
jblaze: a poker ?, but at least dealing with statistics
I'm more interested in the 'how to figure it out' than the actual answer... but if there are 3 unique cards on a flop (as in no pairs on the flop), and it's an Omaha game, where the player has 4 cards in their hand, what are the odds he has 2 pair? Thanks

Dog Hand: Poker Answer
jblaze,
I don't play poker myself, but as I understand the game, the player must use exactly two of his four hole cards in Omaha. Thus, with three unique (unpaired) cards on the flop, the only way the player can have two pair is by having two of his hole cards match the ranks of two of the flop cards. (Note that, if you allow a pair on the flop, then the player could also have two pair by having a pair in his hole cards.)
To make this easier to understand, let's assume the flop is 5, 6, 7. Now if the player has a 5 and a 6 among his hole cards, he'll have two pair. The odds of him having 5, 6, X, X can be found as follows, where X is any card that does not match a flop card (that is, any non(5, 6, 7) card):
First card 5 odds = 3/49, "3" since the deck has three 5's left, and "49" because we know three of the original 52 cards.
Second card 6 odds = 3/48, similar reasoning, except that now we have only 48 unknown cards.
Third card X odds = 40/47, since the deck has 40 non(5, 6, 7) cards (count 'em: 4 each of A's, 2's, 3's, 4's, 8's, 9's, 10's, J's, Q's and K's) of the 47 remaining cards.
Fourth card X odds = 39/46, since the deck now has only 39 non(5, 6, 7) cards in the remaining 46 cards.
Thus, odds of 5, 6, X, X (in that order) are the product of these odds, or
Prob(order) = (3/49)*(3/48)*(40/47)*(39/46) = 0.002761...
Now the four hole cards need not be in the order 5, 6, X, X: 6, X, 5, X would also work just as well. How many ways can we arrange 5, 6, X, X: or, in other words, how many permutations are there?
Ordinarily, if we have 4 objects, say A, B, C, D, we have 4! = 4*3*2*1 = 24 permutations. However, with 5, 6, X, X, two of the objects are the same, so in our case we have only 24/2 = 12 permutations.
Thus, without regard to the order of the hole cards, the odds of pairing the 5 and 6 are
Prob(any order) = 12*Prob(order) = 0.03313...
By similar reasoning, the Prob(any order) of pairing the 5 and 7 is also 0.03313..., and so is the prob of pairing the 6 and 7.
We also have one other prob to consider: what is the prob that the player has 5, 6, 7, X? This is
Prob(5, 6, 7) = 24*(3/49)*(3/48)*(3/47)*(40/46) = 0.005097..., here we have 24 permutations, since all four cards are distinct.
Thus, the overall prob of having two pair after a "unique" flop is 3*Prob(any order) + Prob(5,6,7), or
Prob(2 pair) = 3*(0.03313...) + 0.005097... = 0.1044..., or just over 10% of the time.
Note, though, that this does NOT account for the chance of the player having a BETTER hand, except that we ruled out trips by disallowing a holecard pair that matches a flop card. However, the player could still have a straight, a flush, or a straight flush, either with or without 2 pair. For example, if the flop is 5C, 6C, 7C, and the player's hole cards are 2C, 3C, 8D, JD, then he has a flush but not 2 pair, so that is not included in the prob we calculated above. On the other hand, for the same flop if the hole cards are 2C, 3C, 5H, 6H, then he still has a flush but he also has 2 pair, so that IS included in the prob calculation.
If you want the prob of 2 pair AND NOTHING BETTER, then you have to specify the flop. For example, if the flop contains more than one suit, then a flush is impossible. Similarly, if the ranks of the highest and lowest cards in the flop differ by more than 5, then a straight is impossible. That is, a flop of 5, 7, 9 CAN give a straight, while 5, 6, J cannot.
Hope this helps!
Dog Hand

