? When?
C(4,1) = 4, but 2 suited cards in what? What has 4 cards?
C(4,2) = 6, but 6 pocket pairs in what? What has 4 cards?

weeblewoble · 07-27-2012, 11:23 PM

Quote:

Originally Posted by NewOldGuy

You really need to provide what is being described by 1-3, I don't think anyone can give you an answer with the information provided.

Combo math is pretty simple but it isn't clear what you are asking.

C(3,2) = 3, but 3 sets in what? When?
C(4,1) = 4, but 2 suited cards in what? What has 4 cards?
C(4,2) = 6, but 6 pocket pairs in what? What has 4 cards?

Hi, let me rephrase. I am trying to figure out how the author got his numbers when calculating combos. I think I figured out how to do sets, 2 suited cards, and pocket pairs. Could you tell me what is "n" and "k" in c(n,k) and how did you choose n,k? Thanks.

1) How many combos are there of 2 pairs?
2) How many combos are there of a pair + kicker?
3) How many combos are there of 2 unpaired cards?

NewOldGuy · 07-28-2012, 02:07 AM

Quote:

Originally Posted by weeblewoble

Could you tell me what is "n" and "k" in c(n,k) and how did you choose n,k? Thanks.

That's shorthand for combinations, and the notation just means how many ways can you Choose k things from n things without regard for order. K is any number less than or equal to n. Here's the formula to solve it:

C(n,k) = n! / k!(n-k)!

weeblewoble · 07-29-2012, 04:25 AM

Quote:

Originally Posted by NewOldGuy

That's shorthand for combinations, and the notation just means how many ways can you Choose k things from n things without regard for order. K is any number less than or equal to n. Here's the formula to solve it:

C(n,k) = n! / k!(n-k)!

Thanks for your response. I understand that part. I can't figure out what would n and k are for the following cases.

1) How many combos are there of 2 pairs?
2) How many combos are there of a pair + kicker?
3) How many combos are there of 2 unpaired cards?

**RustyBrooks** · 07-29-2012, 09:05 AM

How many combos of 2 pairs are there given what information? A flop? A whole board? Do we know any of our own cards?

weeblewoble · 07-29-2012, 01:02 PM

Quote:

Originally Posted by RustyBrooks

How many combos of 2 pairs are there given what information? A flop? A whole board? Do we know any of our own cards?

So, for the 2pair, I think we can assume the calculation was made on the flop. Do you know how they got the 9 combos? Lets say, we want to know how many 2pair combos there are for 89 on a Flop: 2h8c9s.

fyi, I am reading the Blueprint and watching vital myths back to math. They don't give answers to the questions you are asking. I am just trying to figure out how they get the basic numbers for their hand combination totals chart.

They just state as fact,
1. there are 1326 combos in NL
2. a set has 3 combos
3. 2pairs have 9 combos
4. pocket pairs have 6 combos
5. a pair + kicker has 12 combos
6. unpaired cards have 16 combos
7. any two specific suited cards have 4 combos
8. any two specific cards have 1 combo

statmanhal · 07-29-2012, 02:17 PM

Quote:

Originally Posted by weeblewoble

So, for the 2pair, I think we can assume the calculation was made on the flop. Do you know how they got the 9 combos? Lets say, we want to know how many 2pair combos there are for 89 on a Flop: 2h8c9s.

fyi, I am reading the Blueprint and watching vital myths back to math. They don't give answers to the questions you are asking. I am just trying to figure out how they get the basic numbers for their hand combination totals chart.

They just state as fact,
1. there are 1326 combos in NL = C(52,3)
2. a set has 3 combos : context????
3. 2pairs have 9 combos : must mean pair each of your hole cards on the flop ignoring the 3rd flop card; 3*3 = 9
4. pocket pairs have 6 combos; e.g. 66; C(4,2)
5. a pair + kicker has 12 combos context ????
6. unpaired cards have 16 combos; e.g. AK; 4 aces x 4 kings = 16 combos
7. any two specific suited cards have 4 combos; e.g., Ah4h, Ad4d, As4s,Ac4c
8. any two specific cards have 1 combo; e.g. Ah5c

Stating some of these results in a book without further description is not very good writing IMO.

**RustyBrooks** · 07-29-2012, 02:32 PM

Quote:

Originally Posted by weeblewoble

1. there are 1326 combos in NL

Yes - C(52,2)

Quote:

2. a set has 3 combos

Any individual set on an unpaired board, yes. Say there is a 5 on the board. There are 3 remaining 5s in the deck and you need 2 of them, so it's C(3,2) which is 3. You could also list them by hand easily.

Quote:

3. 2pairs have 9 combos

Again, on an unpaired board, say, 789, any particular 2 pair means that you need one of 3 cards for your first card and one of 3 (different) cards for your second. 3*3=9

Quote:

4. pocket pairs have 6 combos

Right, C(4,2) because there are 4 to choose from and you need to pick 2.

Quote:

5. a pair + kicker has 12 combos

OK, so this one is wrong I am pretty sure. Consider a flop of 25K and you want to know how often he has a pair of kings and another card. There are 3 kings. Then for a kicker he can have any card except 2, 5 or K so that's 52-4-4-4=40 potential kickers, meaning he can have 120 combos of pair + kicker. Maybe you mistyped and meant 120.

Quote:

6. unpaired cards have 16 combos

Yep, 4*4

Quote:

7. any two specific suited cards have 4 combos
8. any two specific cards have 1 combo

clearly

weeblewoble · 07-29-2012, 03:04 PM

Quote:

Originally Posted by RustyBrooks

Yes - C(52,2)

Any individual set on an unpaired board, yes. Say there is a 5 on the board. There are 3 remaining 5s in the deck and you need 2 of them, so it's C(3,2) which is 3. You could also list them by hand easily.

Again, on an unpaired board, say, 789, any particular 2 pair means that you need one of 3 cards for your first card and one of 3 (different) cards for your second. 3*3=9

Right, C(4,2) because there are 4 to choose from and you need to pick 2.

OK, so this one is wrong I am pretty sure. Consider a flop of 25K and you want to know how often he has a pair of kings and another card. There are 3 kings. Then for a kicker he can have any card except 2, 5 or K so that's 52-4-4-4=40 potential kickers, meaning he can have 120 combos of pair + kicker. Maybe you mistyped and meant 120.

Yep, 4*4

clearly

Case 5 is correct, it is 12 combos according to the book, and I checked again & copied this word for word. How can we solve Case 6 using the formula c(n,k)? I am more interested in the mathematical underpinnings, so I can understand and remember the material better.

Is it c(4,1) * c(4,1) ?

weeblewoble · 07-29-2012, 03:17 PM

I am trying to learn hand combinations, hand reading, equity, and EV. Am I going overboard here on trying to figure the combinatorics? Should I just settle for memorizing the combinations chart?

**RustyBrooks** · 07-29-2012, 05:36 PM

Number 5 literally can not be 12 unless he means a pair with some *specific* kicker, like in my example, exactly K2 in his hand.

Combinations, i.e. C(n,k) are not always the right way to solve a problem. In particular it doesn't work for #6 because we are picking from 2 distinct "buckets" instead of drawing 2 cards from one bucket. And sure, 4 = C(4,1) because X = C(X,1) but that's not really... useful. It's obvious that there are X ways to choose 1 item from a group of X.

weeblewoble · 07-29-2012, 06:19 PM

Thanks everybody.

**RustyBrooks** · 07-30-2012, 12:56 AM

There's nothing wrong with learning combinatorics, I think it's useful in poker and everyday life.