Common Flop Odds

*Last edited by Pyroxene on Fri, 01 Apr 2005, 12:06pm; edited 8 times in total*

There have been several posts asking for flop odds lately. This chart puts most of the important flop odds all together.

Texas Hold'em Flop Odds

Common odds when holding unpaired hole cards:

Where did you get these? I only glanced at a few and they were wrong...

They look about right to me...

They look wrong to me.

Odds of flopping one pair on unpaired hole cards is 41%, I am sure.

I'm certain that it is over 27%.

The key word here is EXACTLY...

Dwarfman I think you are talking about the odds of flopping AT LEAST one pair instead of EXACTLY one pair.

So what is the odds of flopping AT LEAST one pair? Lets see that would be 1 - the probability of neither of your two cards being paired on the flop. The probability of not pairing either of your cards is (44/50)*(43/49)*(42/48), subtract that from 1 and you get 0.3242857 or 32.42857%.

Add the following percentages and see what you get:

As some have questioned the odds I posted, I will walk through the computations. As I stated above, save for the 8 out straight draw numbers, these numbers can all be found in reference manuals. But there is no magic to them so I will demonstrate how to compute them.

There are scads of ways to compute these numbers. Each correct method will arrive at EXACTLY the same number. I will choose certain methods that I think best describe what is going on. You can CERTAINLY arrive at EXACTLY these same numbers using different representations of the problem.

Holding unpaired cards and flopping EXACTLY one pair by matching a hole card.

I walked through that one above. You are looking for one of the 3 cards that match hole card 1 or the 3 cards that match hole card 2. So you are looking for one of 6 cards. Then you are looking for any card that DOES NOT match either of your hole cards (52 cards minus the 2 in your hand minus the 1 card that matched the first time minus the 5 cards that remain that match you hole cards = 44 cards). Then you are looking any card that DOES NOT match your hole cards AND does not match the card that just fell, because that would give you two pair (52 cards minus 2 in your hand minus the 1 card that matched the first time minus the 1 card that fell on the board minus the 8 cards that would match either your hole cards of the card that fell on the board = 40 cards.) Then divide by the total number of combinations, 50 cards * 49 cards * 48 cards. This comes to:

(6 * 44 * 40) / (50 * 49 * 48) = 8.97959%.

This is the chance of flopping one card to your hole cards ON THE FIRST CARD OF THE FLOP, as in:

Hole Cards: Queen of Clubs

Flop:

There is also that same chance to flop the pairing card on the second card, as in:

Flop: Two of Spades

And the same chance to flop the pairing card on the third card, as in:

Flop: Two of Spades

So you have 3 * 8.95959% or 26.939%

Holding unpaired cards and flopping EXACTLY two pair by pairing a hole card AND pairing on the board.

You are looking for one of the 6 cards that pair either of your hole cards, then you are looking for any card that DOES NOT match your hole card (there are 44 remaining), then you are looking for 1 of the 3 cards that will pair that second card. Then divide by the number of combinations. This comes to:

(6 * 44 * 3) / (50 * 49 * 48) = 0.673469%

As with the above example, this is the chance of flopping your hole pair on the first card. There is an equal chance of flopping it on the second card and an equal chance of flopping it on the third card.

So you have 0.673469 * 3 = 2.020%

Holding unpaired cards and flopping EXACTLY two pair by pairing EACH of your hole cards.

You are looking for one of the 3 cards that pair your first hole card, then you are looking for one of the 3 cards that pair your second hole card, then you are looking for any card that DOES NOT pair either of your hole cards (there are 44 remaining). Then divide by the number of combinations. This comes to:

(3 * 3 * 44) / (50 * 49 * 48) = 0.336734%

This is the chance of flopping a match to your first hole card, then a match to your second hole card, then a rag card. Considering that you hold XY, this is the chance of flopping X Y r. There are 5 other ways of flopping what you want, that each have the same chance. They are:

X r Y
Y X r
Y r X
r Y X
r X Y

So you have 0.336734 * 6 = 2.020%

*Last edited by Pyroxene on Fri, 01 Apr 2005, 11:05am; edited 2 times in total*

Holding unpaired cards and flopping EXACTLY trips by flopping two cards to one hole card.

