omaha odds help please
May 18th, 2015 at 6:47:33 PM
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Please, I would like to know the odds of the following:
Game is Omaha.
I hold a pair (out of the 4 cards), the other guy holds a higher pair, and we both flopped the set.
I recently have that occurred to me twice in a short period of time. Each time I was all in and lost both.
Thanks in advance
Game is Omaha.
I hold a pair (out of the 4 cards), the other guy holds a higher pair, and we both flopped the set.
I recently have that occurred to me twice in a short period of time. Each time I was all in and lost both.
Thanks in advance
May 18th, 2015 at 6:52:26 PM
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If you're getting all in with a set in omaha, expecting to win, you have some problems with your game.
May 18th, 2015 at 6:52:49 PM
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I can't help you with odds but my first thought is they are not that high. I have seen this hundreds of times playing Omaha and o8.
Expect the worst and you will never be disappointed.
I AM NOT PART OF GWAE RADIO SHOW
May 18th, 2015 at 7:05:57 PM
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Quote: MaxPenIf you're getting all in with a set in omaha, expecting to win, you have some problems with your game.
yes, if draw to the river, i figured is around 50/50 I would got beaten by a flush or straight.
Thats why I push all in hoping that they would not draw to the river.
May 18th, 2015 at 7:56:29 PM
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Quote: MaxPenIf you're getting all in with a set in omaha, expecting to win, you have some problems with your game.
+100
I wouldn't even call a shove with low trips. Neither should you. Suck it up and fold a few.
Some people need to reimagine their thinking.
May 18th, 2015 at 8:03:18 PM
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I totally understand tilt complaining but you must not play much Omaha if this is shocking to you. Back when I was playing 5000 hands/week of NLHE online this would happen a couple of times each hour, and that was with only two cards per hand.
May 18th, 2015 at 8:10:32 PM
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Quote: 98Clubs+100
I wouldn't even call a shove with low trips. Neither should you. Suck it up and fold a few.
but using a calculator it shows that if you have the top set after the flop you have over 50% chance of winning against a suited connector for both straight and flush draw.
If it is just a flush or straight draw then its almost 2/3.
May 18th, 2015 at 8:15:07 PM
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But you didn't have top set...
May 18th, 2015 at 8:19:55 PM
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Quote: Elrohir44But you didn't have top set...
true. LOL
thats why i want to know the odds of that happening.
Board is dry, not likely flush or straight draw. Only thing that could beat me is another set. How likely is that?
May 18th, 2015 at 8:31:04 PM
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In hold'em, approx. 1/13 hands is a pocket pair. If villain has a PP, and you have bottom set, I believe there would be a better than 1/6 chance he made one of the higher sets. This is assuming you have 22, and the chance rises significantly the higher your pair is. That comes out to 1/78 chance that if you flopped bottom set (deuces), villain has a higher set. This is just quick and very dirty math I'm doing in my head without thinking too hard about it so forgive me if I'm off. In Omaha the chance is much greater though since you each have four cards. This is not a rare occurrence at all.
My advice is to worry less about the odds of frustrating occurrences happening and more about playing a solid game. Going on tilt easily is very -ev.
My advice is to worry less about the odds of frustrating occurrences happening and more about playing a solid game. Going on tilt easily is very -ev.
May 18th, 2015 at 8:34:21 PM
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Odds of a getting pocket pair in hold em is 1 in 17
DUHHIIIIIIIII HEARD THAT!
May 18th, 2015 at 8:40:18 PM
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Oops you're right. It's been a while since I've tackled poker math problems.
May 18th, 2015 at 8:47:40 PM
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Quote: Elrohir44Oops you're right. It's been a while since I've tackled poker math problems.
Easy mistake :-)
DUHHIIIIIIIII HEARD THAT!
May 18th, 2015 at 9:37:55 PM
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Quote: Elrohir44In hold'em, approx. 1/13 hands is a pocket pair. If villain has a PP, and you have bottom set, I believe there would be a better than 1/6 chance he made one of the higher sets. This is assuming you have 22, and the chance rises significantly the higher your pair is. That comes out to 1/78 chance that if you flopped bottom set (deuces), villain has a higher set. This is just quick and very dirty math I'm doing in my head without thinking too hard about it so forgive me if I'm off. In Omaha the chance is much greater though since you each have four cards. This is not a rare occurrence at all.
My advice is to worry less about the odds of frustrating occurrences happening and more about playing a solid game. Going on tilt easily is very -ev.
