My home game has gradually drifted away from hold-em over the years and started playing more mixed games. Recently, we have played several sessions of exclusively PLO8 which is definitely a fun game.

Most in the group are good with math and understand poker probabilities, but nobody (including me) has a firm grasp on what the "normal" distribution of pot winnings are in PLO8. So I volunteered to ask this august forum and undertake a few simple simulations.

Let's start slow and easy. Consider heads-up PLO8 and suppose every deal goes to showdown. What are the expected frequencies of all the possible ways the pot can be split in this case? One player scoops both high and low pots, two players chop the pot (one wins high and the other wins low), one player scoops the entire pot when neither player makes a low (this case is probably not called a scoop), etc.

Since I am fairly new to this game, my intuition is virtually non-existent. I have ideas on how often one or both players make a low, but that is about the extent of my "knowledge".

I guess there are conflicting forces working for and against scoops (supposing the board makes a low available). On one hand aces work both ways so a player with an ace may have a slight edge in scooping. On the other hand, winning the high typically uses high cards so may work against that player winning the low since low cards are needed to win the low pot.

I tried to write a simple HU O8 simulator. I will present the preliminary results below. Since this is a brand new program, there is a chance that I made a mistake in programming somewhere along the line, and since I am such a new O8 player I cannot tell if these results are sensible or not.

Here are the results of a simulation of 100,000 HU O8 deals (both players go to showdown on every deal) broken down by which player won the high pot (including the possibility of a tie) and which player won the low pot (including the possibilities or a tie or neither player having a low).

HIGH WINNER	P1 Wins Low	P2 Wins Low	Tie for Low	Neither has a Low
P1 Wins High	10,531	13,176	774	24,620
P2 Wins High	13,122	10,696	744	24,885
Tie for High	245	254	183	770

Hopefully, everyone can read the above table. Scoops of both high and low pots would be cells [1,1] and [2,2]. Chopped pots where one player wins high and the other player wins low would be cells [1,2] and [2,1]. Scoops (so-called) where neither player has a low would be cells [1,4] and [2,4].

From the table we see that:

(1) In roughly 50% of the pots neither player has a low. Is this believable?

(2) In the other 50% of the pots, roughly 21% are scooped and roughly 26% are chopped (roughly 3% of the pots have either the high or low pot chopped). Again, is this believable?

Any thoughts would be appreciated.

I will kick off a simulation of more deals later tonight, but I wanted to post preliminary results in case they are way off.

Here are the results of a simulation of 1,000,000 deals of HU O8.

HIGH WINNER	P1 Wins Low	P2 Wins Low	Tie for Low	Neither has a Low
P1 Wins High	105,542	131,534	7,487	248,301
P2 Wins High	131,826	105,493	7,361	248,416
Tie for High	2,496	2,423	1,814	7,307

Edited to add the following:

Let's try to summarize a few high-level results:

(1) Pct of deals for which there is No Low = (248,301+248,416+7,307) / 1,000,000 = 50.4024%

(2) Pct of deals for which one player scoops both high and low pots = (105,542+105,493) / 1,000,000 = 21.1035%

(3) Pct of deals for which two different players win high and low pots =(131,826+131,534) / 1,000,000 = 26.3360%

(4) Pct of deals for which both high and low pots exist and there is a tie for one or more halves = (2,496+2,423+7,487+7,361+1,814) / 1,000,000 = 2.1581%

I will attempt to run similar simulations with different numbers of players but the runs take forever on my crappy PC.

Here are the results of 1,000,000 deals for 3-player O8. Note to keep things relatively simple, for each half of the pot I have lumped all ties into one bucket, no matter how many or which players tie.

