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 05-25-2020, 01:00 PM #1 Wehitityesssss journeyman   Join Date: Aug 2018 Posts: 214 Odds of set over set over set Im sure this has been asked, but trying to work out the odds of at a 9 handed table the odds of 3 people flopping set over set over set? Anyone?
 05-25-2020, 01:50 PM #2 heehaww Pooh-Bah     Join Date: Aug 2011 Location: Tacooos!!!! Posts: 4,851 Re: Odds of set over set over set Some problems, like this one, are easier when calculated in reverse order: flop then preflop. P(unpaired flop) = 48*44/(51*50) 9-handed P(set>set>set | unpaired flop) = C(9,3)*3³/(5*3) / C(49,6) There are C(9,3) mutually exclusive ways to pick the 3 players. Each set has 3 combos. There are 5*3 ways to group their hole cards, only one of which is xx|yy|zz. It's 5*3 because there are 5 possible partners for the "first" card, then 3 possible partners for the next. Multiplying those two probabilities: 1 in 111666
 05-26-2020, 06:54 PM #3 whosnext Carpal \'Tunnel     Join Date: Mar 2009 Location: California Posts: 6,253 Re: Odds of set over set over set 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.
 05-26-2020, 08:47 PM #4 heehaww Pooh-Bah     Join Date: Aug 2011 Location: Tacooos!!!! Posts: 4,851 Re: Odds of set over set over set I gave Omaha a try. My answer is a little different (1 in 533 deals), but I'm not supremely confident either, nor have I looked at whosnext's calculations yet. Here are mine, subject to change. Inclusion-exclusion seems to be my hammer (passed on to me by BruceZ of course), so naturally I saw another nail. Given an unpaired flop, each player can have exactly the pocket set pair, or pocket trips of a flop rank. First I'll ignore all but the 6 essential cards. That will triple-count the cases of one player having trips, so I'll double-subtract those, ignoring all but 7 cards. Then I'll have to 4x add back in the cases with two trips, then 8x subtract the cases with 3 trips. For pair|pair|pair the probability of a correct card grouping is still 1/(5*3). For pair|pair|trips the probability is 1 / [C(7,3)*3] For pair|trips|trips and trips|trips|trips it's 1 / [C(8,2)*C(5,2)] P(set>set>set | unpaired flop) = C(9,3)*[ (C(4,2)3)³/(5*3) / C(49,6) - 2*3*4*C(4,2)²3²/C(7,3)/3 / C(49,7) + 4*3*C(4,2)4²3/C(8,2)/C(5,2) / C(49,8) - 8*4³/C(8,2)/C(5,2) / C(49,9) ] = 1 in 441 (1/441) * 48*44/51/50 = 1 in 533
 05-27-2020, 06:52 AM #5 whosnext Carpal \'Tunnel     Join Date: Mar 2009 Location: California Posts: 6,253 Re: Odds of set over set over set I ran a simulation of 100 million 9-player Omaha deals last night. There were 175,375 deals with set-over-set-over-set on the flop. Of course, this is a probability of 0.00175375. Using standard sampling theory to construct a 95% confidence interval, I find that this is likely to occur within [567:573] deals. Thus, I have more confidence in my original derived probability that flopping set-over-set-over-set in 9-player Omaha occurs approximately once every 568 deals. Last edited by whosnext; 05-27-2020 at 07:12 AM.
05-27-2020, 08:04 AM   #6
heehaww
Pooh-Bah

Join Date: Aug 2011
Location: Tacooos!!!!
Posts: 4,851
Re: Odds of set over set over set

Heh, I was just typing about a mistake I thought I found in one of your calcs. Now I don't know what to believe, but here's what I found nonetheless:

Quote:
 Originally Posted by whosnext Case 2A: [2,2,2 with one "extra" card] ...*C(5,1)*C(4,2)*C(2,2)
This is prohblem.

