Flop odds

[IIAce]

If you hold two unpaired cards (ie. 67)

What are the odds of getting trips on the flop including a full house and not including a full house? Also, what are the odds of flopping two pair using both of your cards (ie. flopping 6 7 A while holding 67)?

PS. I'd appreciate it if the odds were expressed as percentages.


Thanks

[LetYouDown]

Assuming you hold 6-7, I come up with 1.35% for trips and only trips and 1.47% for trips, boat or quads on the flop. This was a real quick grunting it out, so someone correct me if I'm wrong.

[BruceZ]

[ QUOTE ]
If you hold two unpaired cards (ie. 67)

What are the odds of getting trips on the flop including a full house and not including a full house?

[/ QUOTE ]

Including full house:

[2*C(3,2)*44 + 2*C(3,2)*3] / C(50,3) =~ 1.44%


Not including full house:

2*C(3,2)*44 / C(50,3) =~ 1.35%


[ QUOTE ]
Also, what are the odds of flopping two pair using both of your cards (ie. flopping 6 7 A while holding 67)?

[/ QUOTE ]

Not including full house:

3*3*44 / C(50,3) =~ 2.02%


Including full house:

[3*3*44 + 2*C(3,2)*3] / C(50,3) =~ 2.11%

[IIAce]

[ QUOTE ]
Including full house:

[2*C(3,2)*44 + 2*C(3,2)*3] / C(50,3) =~ 1.44%


Not including full house:

2*C(3,2)*44 / C(50,3) =~ 1.35%




[/ QUOTE ]

What I did for trips including a full house was
(3/50)*(2/49)*(47/48)*6*2 = .0287755 = 2.88%

If you're wondering where I got the *6*2 it's because there are 6 ways to arrange 3 cards and the last 2 is because you have two cards in your hand so you calculate the probability for getting trips for both of them.

Not including full house:
(3/50)*(2/49)*(44/48)*6*2 = .026938 = 2.69%

I noticed that your values are my values divided by 2 so maybe you forgot to calculate trips for both hole cards? Did I do something wrong? The reason I made this thread was because this calculator: http://www.texasholdempoker.com/calculator.php
and Super System say that's there's a 1.57% chance of getting trips on the flop and .09% of getting a full house..when I add these two numbers I don't get the values I got. What do you think?

Edit: While you're at it...show me how to get the odds of flopping quads? My numbers were different from the calculator's numbers again.

Thanks

[BruceZ]

[ QUOTE ]
What I did for trips including a full house was
(3/50)*(2/49)*(47/48)*6*2 = .0287755 = 2.88%

If you're wondering where I got the *6*2 it's because there are 6 ways to arrange 3 cards and the last 2 is because you have two cards in your hand so you calculate the probability for getting trips for both of them.

Not including full house:
(3/50)*(2/49)*(44/48)*6*2 = .026938 = 2.69%

I noticed that your values are my values divided by 2 so maybe you forgot to calculate trips for both hole cards?

[/ QUOTE ]

No.


[ QUOTE ]
Did I do something wrong?

[/ QUOTE ]

You did something wrong. You should only multiply by 3 instead of 6 because when you multiply 3/50 * 2/49, you are already counting both orders for these two cards, so it only remains to multiply by 3 for the 3 positions of the final card. That is why your numbers are off by a factor of 2.

[ QUOTE ]
The reason I made this thread was because this calculator: http://www.texasholdempoker.com/calculator.php
and Super System say that's there's a 1.57% chance of getting trips on the flop and .09% of getting a full

[/ QUOTE ]

Super System bottom of table XIX shows AK flopping trip aces or kings 1.35%, just as I have calculated, and above in the same table shows a full house is 0.09%, as I have calculated. You can be sure these calculations are correct.

[BruceZ]

I take back what I said about combining the full houses in one term. In many similar cases you need a separate term, but in this case you get the same answer either way.

For quads you need a separate term. There are 2 flops that give quads, so simply add 2/C(50,3) = 0.01%.

[MarkD]

Bruce, I'm sorry, but I want to break down this post as well.

[ QUOTE ]
Including full house:

[2*C(3,2)*44 + 2*C(3,2)*3] / C(50,3) =~ 1.44%


[/ QUOTE ]

C(3,2) = represents the way number of ways for there to be two cards of the same rank on the flop that are also similar to one of the cards in your hand. We multiply this by two because there are two cards of different ranks in our hand since that means there are two ways that the board can contain two of a rank matching a card in our hand, correct? (Oh, that is way too wordy, but I think you will understand what I mean).

We do the same thing in the second term but the 44 becomes a 3 because it has to match one of the cards in our hand.

This is actually becoming simple as I walk through it like a dummy.

So, quads would just be:
2*C(3,3)/C(50,3) right?

[IIAce]

Ok, I'm trying to make this chart with pretty much all the preflop odds. I'm trying to find the probability of getting a two pair and only a two pair on the flop (when you're holding something like A3s) and having all cards be different suites (rainbow). That's how I did it: (3*3*22)/19600 = 1.01% right?