HIGH WINNER	P1 Wins Low	P2 Wins Low	P3 Wins Low	Any Tie for Low	Nobody has a Low
P1 Wins High	51,373	59,222	59,742	10,576	143,382
P2 Wins High	59,609	51,677	59,789	10,909	142,729
P3 Wins High	59,158	60,055	51,616	10,687	143,336
Any Tie for High	3,607	3,695	3,615	3,882	11,341

Let's try to summarize a few high-level results:

(1) Pct of deals for which there is No Low = (143,382+142,729+143,336+11,341) / 1,000,000 = 44.0788%

(2) Pct of deals for which one player scoops both high and low pots = (51,373+51,677+51,616) / 1,000,000 = 15.4666%

(3) Pct of deals for which two different players win high and low pots =(59,222+59,742+59,609+59,789+59,158+60,055) / 1,000,000 = 35.7575%

(4) Pct of deals for which both high and low pots exist and there is a tie for one or more halves = (3,607+3,695+3,615+10,576+10,909+10,687+3,882) / 1,000,000 = 4.6971%

I suppose I can keep track of these high-level summary figures as we vary the number of players at the table.

I probably should have started even before Heads-Up.

Here are two fairly recent threads on how often a Low is Available in O8 (three or more distinct ranks 8 or lower on the board where, of course, Ace is considered low for this).

https://forumserver.twoplustwo.com/2...-plo8-1630125/

https://forumserver.twoplustwo.com/2...ow-o8-1690200/

We found that the probability of a Low being Available in O8 is 60.0905%. Obviously, this is independent of the number of players at the table (considering all deals as we are in this thread). And, furthermore, this means that the probability of a Low being Made in O8 is capped at 60.0905%.

Our intuition and common sense suggest that the probability of a Low being Made will increase (with a cap at 60.0905%) as the number of players at the table increases.

__________

Okay, with those preliminaries taken care of, how about we look at what happens if there is only one player at the table. How often is a Low Made then, again, as above, considering all possible deals (all deals go to showdown)?

I took three different approaches to answer this simple question. (1) Simulation; (2) Combinatorics; (3) Brute-force computer program.

Simulation

Simulating a one-player O8 table is very straightforward when all deals go to showdown.

In a simulation of 1,000,000 deals I found that in 347,589 deals did Hero make a Low. This, of course, is 34.7589%.

The standard formulas for standard errors of a binomial probability show that the true percentage should be around this value +/- 0.095% (as per usual, I report twice the "standard error" to reflect two "standard deviations" in both directions).

Combinatorics

Since this one-player question seems so easy, I thought I could crank out the answer in short order using combinatorics.

From the links posted above, we know that there are 7 different cases as to how a Low can be Available in O8. Let's try to "solve" the first case via combinatorics.

Case 1 of Low Available: Board has 3 Distinct Low Ranks with counts [1,1,1] and 2 High cards

How can a player make a low in this case?

Case 1A: Player has 2 low cards and 2 high cards (low cards are distinct and non-overlapping with low cards on board)

= C(5,2)*C(4,1)*C(4,1)*C(18,2)
= 24,480

Case 1B: Player has 3 low cards with a pair and 1 high card (two distinct low card ranks are non-overlapping with low cards on board)

= C(5,2)*C(2,1)*C(4,2)*C(4,1)*C(18,1)
= 8,640

Case 1C: Player has 3 low cards without a pair and 1 high card (three distinct low card ranks are non-overlapping with low cards on board)

= C(5,3)*C(4,1)*C(4,1)*C(4,1)*C(18,1)
= 11,520

Case 1D: Player has 3 low cards without a pair and 1 high card (one overlapping rank with low cards on board)

= C(5,2)*C(4,1)*C(4,1)*C(3,1)*C(3,1)*C(18,1)
= 25,920

Case 1E: Player has 4 low cards with trips (two distinct low card ranks non-overlapping with low cards on board)

= C(5,2)*C(2,1)*C(4,3)*C(4,1)
= 320

Case 1F: Player has 4 low cards with two pairs (two distinct low card ranks non-overlapping with low cards on board)

= C(5,2)*C(4,2)*C(4,2)
= 360

Case 1G: Player has 4 low cards with 1 pair of overlapping rank and 2 singles non-overlapping ranks

