Odds of flopping 2 sets in a row
GTech1
1/(16*8*16*8)
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Gary Carson
On Oct 19 2007 6:55 PM, GTech1 wrote:
> Got into a small dispute over this. Seems to me that it would just be
> 1/(16*8*16*8)
The probability that someone, at sometime today, will flop two sets in a row, is
mumble, mumble, wave hand, presto, about 1.
To flop to sets in a row you not only have to first be dealt a pocket pair
(gotta calculate that) and somebody else has to have a hand they want to play
(given that you have a pair).
You havn't really specified your question very well.
Gary Carson
http://www.garycarson.com
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GTech1
dealt cards the next two hands and that you will see the next two flops.
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johnnyt
> You know what, you're right. What I'm talking about is assuming you will be
> dealt cards the next two hands and that you will see the next two flops.
More specifically...
A pair in hand, and one of that rank on the flop. And I am guessing, no
quads of full houses either?
Alex
Lost all three hands too.
kimi
> I don't know what the odds are but I've flopped three sets in a row.
>
> Lost all three hands too.
I think the probability of your holding two cards and your getting
your third card and the probability of you holding unlike cards in
your hand and the board giving you a set (two on the board and the one
in your hand) would be different.
I think the probability of getting a set two hands in a row is the
same as the probability of getting the first.
But it's been years since "Discrete Math" so I'd have to brush up,
will let the experts chime in.
Alex
When I say I had three sets in a row, it was three pocket pairs in a
row that flopped a third.
Still lost all three.
Gary Carson
On Oct 19 2007 8:08 PM, kimi wrote:
Yes, you're right, at least almost right, that's the other part he's not
specifiying -- the period of time being observed and the specific person. I'm
sure he's thinking about specifically the next two hands and him as the specific
person.
When calcuating the probability of someone winning a lottery twice in a row
(without specification of who the person is) then it's just the probability of
winning it once for a specific person becuase somebody won it last time, they
become your specific person. (newspaper reporters can always find a math prof
to quote who will get this wrong, so remember that next time some math prof says
something about applied math).
But here he's specifying the person, so it's not that easy. But although he's
talking about the next two hands, that doesn't really make sense. Does he
really think flopping sets in hand 1 and hand 2 is going to be more noteworthy
than flopping sets in hand 2 and hand 3? Does he think it happening to him is
more noteworthy than seeing it happen to the guy in the one seat?
Gary Carson
http://www.garycarson.com
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Lord Turkey Cough
"GTech1" <4307...@recpoker.com> wrote in message
news:1192838143$106...@recpoker.com...
> Got into a small dispute over this. Seems to me that it would just be
> 1/(16*8*16*8)
Assumming you have a pocket pair it is about 64-1? very approximately?
GTech1
you are about to get dealt cards the next two hands. What are the odds that on
both hands, you flop sets (a set being defined as you holding a pocket pair, and
flopping a 3rd card)?
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kl
"GTech1" <4307...@recpoker.com> wrote in message
news:1192838143$106...@recpoker.com...
Batman
Yeah, no one said there'd be any math....
Did you know that 4 out of 3 people have trouble with fractions? :)
Mike
http://thoughtsotheday.blogspot.com/
http://matchesmalone.wordpress.com/
http://www.poker-riot.com
Batman
Nice redirect :)
You can also get there from the last link below....
kurtissimo
Probability of a pocket pair hitting (at least) a set on the flop: 1/8.5
(approx.)
Therefore, the probability of *you* being dealt a pocket pair *and*
flopping a set (or better) on each of the next two hands that you will
play is approx. 1/(17*8.5)^2, which is about 1 in 20,880 (or about
0.0048%).
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Gary Carson
On Oct 21 2007 11:06 PM, kurtissimo wrote:
> Probability of being dealt a pocket pair on any given hand: 1/17 (exact)
> Probability of a pocket pair hitting (at least) a set on the flop: 1/8.5
> (approx.)
>
> Therefore, the probability of *you* being dealt a pocket pair *and*
> flopping a set (or better) on each of the next two hands that you will
> play is approx. 1/(17*8.5)^2, which is about 1 in 20,880 (or about
> 0.0048%).
Now what's the probability that some idiot reads that and the next time they
flop two sets in a row saying, "The odds of that are only 1 in 20,880"?
> > > http://www.garycarson.com/
Gary Carson
http://www.garycarson.com
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