Odds of flopping 2 pair or better with an unpaired hand preflop

[Godfather80]

can somebody tell me what this is and how to calculate it? I know that there are 117,600 possible flops. Given a hand like J [img]/images/graemlins/spade.gif[/img]2 [img]/images/graemlins/diamond.gif[/img], what are my odds of flopping 2 pair or better? And what are my odds with a hand like J[img]/images/graemlins/spade.gif[/img]2[img]/images/graemlins/spade.gif[/img] of flopping 2 pair or better?

Thank you.

[llamaoo7]

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can somebody tell me what this is and how to calculate it? I know that there are 117,600 possible flops. Given a hand like J 2 , what are my odds of flopping 2 pair or better? And what are my odds with a hand like J2 of flopping 2 pair or better. Thank you.

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First off, there are 19600 possible flops. Each of he flops you counted, also have 5 other arrangements that result in the same flop, so you count each possible flop 6 times. If you aren't familiar with combinations, then google for combinatorics 101 or something and you should get a good explanation and look into what combinations are.

The only hands you can make on the flop 2 pair or better are two pair, 3 of a kind, and a full house. For J2o, the flop can be J2x, JJx, 22x are the only way you can flop two pair or better. So you calculate the total number of ways of getting each outcome (watchout on J2x, though) and then add them together and divide by the total possible.

For JJx, there are 3 ways of getting the two remaining jacks from the last 3. This works out to 3 ways and for each of these ways, you have 48 additional cards that can come, so you have 144 ways of this scenario.

For 22x, the calculation is the same as above, so there are 144 ways of this occuring.

For J2x, you have to consider the chances of getting that J and the 2 AND the last card can't be a J or a 2 or you will be overcounting because JJx and 22x has these situations covered. So to get a J and a 2, you have 3 ways of picking a J and 3 ways of picking a 2. Then out of the remaining cards, there are only 46 that will give you two pair and not a boat which has already been counted. so 3*3*46 gives you 276.

Adding all of these together you get 560 ways. (560/19600*100)% is your answer. All of this was mental math so there are probably errors, but I feel the idea is correct.

Edited by llamaoo7 (07/16/05 07:49 PM)

[BruceZ]

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can somebody tell me what this is and how to calculate it? I know that there are 117,600 possible flops. Given a hand like J [img]/images/graemlins/spade.gif[/img]2 [img]/images/graemlins/diamond.gif[/img], what are my odds of flopping 2 pair or better?

[/ QUOTE ]

2-pair J2x: 3*3*44 = 396 (3 jacks, 3 deuces, 44 other)

trips JJx, 22x: 2*3*44 = 264 (3 ways to choose 2 jacks and 44 other, or 3 ways to choose 2 deuces and 44 other)

full house: 2*3*3 = 18 (3 ways to choose 2 jacks times 3 ways to choose 1 deuce, or 3 ways to choose 2 deuces times 3 ways to choose 1 jack)

quads: 2

There are only 19,600 flops if you ignore the order of the cards.

(396 + 264 + 18 + 2) / 19600 =~ 3.5% or 27.8-to-1


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And what are my odds with a hand like J[img]/images/graemlins/spade.gif[/img]2[img]/images/graemlins/spade.gif[/img] of flopping 2 pair or better?

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flush: C(11,3) = 11*10*9/6 = 165

(396 + 264 + 18 + 2 + 165) / 19600 =~ 4.3% or 22.2-to-1

[BruceZ]

llamaoo7:

I originally made these edits to your post by accident, but I put your original post back. That's why yours says it is edited by me.

-Bruce


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For JJx, there are 3 ways of getting the two remaining jacks from the last 3. This works out to 3 ways and for each of these ways, you have 48 additional cards that can come, so you have 144 ways of this scenario.

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This counts JJJ 3 times, so the total should be 142


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For 22x, the calculation is the same as above, so there are 144 ways of this occuring.

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This counts 222 3 times, so the total should be 142.


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For J2x, you have to consider the chances of getting that J and the 2 AND the last card can't be a J or a 2 or you will be overcounting because JJx and 22x has these situations covered. So to get a J and a 2, you have 3 ways of picking a J and 3 ways of picking a 2. Then out of the remaining cards, there are only 46 that will give you two pair and not a boat which has already been counted. so 3*3*46 gives you 276.

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3*3*46 does not equal 276, and anyway it should be 3*3*44 = 396.

[llamaoo7]

Yeah, I just did that in my heads, but there's something I didnt think about. What is the the for comes paired with a jack or three of a king? But I guess the original question is just interested in the hands that involve both whole cards in a pair/trips/quads.

[BruceZ]

[ QUOTE ]
Yeah, I just did that in my heads, but there's something I didnt think about. What is the the for comes paired with a jack or three of a king? But I guess the original question is just interested in the hands that involve both whole cards in a pair/trips/quads.

[/ QUOTE ]

Right, these weren't counted before:

2-pair with pair on flop (e.g. JAA): 2*3*11*6 = 396 (3 ways to pair J times 11*6 ways to pair board, or 3 ways to pair deuce times 11*6 ways to pair board).

3-of-a-kind on flop, not J or 2: 11*4 = 44

Adding these to previous counts gives total of 5.7% or 16.5-to-1 for J2o, and 6.6% or 14.3-to-1 for J2s.

[Godfather80]

[ QUOTE ]
[ QUOTE ]
Yeah, I just did that in my heads, but there's something I didnt think about. What is the the for comes paired with a jack or three of a king? But I guess the original question is just interested in the hands that involve both whole cards in a pair/trips/quads.

[/ QUOTE ]

Right, these weren't counted before:

2-pair with pair on flop (e.g. JAA): 2*3*11*6 = 396 (3 ways to pair J times 11*6 ways to pair board, or 3 ways to pair deuce times 11*6 ways to pair board).

3-of-a-kind on flop, not J or 2: 11*4 = 44

Adding these to previous counts gives total of 5.7% or 16.5-to-1 for J2o, and 6.6% or 14.3-to-1 for J2s.

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Thanks to everyone who posted. Your discussion was very helpful.