= C(3,1)*C(3,2)*C(5,2)*C(4,1)*C(4,1)
= 1,440

Case 1H: Player has 4 low cards with 1 pair of non-overlapping rank, 1 single overlapping rank, and 1 single non-overlapping rank

= C(3,1)*C(3,1)*C(5,2)*C(2,1)*C(4,2)*C(4,1)
= 4,320

Case 1I: Player has 4 low cards with 1 pair of non-overlapping rank, 2 singles non-overlapping ranks

= C(5,3)*C(3,1)*C(4,2)*C(4,1)*C(4,1)
= 2,880

Case 1J: Player has 4 low cards without a pair, 2 overlapping ranks and 2 non-overlapping ranks

= C(3,2)*C(3,1)*C(3,1)*C(5,2)*C(4,1)*C(4,1)
= 4,320

Case 1K: Player has 4 low cards without a pair, 1 overlapping rank and 3 non-overlapping ranks

= C(3,1)*C(3,1)*C(5,3)*C(4,1)*C(4,1)*C(4,1)
= 5,760

Case 1L: Player has 4 low cards without a pair, 4 non-overlapping ranks

= C(5,4)*C(4,1)*C(4,1)*C(4,1)*C(4,1)
= 1,280

Subtotal = 91,240

As you might imagine, slogging through the combinatorics of these subcases loses its appeal after awhile so, given that this was just the first of 7 cases, I looked around and hoped to find an easier way to do this. Which brings us to:

Computer Brute-Force

Brute-forcing every possible deal of even one-person O8 is pushing the limits of my crappy PC, so I didn't like that idea. However, since we know the 7 cases (and their frequencies) in which a board makes a Low Available, brute-forcing those 7 cases one at a time is straightforward and eminently doable.

The following table summarizes these results.

Low Available Case	Description	Number of Boards	Hands One Player Makes Low	Pct of Possible One-Player Hands
1	3 Distinct Low Ranks [1,1,1]	680,960	91,240	51.15%
2	3 Distinct Low Ranks [2,1,1]	322,560	91,240	51.15%
3	3 Distinct Low Ranks [2,2,1]	24,192	91,240	51.15%
4	3 Distinct Low Ranks [3,1,1]	10,752	91,240	51.15%
5	4 Distinct Low Ranks [1,1,1,1]	358,400	126,984	71.19%
6	4 Distinct Low Ranks [2,1,1,1]	107,520	123,776	69.39%
7	5 Distinct Low Ranks [1,1,1,1,1]	57,344	136,127	76.32%

A few comments on the table before I present the key result.

(1) The pcts in the far right column are brand new to me since I am a new O8 player. I was surprised by many of them but don't have anything else to say at this point.

(2) It makes sense to this novice O8 player that the number of hands that make a low would be identical (91,240) in the first four cases above. I think the reason is that in Omaha, of course, you must use exactly three cards from the board and two cards from your hand. So in each of the first four cases above, you must use the same three ranks from the board to make a low. In these cases, then, a "duplicate" of one of those ranks is equivalent to a high card -- it cannot be used to make a low anyway.

(3) On first sight, I was puzzled why the number of hands that make a low in case 5 (126,984) is not identical to the number of hands that make a low in case 6 (123,776). My initial reasoning was the same as above. A "duplicate" of a low card on board is tantamount to a high card. But this reasoning must be wrong since these two cases give different results.

If anyone wants to chime in here and help out a novice O8 player better understand this result, I would appreciate it.

(4) The number of hands that make a low in the first case (91,240) is the value we attained via our tedious combinatoric approach, so I guess I am glad about that. The two approaches reinforce/confirm each other's results.

Okay, now to the key result. Using the information contained in the above table, it is straightforward to obtain the following result:

Pct of One Player making a Low in O8:

= 161,374,983,168 / 463,563,500,400

= 34.8118%

Remember that a Low is Available 60.0905% so, on average, one-player makes a low when a low is available approximately 58% of the time. I suppose experienced O8 players know stuff like this off the top of their head.

