Odds of flopping a set

[iamastud]

If one starts with two pair, what are the odds of flopping a set on the flop?

And what are the odds of flopping 2 sets on the flop?

[Phil153]

Flop single set: C(44,2)*C(4,1)/C(50,3) = 19.3%
Flop double set: C(44,1)*2*2/C(50,3) = 0.89%
Flop quads: [C(44,1)*2+ C(4,1)*1]/C(50,3) = 0.47%

[tvta]

[ QUOTE ]
Flop single set: C(44,2)*C(4,1)/C(50,3) = 19.3%
Flop double set: C(44,1)*2*2/C(50,3) = 0.89%
Flop quads: [C(44,1)*2+ C(4,1)*1]/C(50,3) = 0.47%

[/ QUOTE ]

is this right? i thought it was closer to 25% total.

[franknagaijr]

[ QUOTE ]
Flop single set: C(44,2)*C(4,1)/C(50,3) = 19.3%
Flop double set: C(44,1)*2*2/C(50,3) = 0.89%
Flop quads: [C(44,1)*2+ C(4,1)*1]/C(50,3) = 0.47%

[/ QUOTE ]

I'm pretty sure that:

Chance of flopping quads with a pair in hand
c(2,2) * c(46,1) / c(48,3) = 0.00266

I assume that the chance of flopping either possible quads would be double that. I didn't double check the set numbers, but if you've got 4 cards in your hand, the denominator should be c(48/3).

[Phil153]

I was thinking holdem. You're right about the denominator.

This is wrong though:

[ QUOTE ]
Chance of flopping quads with a pair in hand
c(2,2) * c(46,1) / c(48,3) = 0.00266

[/ QUOTE ]

quads is given by:

(1 non pair card)*(2 quad cards) + (1 pair card)(2 quad cards)

= C(44,1)*2 + C(4,1)*1

So that bit is correct. The mistake in my calc is the denominator. Also note that the set calculation includes a paired board, which gives a full house. I still call it a "flopped set", but others don't.

So the numbers are:

Flopped set on unpaired board: 44*40/2*4/C(48,3) = 20.35%
Flopped set on paired board (i.e. a full house): 11*6*4/C(48,3) = 1.53%
Quads: [C(44,1)*2 + C(4,1)*1]/C(48,3) = 0.53%

[franknagaijr]

Phil - If you have two of the quads in your hand, that leaves two unaccounted for cards, and you need exactly both of them, c(2,2) times the unknown card in the third slot c(46,1). I don't understand your method, and I especially don't understand where the c(4,1) number comes from.

Edit - I just checked wikipedia, and it states that if you've got two pair, the odds of flopping quads is 0.0053191, which is pretty much exactly twice the number I posted. Beer me, my friend.

http://en.wikipedia.org/wiki/Poker_p...ing_high_hands

[Phil153]

Your only error is the C(2,2). Other than that, our numbers agree.

[franknagaijr]

Ok, you've successfully trolled me. You're a twit. Duly noted.

I feel better now.

C(2,2) means there are exactly two cards to fill exactly two slots, and there is one combination to do that, or two permutations, not that we care about permutations as men of science.

[Phil153]

No trolling, just a misunderstanding.