Quote:
Originally Posted by IbelieveinChipKelly
odds of flopping a set 3 straight hands in omaha h/l and not winning any of them?
Sounds like a rough session
Chip - I would love to answer precisely, but this question is not specific enough and it would heavily depend on opponent behavior (folding, betting, etc.).
However, qualitatively - the odds are probably not that tremendously long... losing with the set is relatively easy in this case, it would be the flopping three in a row which would be the longer odds.
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To flop 3 sets in a row in Omaha (if you played just 3 hands for the session & assuming saw flop with every hand)
52*51*50*49/4! hand combos = 270,725
Double paired hands = XXYY
13 ranks XX * 12 ranks YY / 2 orders * 4*3/2 X combos * 4*3/2 Y combos =
78*6*6 =
2,808 double paired hands
To hit a set with a double pair hand:
48*47*46/3! flops = 17,296 flops
Flops that contain one X and no Y(and no pair for flopped FH)
= 2 Xs left * (44 * 40 / 2 orders) = 1,760
(same number for "include one Y and no X")
= 1,760
Flops that contain X and Y (and no quads)
= 2 * 2 * 44 = 176
Odds to flop set with double paired hand
(note: this may also include boards where you also flop a flush)
= (1760 + 1760 + 176) / 17,296 = 21.37%
Odds to get a double paired hand and to flop a set =
2,808 / 270,725 * 3,696 / 17,296 = 0.2216% ~= 1 in 451
Single paired hands = XXYZ
13 ranks for XX * (12 ranks Y* 11 ranksZ / 2 orders * 4 suits Y * 4 suits Z) * 4*3/2 combos XX =
13*6*48*44/2 =
82,368 single paired hands
To hit a set with a single pair hand:
48*47*46/3! flops = 17,296 flops
(without pairing the board)
This is a little tricky because of your YZ cards in XXYZ hand:
Flop can be XYZ, XYA, XZA, XAB
XYZ = 2 * 3 * 3 = 18
XYA or XZA = 2 * 6 * 40 = 480
XAB = 2 * 40 * 36 / 2 = 1440
(1440 + 480 + 18) / 17,296 = 11.2% ~= 1 in 8.92
Odds to get a single paired hand and to flop a set =
82,368 / 270,725 * 1938/17,926 = 3.41% ~= 1 in 29.3
So overall odds of being dealt a paired hand in Omaha followed by flopping a set in given hand =
(3.41% + 0.2216%)^3 = 1 in 20,893
.. and to address "losing with a set in Omaha" - I almost feel it would be more amazing to win with a flopped set in an Omaha H/L game
We could get an approximation by making this assumption that any improvement on the hand (a FH or quads) will win you the pot, while any non-improvement will result in a loss. While clearly not perfect (sometimes the set wins, sometimes the FH loses, etc)- I will call it good enough...
(I am also going to pretend your other two cards do not exist and ignore the double paired situations..etc.. at this point we are approximating very inaccurately anyways)
You must miss 7 turn "outs" and then 10 river "outs" to lose =
40/47 * 36/46 = 66.6% unimproved
So you lose 66.6% of the time and win 33.4% of the time in my simplified scenario. Compound the odds above (about flopping set) with the odds to lose that hand @ 66.6% =
(66.6%) ^ 3 * 1/20,893 = 1 in 70,726
wow - that sucks Chip.
(last note: again this is as if you only played 3 hands and the result was as noted... if you played 150 hands that session you would have ~150X the odds of this happening, so you get into the 1 in several hundred range)