You are looking any of the 6 cards that will pair one of your hold cards, then you are looking any of the 2 remaining cards that will trip that hole card, then you are looking for any card that DOES NOT quad up your hole card (that would give you 4 of a kind) and DOES NOT pair up your other hole card (that would give you a full house). Then divide by the number of combinations. That gives you:

(6 * 2 * 44 ) / (50 * 49 * 48) = 0.4489796%

That is the chance of flopping trips on the first two cards of the flop and a rag on the third. There are two other ways to flop trips that have an equal chance: (1) matching the first and third cards, (2) matching the second and third cards.

So, you have 3 * 0.4489796 = 1.347%.

Holding unpaired cards and flopping EXACTLY a full house, trips of one hole card and pairing the other.

You are looking any of the 6 cards that will pair one of your hold cards, then you are looking any of the 2 remaining cards that will trip that hole card, then you card looking for any of the 3 cards that will pair your other hole card. Then divide by the number of combinations. That gives you:

(6 * 2 * 3 ) / (50 * 49 * 48) = 0.03061224%

That is the chance of flopping trips on the first two cards of the flop and pairing on the third. There are two other ways to flop this type of full house that have an equal chance: (1) tripping the first and third cards while pairing the second, (2) tripping the second and third cards while pairing the first.

So, you have 3 * 0.03061224 = 0.0918%.

Holding unpaired cards and flopping EXACTLY four of a kind, three cards to one of your hole cards.

You are looking for any of the 6 cards that will pair one of your hold cards, then you are looking any of the 2 remaining cards that will trip that hole card, then you card looking for the 1 card that will quad that hole card. Then divide by the number of combinations. That gives you:

(6 * 2 * 1 ) / (50 * 49 * 48) = 0.0102%

*Last edited by Pyroxene on Fri, 01 Apr 2005, 10:11pm; edited 2 times in total*

Holding paired cards and flopping EXACTLY two pair by pairing the board.

You are looking for any of the 48 cards that DOES NOT match your hole card. Then you are looking one of the 3 cards that match that card. Then you are looking for a card that DOES NOT match your hole pair and DOES NOT trip up the first two cards of the flop, there are 44 such cards. Then divide by the number of combinations. That gives you:

(48 * 3 * 44 ) / (50 * 49 * 48) = 5.387755%

That is the chance of flopping the board pair on the first two cards. There are two other ways to flop a board pair that have an equal chance: (1) pairing the first and third cards, (2) pairing the second and third cards.

So, you have 3 * 5.387755% = 16.163%

Holding paired cards and flopping EXACTLY trips by flopping a set for your pocket pair.

You are looking for any of the 2 cards that pair your hole cards. Then you are looking for any card that DOES NOT quad up your hole card, there are 48. Then you are looking for a card that DOES NOT quad up your hole card AND DOES NOT match the other flop card (that would give you a full house), there are 44 such cards. Then divide by the number of combinations. That gives you:

(48 * 3 * 44 ) / (50 * 49 * 48) = 3.59184%

That is the chance of flopping the set on the first card. There are two other ways to flop a set that have an equal chance: (1) tripping on the second card, (2) tripping on the third card.

So, you have 3 * 3.59184% = 10.775%

Holding paired cards and flopping EXACTLY a full house, a set to your hole pair and pairing the board.

You are looking for any of the 2 cards that pair your hole cards. Then you are looking for any card that DOES NOT quad up your hole card, there are 48. Then you are looking for a card that pairs up that board card, of which there are 3. Then divide by the number of combinations. That gives you:

(2 * 48 * 3 ) / (50 * 49 * 48) = 0.2448979%

That is the chance of flopping the set on the first card and pairing the second and third. There are two other ways to flop this type of full house that have an equal chance: (1) tripping on the second card, (2) tripping on the third card.