jblaze: Thanks, I will try to follow up
Thanks for your answer. I was able to follow it well. My next question is what's the probability someone has a flush draw on a 2 tone board? I will try to answer this one myself using what I learned from your response and repost when I have some time.
The overall question is in determining a range, based on a probabilities alone, what is the player more likely to have...
> jblaze,
> I don't play poker myself, but as I understand the
> game, the player must use exactly two of his four hole
> cards in Omaha. Thus, with three unique (unpaired)
> cards on the flop, the only way the player can have
> two pair is by having two of his hole cards match the
> ranks of two of the flop cards. (Note that, if you
> allow a pair on the flop, then the player could also
> have two pair by having a pair in his hole cards.)
> To make this easier to understand, let's assume the
> flop is 5, 6, 7. Now if the player has a 5 and a 6
> among his hole cards, he'll have two pair. The odds of
> him having 5, 6, X, X can be found as follows, where X
> is any card that does not match a flop card (that is,
> any non(5, 6, 7) card):
> First card 5 odds = 3/49, "3" since the deck
> has three 5's left, and "49" because we know
> three of the original 52 cards.
> Second card 6 odds = 3/48, similar reasoning, except
> that now we have only 48 unknown cards.
> Third card X odds = 40/47, since the deck has 40
> non(5, 6, 7) cards (count 'em: 4 each of A's, 2's,
> 3's, 4's, 8's, 9's, 10's, J's, Q's and K's) of the 47
> remaining cards.
> Fourth card X odds = 39/46, since the deck now has
> only 39 non(5, 6, 7) cards in the remaining 46 cards.
> Thus, odds of 5, 6, X, X (in that order) are the
> product of these odds, or
> Prob(order) = (3/49)*(3/48)*(40/47)*(39/46) =
> 0.002761...
> Now the four hole cards need not be in the order 5, 6,
> X, X: 6, X, 5, X would also work just as well. How
> many ways can we arrange 5, 6, X, X: or, in other
> words, how many permutations are there?
> Ordinarily, if we have 4 objects, say A, B, C, D, we
> have 4! = 4*3*2*1 = 24 permutations. However, with 5,
> 6, X, X, two of the objects are the same, so in our
> case we have only 24/2 = 12 permutations.
> Thus, without regard to the order of the hole cards,
> the odds of pairing the 5 and 6 are
> Prob(any order) = 12*Prob(order) = 0.03313...
> By similar reasoning, the Prob(any order) of pairing
> the 5 and 7 is also 0.03313..., and so is the prob of
> pairing the 6 and 7.
> We also have one other prob to consider: what is the
> prob that the player has 5, 6, 7, X? This is
> Prob(5, 6, 7) = 24*(3/49)*(3/48)*(3/47)*(40/46) =
> 0.005097..., here we have 24 permutations, since all
> four cards are distinct.
> Thus, the overall prob of having two pair after a
> "unique" flop is 3*Prob(any order) +
> Prob(5,6,7), or
> Prob(2 pair) = 3*(0.03313...) + 0.005097... =
> 0.1044..., or just over 10% of the time.
> Note, though, that this does NOT account for the
> chance of the player having a BETTER hand, except that
> we ruled out trips by disallowing a holecard pair
> that matches a flop card. However, the player could
> still have a straight, a flush, or a straight flush,
> either with or without 2 pair. For example, if the
> flop is 5C, 6C, 7C, and the player's hole cards are
> 2C, 3C, 8D, JD, then he has a flush but not 2 pair, so
> that is not included in the prob we calculated above.
> On the other hand, for the same flop if the hole cards
> are 2C, 3C, 5H, 6H, then he still has a flush but he
> also has 2 pair, so that IS included in the prob
> calculation.
> If you want the prob of 2 pair AND NOTHING BETTER,
> then you have to specify the flop. For example, if the
> flop contains more than one suit, then a flush is
> impossible. Similarly, if the ranks of the highest and
> lowest cards in the flop differ by more than 5, then a
> straight is impossible. That is, a flop of 5, 7, 9 CAN
> give a straight, while 5, 6, J cannot.
> Hope this helps!
> Dog Hand

jblaze: Flush draw, is this correct?
OK
Given flop CCX (club, club, nonclub)
Prob of flush draw with 2 of your cards being clubs (CCXX):
[11/49 * 10/48 * 38/47 * 37/46] * [ 4 choose 2] = .182
Prob of flush draw with 3 of your cards being clubs (CCCX):
[11/49 * 10/48 * 9/47 * 38/46] * [4 choose 3] = .03
Prob of flush draw with 4 of your cards being clubs (CCCC):
[11/49 * 10/48 * 9/47 * 8/46] * [4 choose 4] = ~0 (not rounded in final addition)
adding all scenarios, prob of flush draw give 2 tone board = 21.36%
Is this right? Sounds kind of high...

Dog Hand: Looks right to me... good job!
jblaze,
One caveat though... we've been assuming that none of the players folds before the flop. That would have some effect on the final percentages, of course.
Dog Hand
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