Thanks
But I am interested in flopping both sets (since he won't shove it if he didn't flop his and i won't shove it if I didn't flop mine)
So following your line of thoughts (with hold'em):
chance of pocket pair is 1/17 (as revised in later posts);
I figured the chance of him making his set in the other 2 cards (I used up 1 quota to make my set) is around 8% (1/12.5)
So his chance of flopping higher set than mine is around 1/200
and when my pair is anything higher than deuces, chance reduces exponentially, i will arbitrary assigned a factor of 3, so on average i will say around 1 in 600
I really had no idea how to convert this to omaha. With a wild guess I will just use a factor of 2C4 = 6 so I will say the chance of me and the villain both flopping the set with his set beating mine is 1/100 ??
May 18th, 2015 at 10:21:07 PM
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Approximately 1/3 of the time your 4 card hand will contain a pair.
You would need to know the exact number of players in the game to get the answer you seek.
I think your 1/100 is close probably closer to 1/80.
However, that is not the concern. It quite often takes more than a set to win in Omaha. Omaha is about the draw. You sound like a holdem player trying to switch to omaha. Are you going to shove when you flop middle pair or bottom pair on the flop in holdem? Would you call an all in with that? It's the same thing.
You have to get used to folding bottom and middle set in Omaha. If not you will continue to have a serious leak.
This is all my opinion of course.
You would need to know the exact number of players in the game to get the answer you seek.
I think your 1/100 is close probably closer to 1/80.
However, that is not the concern. It quite often takes more than a set to win in Omaha. Omaha is about the draw. You sound like a holdem player trying to switch to omaha. Are you going to shove when you flop middle pair or bottom pair on the flop in holdem? Would you call an all in with that? It's the same thing.
You have to get used to folding bottom and middle set in Omaha. If not you will continue to have a serious leak.
This is all my opinion of course.
May 18th, 2015 at 10:33:05 PM
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Quote: MaxPen
You have to get used to folding bottom and middle set in Omaha. If not you will continue to have a serious leak.
This is all my opinion of course.
Thanks and sure will remember that
May 18th, 2015 at 11:19:36 PM
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Quote: MaxPenApproximately 1/3 of the time your 4 card hand will contain a pair.
You would need to know the exact number of players in the game to get the answer you seek.
I think your 1/100 is close probably closer to 1/80.
However, that is not the concern. It quite often takes more than a set to win in Omaha. Omaha is about the draw. You sound like a holdem player trying to switch to omaha. Are you going to shove when you flop middle pair or bottom pair on the flop in holdem? Would you call an all in with that? It's the same thing.
You have to get used to folding bottom and middle set in Omaha. If not you will continue to have a serious leak.
This is all my opinion of course.
But a set is a full house draw, no?
Give me top set all the time, and I will love life as worst case scenario, I'm a mild dog and usually a favorite on non-suited/non-connected boards.
Middle set...meh
Bottom set...beware!
May 18th, 2015 at 11:49:14 PM
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Well a set of queens on a QJ5 flop should be played agressively. OP is the guy calling with JJ or 55, wondering what the hell happened when he gets felted.
May 19th, 2015 at 1:55:26 AM
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Quote: MaxPenWell a set of queens on a QJ5 flop should be played agressively. OP is the guy calling with JJ or 55, wondering what the hell happened when he gets felted.
guess what, he goes to the wizard of odds forum to ask.
May 19th, 2015 at 8:13:21 AM
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Quote: andysifPlease, I would like to know the odds of the following:
Game is Omaha.
I hold a pair (out of the 4 cards), the other guy holds a higher pair, and we both flopped the set.
About 1 time in 1984
("Doesn't that include times when either or both of you have a full house?" Not from the flop, as all three cards are different, so in order to have a full house, you would need to use three cards from your hand, which is not allowed in Omaha.)
Of course, this assumes heads-up; the more players at the table, the more likely it is that somebody besides you is dealt a pair.
As usual, (A)C(B) = ACB = the number of combinations of A things taken B at a time.
There are (52)C(4) = 270,725 possible Player 1 hands.
If the hand is a pair (and not two pair or 3/4OAK), there are 13 possible values (e.g. Aces) for the pair, 6 possible sets of suits (e.g. Ah & As) for the pair, 48 possible cards for the first single, and 44 possible cards for the second single, but divide that by 2 as, e.g., it counts both (2s, 3h) and (3h, 2s); there are 82,368 4-card hands that include one pair.
