Re: what are the odds of....
- From: "Stephen Touset" <stephen@xxxxxxxxxx>
- Date: Fri, 09 Mar 2007 08:03:40 -0800
On Mar 9 2007 7:55 AM, xyious wrote:
making a straight flush with suited connectors by the river ?
The probability P(SF) of getting a straight flush with suited connectors
is the probability that you hit one of four combinations of cards by the
river: the under-straight-flush, the over-straight-flush, and the two
inbetween. Given that there are four ways of doing this out of 50-choose-5
(2,118,760) combinations of community cards
P(SF) = 4 / 2,118,760
= 0.000001888
= 0.00018879%
= 1:5297
making a 2nd best straight flush with suited connectors by the river ?
This is a much more difficult situation, since the opponent could have,
for instance, T8c, while you hold 43c and the board is 9c7c5c6c. I'll be
happy to do it for the trivial case of you having the bottom end of the
board three-straight and he has the top end. Any edge cases' probabilities
will simply be added to the probability of this occurrence, so if someone
wants to pick up on this later and do the rest, they'll need this part
anyway.
Assuming a heads-up match and assuming 98 suited or less (since it's
impossible to lose with any higher straight flush hole cards), the
probability of having the 2nd best straight flush P(2SF) is the odds he
has the exact suited connectors 3 higher than you P(3SC), and hitting
exactly one possible set of community cards by the river P(SFR).
P(2SF) = P(3SC and SFR)
= P(3SC) * P(SFR | 3SC)
The probability he has the exact two cards to allow the three middle cards
to hit both of your straight flush P(3SC) is one out of 52-choose-2 hole
cards (1,326). In other words, there's only one possible set of hole cards
to give him the prerequisites needed for this outcome.
P(3SC) = 1 / 1,326
The probability of the five community cards showing the three suited cards
inbetween your two by the river, given that he already has those two hole
cards is also simple to find. There's one satisfying combination out of
48-choose-5 possible sets of community cards, so
P(SFR | 3SC) = 1 / Bin2(48,5)
= 1 / 1,712,304
Therefore,
P(2SF) = P(3SC) * P(SFR | 3SC)
= 1 / 1,326 * 1 / 1,712,304
= 0.000000004
= 1:250,000,000
Adding the probabilities of other combinations won't change this by more
than an order of magnitude or so, so I'd estimate about a one in
one-hundred million shot.
making 2nd best quads by the river ?
Too lengthy for me to work out right now; I'm at work. It's either you
have an underpair to his overpair and both make quads, or he has one over
to your pair and the board boats to give you both quads, or vice versa, or
he has two overs to your pair and the board boats to give you quads, or
vice versa.
Even figuring out the odds you've got an underpair will probably be kind
of tetchy, I think. Maybe later tonight I'll give it a whirl.
--
Stephen Touset <stephen@xxxxxxxxxx>
Full Tilt Poker: stouset
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