Your denominator is C(49,12)*C(11,3)*C(7,3). Of interest here is the C(11,3)*C(7,3) which counts the ways to assign the 12 cards into 4 unordered groups. Your numerators will seek to do the same. So far so good.

We need to count the unordered groupings of AAxK|KKxx|QQxx, allowing the extra K to be with the AA or the QQ.

You counted the ways to group the x's and you correctly counted order of groups for that, since it matters which x cards the AKQ cards are grouped with. Your fraction is 5*C(4,2)/C(11,3)/C(7,3) = 2/385

But you didn't count the the 2 choices for whether it's in the AA group or the QQ group, nor the 3 choices for which K is the loner. It should be:

2*3*5*C(4,2)/C(11,3)/C(7,3) = 12/385

To check that, multinomial arrangements are another way to find the grouping probability.

There are C(12,2)*C(10,2)*C(8,3) ways to arrange AAKKKQQxxxxx

There are 3! ways to arrange the 4-card regions. There are 2 regions in which to put the extra K. Suppose it goes with the AA. Then there are 4*3 ways to arrange AAKx and C(4,2) ways to arrange KKxx and QQxx.

2*12*6³ / (66*45*56) = 12/385

Haven't checked your other tallies yet.

05-27-2020, 12:32 PM   #7
whosnext
Carpal \'Tunnel

Join Date: Mar 2009
Location: California
Posts: 6,253
Re: Odds of set over set over set

Quote:
 Originally Posted by whosnext 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
Here is what I thought I did:

Wlog say the flop is AsKsQs.

The first three terms (C(3,2)*C(3,2)*C(3,2)) capture which AKQ's form the pairs in the three players' hands. Wlog say the pairs are AhAc, KhKc, QhQc.

The C(3,1) captures which of the "leftover" cards (Ad,Kd,Qd) is the "extra" card. Wlog say it is the Kd.

The C(2,1) captures which of the "other" players (with Aces or Queens) does the Kd go with. Wlog say it is the Aces hand.

The C(40,5) captures which 5 of the 40 non-AKQ cards appear in these hands.

The C(5,1) captures which 1 of the 5 cards goes with the Kd hand (Aces).

The C(4,2) captures which 2 of the remaining 4 cards goes with the Kings hand.

The C(2,2) captures which 2 of the remaining 2 cards goes with the Queens hand.

As you can see, I used this "method" to derive the tallies for each of the subcases. So if this is wrong here, then all the tallies for any subcase with "extra" cards is undoubtedly wrong too.

 05-27-2020, 01:44 PM #8 heehaww Pooh-Bah     Join Date: Aug 2011 Location: Tacooos!!!! Posts: 4,851 Re: Odds of set over set over set Whoops, somehow I hadn't noticed the part you bolded. So then there is no missing factor of 2*3 Apologies for my awful reading earlier. Later I'll have to look at the rest of your calcs and check my own again too. I'll start with my own since there's less to proofread and no sim supporting it atm.
 05-27-2020, 02:14 PM #9 whosnext Carpal \'Tunnel     Join Date: Mar 2009 Location: California Posts: 6,253 Re: Odds of set over set over set No need to apologize. My original presentation arrived with virtually zero explanation so it is my fault if any misinterpretations occur. Later today I will kick off another simulation of 100 million deals and this time keep track of the "subcase" tallies. I probably should have done that last night on the first simulation. Of course, in addition to answering an unasked question (prob of flopping set-over-set-over-set in Omaha), this is largely an exercise in combinatorics. And, as has occurred several times in the past in this forum, seeing PIE vs "direct" methods (strengths, weaknesses, etc.). Thanks again.
05-27-2020, 08:28 PM   #10
heehaww
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Join Date: Aug 2011
Location: Tacooos!!!!
Posts: 4,851
Re: Odds of set over set over set

Thank you for writing a sim.