So, you have 3 * 0.2448979% = 0.735% (reviewing my initial post, I transcribed this number incorrectly from my notes. I have fixed the initial post.)

Holding paired cards and flopping EXACTLY a full house, by the board tripping up.

You are looking for any card that does not match your hole pair, there are 48. Then you are looking for any of the 3 cards that will pair up the board. Then you are looking for any of the 2 cards that will trip up the board. Then divide by the number of combinations. That gives you:

(48 * 3 * 2) / (50 * 49 * 48) = 0.2448979%

Holding paired cards and flopping EXACTLY four of a kind, two cards to your hole pair.

You are looking for any of the 2 cards that will trip your hole pair. Then you are looking for the remaining card that will quad your hole pair. Then you are looking for any of the remaining 48 cards. Then divide by the number of combinations. That gives you:

(2 * 1 * 48) / (50 * 49 * 48) = 0.0816327%

That is the chance of flopping the quads on the first and second flop cards. There are two other ways to flop quads that have an equal chance: (1) tripping on the first card and quading on the third, (2) tripping on the second card and quading on the third.

So, you have 3 * 0.0816327% = 0.245%

*Last edited by Pyroxene on Fri, 01 Apr 2005, 11:46am; edited 1 time in total*

Holding two unsuited cards and flopping a four flush.

You are looking for any of the 24 cards that are the same suit as either of the two cards that you hold. Then you are looking for any of the 11 cards that are the same suit as the first flop card. Then you are looking for any of the 10 cards that are the same suit as the first two flop cards. Then divide by the number of combinations. That gives you:

(24 * 11 * 10) / (50 * 49 * 48) = 2.245%

Holding two suited cards and flopping a flush.

You are looking for any of the 11 cards that are the same suit as the two cards that you hold. Then you are looking for any of the 10 cards that are the same suit as the first flop card. Then you are looking for any of the 9 cards that are the same suit as the first two flop cards. Then divide by the number of combinations. That gives you:

(11 * 10 * 9) / (50 * 49 * 48) = 0.842%

Holding two suited cards and flopping a four flush.

You are looking for any of the 11 cards that are the same suit as the two cards that you hold. Then you are looking for any of the 10 cards that are the same suit as the first flop card. Then you are looking for any card that DOES NOT match the suit of the first two cards, there are 39. Then divide by the number of combinations. That gives you:

(11 * 10 * 39) / (50 * 49 * 48) = 3.648%

That is the chance of flopping the four flush on the first two cards of the flop. There are two other ways to flop a four flush that have an equal chance: (1) matching the suit of the first and third cards, (2) matching the suit of the second and third cards.

So, you have 3 * 3.648% = 10.944%

The straights are tricky. I prefer to think of them in a certain way and I acknowledge that not everyone may think about them the same as I. A mental model that works well for me and simplifies the math is computing the chance of EXACTLY 1 of EACH of 3 ranks falling on the flop when you do not hold any of those ranks in the hole. There are a lot of ways to compute that number. I am going to stick with the method I have used throughout the other explanations.

If you want the flop to contain ranks X, Y, Z when your hole cards do not contain X, Y, or Z then:

You are looking for 1 of the 4 X. Then you are looking for 1 of the 4 Y. Then you are looking for 1 of the 4 Z. Then divide by the number of combinations. That gives you:

(4 * 4 * 4) / (50 * 49 * 48) = 0.054422%

That is the chance of EXACTLY flopping X, then Y, then Z. There are 5 other ways of flopping X, Y, Z that all have an equal chance:

X Z Y
Y X Z
Y Z X
Z X Y
Z Y X

So, the odds of flopping EXACTLY one each of ranks X, Y, and Z when you do not hold either X, Y, or Z on the hole are:

6 * 0.054422 = 0.3265%

We are going to call that number S, and we are going to be using it a lot.