Ignore suits, and assume the hand is AA23. (Also ignore the fact that your pair of aces is going to be higher than the other player's pair. We'll take that into account later.)
The possible Player 2 hands are:
4456 (none of your cards match any of his)
There are 10 possible values for the rank of his pair (4-K), and 6 pairs of suits for each rank
For each of the 60 pairs, there are 36 possible cards for the first single, and 32 for the second, but again divide by 2 as you are counting, e.g., both (5s, 6h) and (6h, 5s).
The flop contains one of the 2 remaining cards to give you a set, and one of the 2 remaining cards to give him a set.
The third card can be:
(a) one of the 28 cards with a rank not in either of your hands;
(b) one of the 12 cards (three 2s, three 3s, three 5s, and three 6s) with a rank in one of your hands.
There are 10 x 6 x (36 x 32 / 2) x 2 x 2 x (28 + 12) = 5,529,600 different hand/flop combinations.
4425 (one of your singles matches one of his):
10 x 6 x 2 (the value of the matching single - in this case, 2 or 3) x 3 (the number of remaining cards of that value) x 36 (the number of cards that can be the fourth card in this hand) x 2 x 2 x (8 x 4 (cards not matching either hand) + 1 x 2 (cards matching the single in both hands) + 2 x 3 (cards matching the single in one hand)) = 2,073,600
4423 (both of your singles match both of his):
10 x 6 x 3 (remaining cards of the lower single) x 3 (remaining cards of the higher single) x 2 x 2 x (9 x 4 + 2 x 2) = 86,400
44A5 (one of his singles matches your pair; the other is not in your hand)
10 x 6 x 2 (remaining cards of your pair) x 36 (cards not in your hand or his pair) x 2 x 1 (there's only one card left in the deck that can give you a set) x (8 x 4 + 3 x 3) = 354,240
44A2 (one of his singles matches your pair; the other matches one of your singles)
10 x 6 x 2 (remaining cards of your pair) x 6 (remaining cards in either of your singles) x 2 x 1 x (9 x 4 + 1 x 2 + 1 x 3) = 59,040
2245 (his pair matches one of your singles; the singles are not in your hand)
2 (possible values for the pair - in this case, 2 and 3) x 3 (number of pairs of that value still in the deck) x (40 x 36 / 2) (number of sets of 2 different cards that don't match any your hand or his pair) x 2 (number of cards that make your set) x 1 (number that make his) x (8 x 4 + 3 x 3) = 354,240
2234 (his pair matches one of your singles, and one of his singles matches the other one)
2 x 3 x 3 (number of cards that can match the other single) x 40 (number that don't match anything) x 2 x 1 x (9 x 4 + 1 x 3 + 1 x 2) = 59,040
22A4 (his pair matches one of your singles; one of his singles matches your pair, and the other does not match anything)
2 x 3 x 2 (number of cards that match your pair) x 40 (number of cards that don't match anything) x 1 x 1 (there is only one card of each pair still in the deck) x (9 x 4 + 2 x 3) = 20,160
22A3 (his pair matches one of your singles; one of his singles matches your pair, and the other matches your other single)
2 x 3 x 2 x 3 (number of cards that match your other single) x (10 x 4 + 1 x 2) = 1512
The total number of second hand/flop combinations where both sets are made = 8,537,832
There are (48)C(4) x (44)C(3) = 2,577,017,520 second hand/flop combinations
Now, remember what I said about your pair always being higher? Every deal can be divided into pairs of two, where the flops are the same but the hands are switched; in each case, in one hand, your pair is higher, and in the other, his is. We are only interested in hands where his is higher, so divide the total by two.
The probability of this happening = 1/2 x 82,368 / 270,725 x 8,537,832 / 2,577,017,520 = 1 / 1984.1276
May 19th, 2015 at 8:35:40 AM
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Here's a tough Omaha beat. And fittingly against Phil Hellmuth.
https://youtu.be/KHqLGXu63jU
https://youtu.be/KHqLGXu63jU
DUHHIIIIIIIII HEARD THAT!