I found my mistake! Same mistake twice:
Quote:
 Originally Posted by heehaww For pair|pair|trips the probability [of a correct card grouping] is 1 / [C(7,3)*3] For pair|trips|trips ... it's 1 / [C(8,2)*C(5,2)]
Those numerators should be 3 (and it has to be the numerators because the 2nd denominator isn't divisible by 3). It's easier to see with multinomial arrangements:

AAABBCC can be arranged C(7,2)*C(5,2) ways, of which 3! are valid for our purposes. (The valid arrangements have identical letters clustered together, so treat the letter blocks as 3 single elements.)
p = 1/35 = 3/105
AAABBBCC can be arranged C(8,2)*C(6,3) ways, again 3! of which are valid. p = 3/280

But I'm not sure how to explain it within the framework of the first method. Maybe we're not even supposed to use the first method on groups of unequal size, due to the oddball group being automatically distinguishable. We could list every possible partition and there would in fact be only 1 valid partition out of 105, but 1/105 is wrong. Today I learned! It reminds me of what happens when you try to compute a dice probability using multiset combinations. One can correctly list all 21 combos, but it would be wrong to conclude that the chance of double-six is 1/21.

Perhaps we can cheat using a heuristic: we know there is one oddball group out of 3, so multiply the probability by 3 "just because". We can pretend we're multiplying by 3 ways to pick which group (1st, 2nd or 3rd) is the uniquely sized one, and maybe that mental model will always lead to the right answer, but IMO it's not a true explanation of what's happening.

Let's test this cheater method: how many regroupings of AA|BBB|CCC|DDDD are grouped like that?

Via multinomial arrangements: 4! / C(12,2)*C(10,3)*C(7,3) = 1/11550

Via partner assignments:
Initial step: 1 / C(12,2)*C(10,4)*C(5,2)
Cheating step: multiply by the # of ways to arrange XYZZ, which is 4*3, for a fraction that reduces to 1/11550.

How about that, it appears to work. Maybe some time I can think of a satisfactory explanation.

Returning to the Omaha problem, fixing my mistake looks like this:

P(set>set>set | unpaired flop) =
C(9,3)*[
(C(4,2)3)³/(5*3) / C(49,6) -
2*3*4*C(4,2)²3²/ C(7,3) / C(49,7) +
4*3*C(4,2)4²3²/C(8,2)/C(5,2) / C(49,8) -
8*4³/C(8,2)/C(5,2) / C(49,9)
]

Then, multiplying by 48*44/51/50 we get .0017600549 which is approximately...drumroll...1 in 568

Side note, take a moment to appreciate the beauty of inclusion-exclusion. It allows us to ignore all the cases where an extra card of a set's rank are in a different player's hand, as well as ignore the non-AKQ cards.

If we wanted the chance of set>set>set without one of the players having trips of the set rank, we'd replace the leading coefs 2,4,8 with 3,9,27.

05-28-2020, 01:10 AM   #11
whosnext
Carpal \'Tunnel

Join Date: Mar 2009
Location: California
Posts: 6,253
Re: Odds of set over set over set

Good news. Sounds like we have agreement.

Not that it is very relevant at this point, but my second simulation of 100 million 9-player Omaha deals has completed. I kept track of tallies of the subcases since, at the time I kicked off the simulation, I thought it could be helpful to have information at the subcase level.

Here are the results by subcase.

# of total AKQ cards in 3 hands# of AKQ in 3 handsSimulation Tally of SubcaseTally extrapolated from Analytical Derivation
6
[2,2,2]
121,039
121,788
7
[2,2,2]
41,907
41,756
7
[3,2,2]
6,896
6,959
8
[2,2,2]
4,186
4,060
8
[3,2,2]
1,167
1,160
8
[3,3,2]
112
129
9
[2,2,2]
112
104
9
[3,2,2]
47
42
9
[3,3,2]
8
7
9
[3,3,3]
1
1
.
TOTAL

175,475
176,005

Anyway, all the above subcase tallies match the analytically-derived figures within sampling error, so no red flags are raised.

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