heehaww's great answer pertained to NLHE (which undoubtedly is what OP was looking for). I wanted to see how the answer changes if it is 9-player Omaha. Of course, we expect the probability of flopping set-over-set-over-set to be higher in Omaha than in NLHE for obvious reasons.
Below I give my derivation. Note that I expect there to be easier methods than what I used. My method was painstaking and, perhaps, fraught with mistakes, so I am not at all confident that the derived answer is correct. Others are encouraged to comment/confirm/refute.
Here is the recipe:
Step 1. Flop is unpaired
Step 2. Tally three specific players each flopping a set
Step 3. Calculate how many three-player hands are possible
Step 4. Consider that there are 9 players at the table
Step 5. Put it all together for a final result
Step 1. Flop is unpaired (see heehaww's post above). Wlog suppose flop is AKQ.
Step 2. Tally three specific players each flopping a set. Wlog suppose players are Frank, George, and Harry (FGH). We will break down tallies by case.
Case 1. Exactly 6 AKQ appear in FGH hands.
Of course, for each to flop a set the AKQ's must be [2,2,2] in the three hands.
Tally = C(3,2)*C(3,2)*C(3,2)*C(40,6)*C(6,2)*C(4,2)*C(2,2)
= 9,327,263,400
Case 2. Exactly 7 AKQ appear in FGH hands
Case 2A: [2,2,2 with one "extra" card]
Tally = C(3,2)*C(3,2)*C(3,2)*C(3,1)*C(2,1)*C(40,5)*C(5,1)* C(4,2)*C(2,2)
= 3,197,918,880
Case 2B: [3,2,2]
Tally = C(3,1)*C(3,3)*C(3,2)*C(3,2)*C(40,5)*C(5,1)*C(4,2)* C(2,2)
= 532,986,480
Subtotal for Case 2 = 3,730,905,360
Case 3: Exactly 8 AKQ appear in FGH hands
Case 3A: [2,2,2 with two extras]
Case 3A1: two extras are in two diff hands
Tally = C(3,2)*C(3,2)*C(3,2)*C(3,2)*3*C(40,4)*C(4,1)*C(3,1 )*C(2,2)
= 266,493,240
Case 3A2: two extra cards are in same hand
Tally = C(3,2)*C(3,2)*C(3,2)*C(3,2)*1*C(40,4)*C(4,2)*C(2,2 )
= 44,415,540
Case 3B: [3,2,2 with one extra]
Case 3B1: one extra card goes with a pair
Tally = C(3,1)*C(3,3)*C(3,2)*C(3,2)*C(2,1)*C(1,1)*C(40,4)* C(4,1)*C(3,1)*C(2,2)
= 59,220,720
Case 3B2: one extra card goes with trips
Tally = C(3,1)*C(3,3)*C(3,2)*(C3,2)*C(2,1)*C(1,1)*C(40,4)* C(4,2)*C(2,2)
= 29,610,360
Case 3C: [3,3,2]
Tally = C(3,2)*C(3,3)*C(3,3)*C(3,2)*C(40,4)*C(4,1)*C(3,1)* C(2,2)
= 9,870,120
Subtotal for Case 3 = 409,609,980
Case 4: Exactly 9 AKQ appear in FGH hands
Case 4A: [2,2,2 with three extras]
Case 4A1: three extra cards are in three diff hands
Tally = C(3,2)*C(3,2)*C(3,2)*2*C(40,3)*C(3,1)*C(2,1)*C(1,1 )
= 3,201,120
Case 4A2: three extra cards are in two diff hands
Tally = C(3,2)*C(3,2)*C(3,2)*C(3,1)*C(2,1)*1*C(40,3)*C(3,2 )*C(1,1)
= 4,801,680
Case 4B: [3,2,2 with two extras]
Case 4B1: two extra cards are in two diff hands, both pairs
Tally = C(3,1)*C(3,3)*C(3,2)*C(3,2)*1*C(40,3)*C(3,1)*C(2,1 )*C(1,1)
= 1,600,560
Case 4B2: two extra cards are in two diff hands, one trips and one pair
Tally = C(3,1)*C(3,3)*C(3,2)*C(3,2)*C(2,1)*C(1,1)*C(40,3)* C(3,1)*C(2,2)
= 1,600,560
Case 4B3: two extra cards are in one hand
Tally = 0 (impossible)
Case 4C: [3,3,2 with one extra]
Tally = C(3,2)*C(3,3)*C(3,3)*C(3,2)*C(2,1)*C(40,3)*C(3,1)* C(2,2)
= 533,520
Case 4D: [3,3,3]
Tally = C(3,3)*C(3,3)*C(3,3)*C(40,3)*C(3,1)*C(2,1)*C(1,1)
= 59,280
Subtotal for Case 4 = 11,796,720
GRAND TOTAL = 13,479,575,460
Step 3: The above Grand Total is the numerator of some fraction. We now derive the denominator, the total number of 3-player Omaha deals (where, as throughout this post, the order of the three players is irrelevant).
Of course, in this derivation (as in heehaww's above), we assume that the flop is dealt first. Thus the players' hole cards are dealt from a 49-card deck-stub.
= C(49,12)*C(12,4)*C(8,4)*C(4,4)/3!
= 532,823,068,677,900
Step 4: There are 9 players at the table. Above we found the tally for three players to make set-over-set-over-set on the flop. Of course, the other six players do not flop a set. So our tally applies to each possible triple of 3-players out of the total of 9 players.
There are clearly C(9,3) such triples (as in heehaww's post above).
= 84
Step 5. Putting it all together.
Taking everything into account, I find that the probability of three players flopping set-over-set-over-set in a 9-player Omaha game is:
= [(48*44)/(51*50)] * 13,479,575,460 * 84 / 532,823,068,677,900
= 0.001760055
which is approx once in every 568 deals.
To be perfectly honest, when I first derived this probability I was very skeptical since it seems too large. That is, I thought that flopping set-over-set-over-set in 9-player Omaha would be less likely than once in every 568 deals (all deals see a flop). I have double-checked my derivation, which itself was not easy, and I have not identified any mistakes. Of course, this is not very convincing for obvious reasons.
Comments welcome.
Last edited by whosnext; 05-26-2020 at 07:13 PM.