May 19th, 2015 at 7:32:21 PM
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If you both hold a pair and don't block each others cards then it's 1.2%
PLAYER_1 AA47
PLAYER_2 JJ23
800000 trials (randomized)
How often do(es)( PLAYER_1 flop hand category is set AND PLAYER_2 flop hand category is set)
1.2085%
PLAYER_1 AA47
PLAYER_2 JJ23
800000 trials (randomized)
How often do(es)( PLAYER_1 flop hand category is set AND PLAYER_2 flop hand category is set)
1.2085%
May 19th, 2015 at 7:36:57 PM
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Quote: ThatDonGuyAbout 1 time in 1984
("Doesn't that include times when either or both of you have a full house?" Not from the flop, as all three cards are different, so in order to have a full house, you would need to use three cards from your hand, which is not allowed in Omaha.)
Of course, this assumes heads-up; the more players at the table, the more likely it is that somebody besides you is dealt a pair.
As usual, (A)C(B) = ACB = the number of combinations of A things taken B at a time.
There are (52)C(4) = 270,725 possible Player 1 hands.
If the hand is a pair (and not two pair or 3/4OAK), there are 13 possible values (e.g. Aces) for the pair, 6 possible sets of suits (e.g. Ah & As) for the pair, 48 possible cards for the first single, and 44 possible cards for the second single, but divide that by 2 as, e.g., it counts both (2s, 3h) and (3h, 2s); there are 82,368 4-card hands that include one pair.
Ignore suits, and assume the hand is AA23. (Also ignore the fact that your pair of aces is going to be higher than the other player's pair. We'll take that into account later.)
The possible Player 2 hands are:
4456 (none of your cards match any of his)
There are 10 possible values for the rank of his pair (4-K), and 6 pairs of suits for each rank
For each of the 60 pairs, there are 36 possible cards for the first single, and 32 for the second, but again divide by 2 as you are counting, e.g., both (5s, 6h) and (6h, 5s).
The flop contains one of the 2 remaining cards to give you a set, and one of the 2 remaining cards to give him a set.
The third card can be:
(a) one of the 28 cards with a rank not in either of your hands;
(b) one of the 12 cards (three 2s, three 3s, three 5s, and three 6s) with a rank in one of your hands.
There are 10 x 6 x (36 x 32 / 2) x 2 x 2 x (28 + 12) = 5,529,600 different hand/flop combinations.
4425 (one of your singles matches one of his):
10 x 6 x 2 (the value of the matching single - in this case, 2 or 3) x 3 (the number of remaining cards of that value) x 36 (the number of cards that can be the fourth card in this hand) x 2 x 2 x (8 x 4 (cards not matching either hand) + 1 x 2 (cards matching the single in both hands) + 2 x 3 (cards matching the single in one hand)) = 2,073,600
4423 (both of your singles match both of his):
10 x 6 x 3 (remaining cards of the lower single) x 3 (remaining cards of the higher single) x 2 x 2 x (9 x 4 + 2 x 2) = 86,400
44A5 (one of his singles matches your pair; the other is not in your hand)
10 x 6 x 2 (remaining cards of your pair) x 36 (cards not in your hand or his pair) x 2 x 1 (there's only one card left in the deck that can give you a set) x (8 x 4 + 3 x 3) = 354,240
44A2 (one of his singles matches your pair; the other matches one of your singles)
10 x 6 x 2 (remaining cards of your pair) x 6 (remaining cards in either of your singles) x 2 x 1 x (9 x 4 + 1 x 2 + 1 x 3) = 59,040
2245 (his pair matches one of your singles; the singles are not in your hand)
2 (possible values for the pair - in this case, 2 and 3) x 3 (number of pairs of that value still in the deck) x (40 x 36 / 2) (number of sets of 2 different cards that don't match any your hand or his pair) x 2 (number of cards that make your set) x 1 (number that make his) x (8 x 4 + 3 x 3) = 354,240
2234 (his pair matches one of your singles, and one of his singles matches the other one)
2 x 3 x 3 (number of cards that can match the other single) x 40 (number that don't match anything) x 2 x 1 x (9 x 4 + 1 x 3 + 1 x 2) = 59,040
22A4 (his pair matches one of your singles; one of his singles matches your pair, and the other does not match anything)
2 x 3 x 2 (number of cards that match your pair) x 40 (number of cards that don't match anything) x 1 x 1 (there is only one card of each pair still in the deck) x (9 x 4 + 2 x 3) = 20,160
22A3 (his pair matches one of your singles; one of his singles matches your pair, and the other matches your other single)
2 x 3 x 2 x 3 (number of cards that match your other single) x (10 x 4 + 1 x 2) = 1512
The total number of second hand/flop combinations where both sets are made = 8,537,832
There are (48)C(4) x (44)C(3) = 2,577,017,520 second hand/flop combinations
Now, remember what I said about your pair always being higher? Every deal can be divided into pairs of two, where the flops are the same but the hands are switched; in each case, in one hand, your pair is higher, and in the other, his is. We are only interested in hands where his is higher, so divide the total by two.