Holding connectors 54 through JT and flopping a straight.

These connectors have room on both sides to form a number of straights. Graphically, if you are holding cards AB, then you could flop the following straights:

X Y Z A B - - -
- X Y A B Z - -
- - X A B Y Z -
- - - A B X Y Z

So there are 4 sets of X, Y, Z that will give you a straight. So you have a 4 times S chance of flopping a straight.

4 * 0.3265 = 1.306%

Holding one gapped connectors 53 through QT and flopping a straight.

These connectors have room on both sides to form a number of straights. Graphically, if you are holding cards AB, then you could flop the following straights:

X Y A Z B - -
- X A Y B Z -
- - A X B Y Z

So there are 3 sets of X, Y, Z that will give you a straight. So you have a 3 times S chance of flopping a straight.

3 * 0.3265 = 0.980%

Holding two gapped connectors 52 through KT and flopping a straight.

These connectors have room on both sides to form two straights. Graphically, if you are holding cards AB, then you could flop the following straights:

X A Y Z B -
- A X Y B Z

So there are 2 sets of X, Y, Z that will give you a straight. So you have a 2 times S chance of flopping a straight.

2 * 0.3265 = 0.653%

Holding three gapped connectors A5 through AT and flopping a straight.

These connectors can only flop one straight. Graphically, if you are holding cards AB, then you could flop the following straight:

A X Y Z B

So there are 1 set of X, Y, Z that will give you a straight. So you have a 1 times S chance of flopping a straight.

1 * 0.3265 = 0.327%

Lastly, the 8 out straight draws. For all the other odds, I have confirmed my results against reference material and they match. I have not found any references for these 8 out straight draws. I may have errors and I invite discussion.

The 5 card, 8 out straight draws fall into two categories. First, there is flopping 4 consecutive cards such as:

- A B C D -

So there are 4 outs to the left of A and 4 outs to the right of D.

We will call this a TYPE-1 8 out draw.

Secondly, there are double barreled gut shots (double bellied gut shots) such as:

A - C D E - G

So there are 4 outs to the left of C that will give you an A - E straight and there are 4 outs to the right of E that will give you a C - G straight.

We will call this a TYPE-2 8 out draw.

Holding connectors 54 through JT and flopping an 8 out draw.

First, we will compute the TYPE-1 draws. Given that our hole cards are F and G and they are connected, graphically here are the three TYPE-1 8 out draws:

D E F G - -
- E F G H -
- - F G H I

For each of these we are looking 1 of the 4 left most flop cards, then 1 of the 4 right most flop cards, then any card that DOES NOT give us a straight (there are 40). Then divide by the number of combinations. That gives you:

(4 * 4 * 40) / (50 * 49 *48) = 0.54422%

That is the chance of flopping the cards in exactly the order: L R x. There are 5 other combinations that have an equal chance:

L x R
R L x
R x L
x R L
x L R

So, the odds of flopping any one of those TYPE-1 draws would be:

6 * 0.54422% = 3.265%

In turn, there are 3 of those TYPE-1 draws, so the odds of flopping any of the TYPE-1 draws would be:

3 * 3.265% = 9.796%

Second, we will compute the TYPE-2 draws. Given that our hole cards are F and G and they are connected, graphically here are the two TYPE-2 8 out draws:

C - E F G - I -
- D - F G H - J

This turns out to be a problem where we are looking for a flop containing EXACTLY an X, Y and Z rank. We already computed that chance, it was S. There are two sets of X, Y and Z that would give us two TYPE-2 draws so the chance is 2 * S or:

2 * 0.3265% = 0.653%

Lastly, the chance of any 8 out draw would be the sum of the chance of the TYPE-1 draws and the TYPE-2 draws. So,

9.796% + 0.653% = 10.449%

Holding one gapped connectors 53 through QT and flopping an 8 out draw.

