Probability to flop a set?

[Relentless]

I'm not sure on this so if someone could clear this up it'd be great.

What's the probablity to flop a set when you have a pocket pair as your hole cards?

[gaming_mouse]

(2 * (48 choose 2)+48) / (50 choose 3) = 0.1175

Or, in odds form: 7.51:1 against.

This includes the chance that you flop quads.

gm

[CCraft_42]

7 to 1 on the flop 16 to 1 on the turn

[binions]

Assume you have a pair of 8s.

There are 50 other cards in the deck. Accordingly, there are 50*49*48/3*2*1 possible flops = 19,600

2 of those cards are 8s, and 48 are not.

So, to flop a set, we need to flop 8xy, where xy is not a pair (since that would be flopping a full house, not a set).

The total number of xy combinations are 48*47/2*1 = 1128.
The total number of xy combinations that can be pairs is 72 (each pair has 6 possible combinations. There are 13 ranks, but we have taken the 8s out of the deck for now. 12*6=72).

So, there are 1128-72=1056 possible nonpair xy combos. There are two 8s left in the deck that can combine with these xy combos, so we multiply 1056*2=2112.

2112 possible flops where we flop a set of 8s (but not a full house or quads).

2112/19600 = 10.8%, or 8.3:1

If you count flopping boats 8s full and quad 8s, the odds are 7.5:1 against flopping a set OR BETTER.

If you go all in preflop, and get to see all 5 cards, your odds are 4.2:1 of making your set or better by the river.

[Relentless]

Thanks for the detailed reply.

[Zapped]

[ QUOTE ]
(2 * (48 choose 2)+48) / (50 choose 3) = 0.1175
This includes the chance that you flop quads.

[/ QUOTE ]

I was able to follow Binion's detailed reply, but I'm too new to absorb gaming_mouse's terse answer.

I understand that gaming_mouse is calculating the probability that the flop is 8xx where it doesn't matter if either x is also an 8 (for quads), or both x's match for the boat. I just can't for the life of me figure out how "(2 * (48 choose 2)+48)" represents that...

[gaming_mouse]

[ QUOTE ]
I understand that gaming_mouse is calculating the probability that the flop is 8xx where it doesn't matter if either x is also an 8 (for quads), or both x's match for the boat. I just can't for the life of me figure out how "(2 * (48 choose 2)+48)" represents that...

[/ QUOTE ]

If we assume that we have pocket 88's, then there are two 8's left, call the 8 #1 and 8#2. First ask, How many sets (not quads) can we make with 8 #1?

Well, the possible 3 card combos will all include 8 #1, so how many other cards (excluding the remaining 8) can we bet together with it. Well, after removing the remaining 8, there are 48 cards left. So we can put 8 #1 together with (48 choose 2) other cards. Sam argument for 8 #2. So there are 2*(48 choose 2) ways to flop a set WITHOUT flopping quads.

Now ask, How many ways can we flop quads. Well obviously we need both 8's, so there is only 1 card remaining to put together with those 8's. That one card can take on 48 different values. So there are 48 ways to flop quads.

gm

[CrazyN8]

Binions, do you have a book or something I could buy /images/graemlins/grin.gif

That was a very good explaination!

[gamble4pro]

The odds of flopping a set (not at least a set):
pocket pair (PP). The favorable 3-card combinations for the flop: (Pxy), with x and y different from P and different from eachother. The probability for such combination to come is 2*(C(48,2) - 12*C(4,2))/C(50,3) = 10.775%

[gamble4pro]

This also includes the chance of flopping full house.

[kowboy]

Thanks for letting the non-math inclined learn. /images/graemlins/smile.gif

[Zapped]

Thanks g.m., that's very clear now!