The probability of this happening = 1/2 x 82,368 / 270,725 x 8,537,832 / 2,577,017,520 = 1 / 1984.1276
thank you very much. thats one hell of a job. it will take me some time to go over it.
May 19th, 2015 at 8:30:09 PM
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Quote: BTLWIIf you both hold a pair and don't block each others cards then it's 1.2%
PLAYER_1 AA47
PLAYER_2 JJ23
800000 trials (randomized)
How often do(es)( PLAYER_1 flop hand category is set AND PLAYER_2 flop hand category is set)
1.2085%
thanks. is this some program that you write yourself or is it available online
May 22nd, 2015 at 11:31:27 AM
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It's from Odds Oracle which costs about $90 from www.propokertools.com That website has a Simulator tab which allows you to run hand vs. hand matchups for all forms of poker but to do english like queries you need to download Odds Oracle. I think it has a 7-day free trial. Pretty high learning curve for all of it's capabilities but I've had it for 2 years now and am fairly proficient.
May 28th, 2015 at 8:32:42 AM
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Quote: andysiftrue. LOL
thats why i want to know the odds of that happening.
Board is dry, not likely flush or straight draw. Only thing that could beat me is another set. How likely is that?
No math required... When I've taught a couple different people how to play Omaha, I give them the golden rule: If it's possible, someone probably has it.
Playing it correctly means you've already won.
May 28th, 2015 at 9:29:51 AM
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Quote: andysifthanks. is this some program that you write yourself or is it available online
Actually, you don't need to simulate it under those conditions; it can be calculated.
BTLWI assumes that both players have different pairs, none of the cards in either player's hand are in the other's, and is calculating the probability that the flop will not have two cards from either pair but the third card can be anything, including a card that pairs another card in either player's hand. (My number (1 / 1984) is the probability without already knowing that both players have pairs, and taking the possibility of "overlap" of the hole cards into account.)
In this case, of the 44 cards still in the deck, 2 are Aces, 2 are Jacks, and 40 are others, so there are 2 x 2 x 40 = 160 possible flops that make both sets, out of a total of (44)C(3) = 13,244 flops, so the probability = 160 / 13,244 = about 1.2081%
May 28th, 2015 at 9:34:36 AM
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Quote: RomesNo math required... When I've taught a couple different people how to play Omaha, I give them the golden rule: If it's possible, someone probably has it.
That is so, so true, Romes. I kept getting beat up on Omaha 8 (online) until I realized that my 4OAK's were always going to get rivered by a SF, and the FH I folded would have beaten the 2pair that won, but the FH I shoved on was going to lose. Etc. ad nauseum. Frustrating game. I still like it, but geez.... and if you didn't have a perfect low, it was deadly to stay. Very counter-intuitive, at least for me.
If the House lost every hand, they wouldn't deal the game.
June 9th, 2015 at 8:15:42 AM
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Quote: beachbumbabsThat is so, so true, Romes. I kept getting beat up on Omaha 8 (online) until I realized that my 4OAK's were always going to get rivered by a SF, and the FH I folded would have beaten the 2pair that won, but the FH I shoved on was going to lose. Etc. ad nauseum. Frustrating game. I still like it, but geez.... and if you didn't have a perfect low, it was deadly to stay. Very counter-intuitive, at least for me.
Well, recognizing the nuts is one thing (and the easy thing), but the other big part of the game is recognizing what the other players have (all the same as hold 'em). Bluffing, in Omaha, is an art in my opinion. That's also what sets you up to get paid off when you do have the nuts. In hold 'em, you can just decide "I'm going to bluff this hand and bet every street" and have a decent shot of someone not making a strong enough hand to call. However, in Omaha, you'll run in to the made hand a lot more often so you can't just blast away. I like that and think it's more intriguing/thoughtful. I played far too much Omaha for far too long =P. That's probably my highest edge game of anything I've ever played in my life, I just pealed away from it when I kept getting 2 outed after being all in on the turn. I kept thinking "no one's THAT unlucky" but the cards will teach you quickly.
Playing it correctly means you've already won.
