Probability of flopping 2pr?

[JackAll]

What's the probability of flopping 2pr in nlhe?

[mykey1961]

Assuming you don't have a pair in your hand to start with:

You will flop exactly 2 pair:

(44/50*6/49*3/48)*3 = 99/4600 = 0.0215

[Poker monkey]

This is correct - 2.2% - I looked it up in Super System to check...

[BruceZ]

[ QUOTE ]
This is correct - 2.2% - I looked it up in Super System to check...

[/ QUOTE ]

If you looked it up in Super System, then you should have caught his error in arithmetic. His formula is correct, but it comes to 99/4900 =~ 2.02%, not 99/4600 =~ 2.15%. Super System reports this correctly (original version, table XIX, AKs flops A-K and smaller card).

Note that this does not include 2-pair made with a pair on the flop (like AK flops A22). If we include these, it would add another 2.02% as the probability is exactly the same for this type of 2-pair.

[mykey1961]

yep somehow when putting 50*49*48 into the calculator I hit 50*46*48 giving 110400 instead of the proper 117600

[helter skelter]

[ QUOTE ]
Assuming you don't have a pair in your hand to start with:

You will flop exactly 2 pair:

(44/50*6/49*3/48)*3 = 99/4600 = 0.0215

[/ QUOTE ]

What do these numbers represent? I was trying to figure this out using the poker odds calculator and subtracting out trips and full houses and came up with a different number.

This appears simpler and more precise anyway, so I would like to know how to construct this type of formula for 2 pair, trips, and full house.

[BruceZ]

[ QUOTE ]
[ QUOTE ]
Assuming you don't have a pair in your hand to start with:

You will flop exactly 2 pair:

(44/50*6/49*3/48)*3 = 99/4600 = 0.0215

[/ QUOTE ]

What do these numbers represent? I was trying to figure this out using the poker odds calculator and subtracting out trips and full houses and came up with a different number.

This appears simpler and more precise anyway, so I would like to know how to construct this type of formula for 2 pair, trips, and full house.

[/ QUOTE ]

Suppose that we hold QJ (the calculation is the same for any non-pair). There are 3 Jacks, 3 Queens, and 44 other cards out of the 50 cards remaining. The probability that the flop comes xJQ or xQJ is the probability that the first card is one of the x other cards (44/50), times the probability that the second card is a J or Q (6/49), times the probability that the last card makes 2-pair (Q when 2nd card was J, or J when 2nd card was Q, 3/48). Then multiply all of this by 3 since the x can come in any of 3 positions.

See this post (http://forumserver.twoplustwo.com/showthreaded.php?Cat=0&Number=2948434&page=&vc=1) and its links for more on trips, full house, etc.

[helter skelter]

Thanks for the explanation and the links. So, using J2, if I then want a subset of the 2 pair flops where my J will be the top pair, then I can take the A,K,Q's out and I get

32/50*6/49*3/48 = .0147

Is this correct?

[BruceZ]

[ QUOTE ]
Thanks for the explanation and the links. So, using J2, if I then want a subset of the 2 pair flops where my J will be the top pair, then I can take the A,K,Q's out and I get

32/50*6/49*3/48 = .0147

Is this correct?

[/ QUOTE ]

Multiply that expression by 3, but you already did that since the number is correct.