[BruceZ]

[ QUOTE ]
[ QUOTE ]
Including full house:

[2*C(3,2)*44 + 2*C(3,2)*3] / C(50,3) =~ 1.44%


[/ QUOTE ]

C(3,2) = represents the way number of ways for there to be two cards of the same rank on the flop that are also similar to one of the cards in your hand. We multiply this by two because there are two cards of different ranks in our hand since that means there are two ways that the board can contain two of a rank matching a card in our hand, correct?

[/ QUOTE ]

Exactly.


[ QUOTE ]
We do the same thing in the second term but the 44 becomes a 3 because it has to match one of the cards in our hand.

[/ QUOTE ]

Right, and we could just as well have added 3 to the first term to get 47 instead of 44. There are many cases where you would need the separate term to avoid over counting, so I think it is best not to get in the habit of combining terms like this until you are sure you understand when it is appropriate.


[ QUOTE ]
So, quads would just be:
2*C(3,3)/C(50,3) right?

[/ QUOTE ]

Right, and this is the same as 2/C(50,3) since there are only 2 flops that give quads. Note that here we need the extra term. We cannot just add 1 to the 44 in the first term to get 2*C(3,2)*45 as this would have the effect of counting the quads 3 times. This problem comes about when the cards we are adding are the same rank as some cards already on the board. If you can just understand that, you'll be light years ahead of most people.

[BruceZ]

[ QUOTE ]
Ok, I'm trying to make this chart with pretty much all the preflop odds. I'm trying to find the probability of getting a two pair and only a two pair on the flop (when you're holding something like A3s) and having all cards be different suites (rainbow). That's how I did it: (3*3*22)/19600 = 1.01% right?

[/ QUOTE ]

No, if you want 3 different suits on the flop it would have to be 3*2*11/19600 = 0.34%.

[IIAce]

[ QUOTE ]
No, if you want 3 different suits on the flop it would have to be 3*2*11/19600 = 0.34%.

[/ QUOTE ]
Ok I see why you have 2 instead of 3 but I don't get the 11 part. Let's say you have 67 of clubs, the flop could contain clubs, diamonds, or hearts. 3 ways to pick the first card (assume it happens to be a diamond), 2 ways to pick the second card (assume it happens to be a heart), 22 ways to pick the last card (all other cards that are not 6, 7, diamonds, or hearts). Right?

[BruceZ]

[ QUOTE ]
[ QUOTE ]
No, if you want 3 different suits on the flop it would have to be 3*2*11/19600 = 0.34%.

[/ QUOTE ]
Ok I see why you have 2 instead of 3 but I don't get the 11 part. Let's say you have 67 of clubs, the flop could contain clubs, diamonds, or hearts. 3 ways to pick the first card (assume it happens to be a diamond), 2 ways to pick the second card (assume it happens to be a heart), 22 ways to pick the last card (all other cards that are not 6, 7, diamonds, or hearts). Right?

[/ QUOTE ]

I thought you wanted 3 different suits on the flop that were different from the suit in your hand. Then if you have 67c in your hand, the flop would be spades, hearts, and diamonds. 3 choices for the six (say spades), 2 choices for the 7 (say hearts), and 11 choices for the last suit (diamonds) since it can't be 6 or 7.

If you just want a rainbow flop which can match your hole cards, then the last card could be 11 diamonds or 11 clubs, and the probability would be 3*2*22/19600 = 0.67%.

[MarkD]

[ QUOTE ]
Note that here we need the extra term. We cannot just add 1 to the 44 in the first term to get 2*C(3,2)*45 as this would have the effect of counting the quads 3 times. This problem comes about when the cards we are adding are the same rank as some cards already on the board. If you can just understand that, you'll be light years ahead of most people.


[/ QUOTE ]

You lost me here. Sorry, but I'm not sure what you are talking about.

[ QUOTE ]
Right, and we could just as well have added 3 to the first term to get 47 instead of 44. There are many cases where you would need the separate term to avoid over counting, so I think it is best not to get in the habit of combining terms like this until you are sure you understand when it is appropriate.


[/ QUOTE ]
Ok, I think I might understand a bit now. I really do prefer to break out the individual terms. I like doing that in algebra as well - I always solved problems in a lot of steps (but very fast).

[BruceZ]

[ QUOTE ]
[ QUOTE ]
Note that here we need the extra term. We cannot just add 1 to the 44 in the first term to get 2*C(3,2)*45 as this would have the effect of counting the quads 3 times. This problem comes about when the cards we are adding are the same rank as some cards already on the board. If you can just understand that, you'll be light years ahead of most people.


[/ QUOTE ]

You lost me here. Sorry, but I'm not sure what you are talking about.

[/ QUOTE ]

I'm just saying that when we counted the number of flops that made trips, we got 2*C(3,2)*44. Now suppose we want to add the flops that make quads. Many would be tempted to simply add the 1 extra card that makes quads on to the 44, to get 2*C(3,2)*45. That's a big no-no. Look how many flops we added, 2*C(3,2)*1 = 6. How many should we have added? Only 2. We are actually counting the quad flops 3 times! This will happen whenever we have a situation like this where the board is paired, and instead of handling this as a separate case, we try to handle the paired rank like it is any other rank. In this example it is ridiculously easy to see what the problem is, but even advanced people make exactly this kind of error all the time in cases where it is not so obvious, and sometimes they can't see it even when it is pointed out to them. A popular book on hold'em odds by Mike Petriv if full of these types of errors.