First, we will compute the TYPE-1 draws. Given that our hole cards are F and H and they are one gapped, graphically here are the two TYPE-1 draws:

- - E F G H - - -
- - - F G H I - -
C D E F - H - - -
- - - F - H I J K

The first two require two cards to form up the 4 sequence. The second two require 3 cards to form up the sequence.

Using the numbers we figured out for the step above that give us:

2 * 3.265% + 2 * 0.3265 = 7.18%

Second, we will compute the TYPE-2 draws. Given that our hole cards are F and H and they are one gapped, graphically here are the three TYPE-2 draws:

B - D E F - H - - - -
- - D - F G H - J - - -
- - - - F - H I J - L

That would give us three TYPE-2 draws so the chance is 3 * S or:

3 * 0.3265% = 0.980%

Lastly, the chance of any 8 out draw would be the sum of the chance of the TYPE-1 draws and the TYPE-2 draws. So,

7.18% + 0.980% = 8.08% (I forgot to include two of the TYPE-1 draws in my initial posting. I have corrected the numbers.)

As a final note, the 53 and QT combinations cannot form one of the TYPE-2 draws as there is not enough room to the left and right respectively so their odds are slightly lower.

Holding two gapped connectors 52 through KT and flopping an 8 out draw.

First, we will compute the TYPE-1 draws. Given that our hole cards are F and I and they are two gapped, graphically there are five TYPE-1 draws:

- - - F G H I - - -
- D E F H - I - - -
C D E F - - I - - -
- - - F - H I J K -
- - - F - - I J K L

The first requires two cards to form up the 4 sequence. The remaining 4 require 3 cards to form up the sequence.

Using the numbers we figured out for the step above that give us:

1 * 3.265% + 4 * 0.3265 = 4.571%

Second, we will compute the TYPE-2 draws. Given that our hole cards are F and I and they are two gapped, graphically here are the two TYPE-2 draws:

- C - E F G - I - - -
- - - - F - H I J - K -

That would give us two TYPE-2 draws so the chance is 2 * S or:

2 * 0.3265% = 0.653%

Lastly, the chance of any 8 out draw would be the sum of the chance of the TYPE-1 draws and the TYPE-2 draws. So,

4.571% + 0.653% = 5.224% (I forgot to include four of the TYPE-1 draws in my initial posting. I have corrected the numbers.)

As a final note, the 52 and KT combinations cannot form one of the TYPE-2 draws as there is not enough room to the left and right respectively so their odds are slightly lower.

Holding three gapped connectors A5 through AT and flopping an 8 out draw.

First, we will compute the TYPE-1 draws. Given that our hole cards are F and J and they are three gapped, graphically here are TYPE-1 draw:

C D E F - - - J - - -
- D E F G - - J - - -
- - E F G H - J - - -
- - - F - H I J K - -
- - - F - - I J K L -
- - - F - - - J K L M

These all require 3 cards to form up the sequence.

Using the numbers we figured out for the step above that give us:

6 * 0.3265 = 1.959%

Second, we will compute the TYPE-2 draws. Given that our hole cards are F and J and they are three gapped, graphically here are the two TYPE-2 draws:

D - F G H - J - -
- - F - H I J - L

That would give us two TYPE-2 draws so the chance is 2 * S or:

2 * 0.3265% = 0.653%

Lastly, the chance of any 8 out draw would be the sum of the chance of the TYPE-1 draws and the TYPE-2 draws. So,

1.959% + 0.653% = 2.612% (I forgot to include scads of the TYPE-1 draws in my initial posting. I have corrected the numbers.)

As a final note, the actual numbers vary all over the place for three gappers. Only the 95 and T6 three gappers can form all of the combinations listed. As the three gappers get closer to the edge, fewer combinations are possible. In short, do not expect much from three gappers.

My Head hurts

Pyroxene, PM me; I'm working on sketches for bunch of elaborate odds charts/software, and seriously could use a partner

(I'm not joking)

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