Note that the result from the modified brute-force approach that 34.8118% of all deals of one-player O8 have a low is consistent with the result from our simulation of 1,000,000 deals (34.7589%).

To match the way the results of the multiple-player tables have been presented in the thread, this means that there is No Low Made in 65.1882% of deals in one-player O8.


As I stated in an earlier post in this thread, since I am an O8 novice, I could easily present results in this thread that are completely wrong, but I would not even know they are wrong. So feel free to confirm/refute any results presented herein.

I am not interested in adapting my hold'em evaluator for this right now, but I can help with one thing. I am probably not much less of a novice O8 player than you, but I believe I can explain observation (3). Case 7 has the same number of 2-low combinations as case 6. However, case 7 has more unpaired 2-low combinations than Case 6. Consider the trivial case of a four card deck, {1, 1, 2, 2}. If the board is {2, 2} we can only have {1, 1} in our hand, and a pair can't qualify for low. If the board is {1, 2} then we have {1, 2} in our hand, which is a two card low.

Yes, I guess it is a type of card removal.

In the four cases where there are exactly three distinct low cards on board (Cases 1-4), here was my thinking.

Suppose you have a list of the 91,240 hands that make a low in Case 1 [1,1,1], three low singletons on board. Then consider Case 2 [2,1,1] where one of the three low ranks on board is paired.

For concreteness, suppose Board 1 is 8s 6s 4s Kh Qh and Board 2 is 8s 8h 6s 4s Kh (swapping the Qh for the 8h). Then if you simply replace the 8h with the Qh wherever it appears in the list of 91,240 hands which make a low on Board 1, you will have automatically have a list of 91,240 hands which make a low on Board 2. There is a one-to-one mapping between the two lists, so, of course, the total number in the two lists is identical.

As browni3141 suggests, the same "mapping" idea does not work when there are more than 3 low ranks on board as is the case for Cases 5 and 6.

Again, for concreteness, suppose Board 5 is 8s 6s 4s 2s Kh and Board 6 is 8s 8h 6s 4s 2s (swapping the Kh for the 8h). Then simply replacing the 8h with the Kh does not automatically give you a hand that makes a low on Board 6.

I suppose this is obvious. Consider the hand Qd Jc 8h 3c in the list of hands that make a low on Board 5. If you replace the 8h with the Kh you now have the hand Qd Jc Kh 3c which clearly does NOT make a low on Board 6.

I have attempted to construct other arguments related to this question but they all encounter this same difficulty. An additional duplicated low card on board surely reduces the number of hands that make a low whenever there is "slack" in the system (i.e., 4 low ranks on board) since there are hands that rely upon that rank to make the low. So removing one of that rank reduces the number of possible ways that could happen in a version of card removal.

This is not the situation of Case 1 vs. Case 2 since the 8h can never be critical to making a low in either case. If the 8h appears in a hand which makes a low on Board 1, then it must be true that there are at least 2 other low ranks (other than 8) appearing in the hand since there are exactly 3 low ranks on board (one of them being an 8). So removing the 8h from that hand (via the card removal/swapping mapping) can never turn a hand from making a low into one that no longer makes a low.

Despite my reluctance, I tried to derive the other case tallies via combinatorics. Recall that the question is: given a specific board type, how many 4-card Omaha hands make a Low with that Board (using the normal Omaha rules where you must use exactly 3 cards from the board and exactly two cards from your hand).

The following is extremely tedious and may very well not be the most efficient way to proceed. Also, another possible approach in this type of question is to apply the Principle of Inclusion-Exclusion (PIE) of which I am merely a dabbler.

Case 5: Board = 4 Distinct Low Cards [1,1,1,1] plus 1 High Card

Case 5A. 2 low cards and 2 high cards (2 non-overlapping ranks from low ranks on board)
= C(4,2)*C(4,1)*C(4,1)*C(19,2)
= 16,416

Case 5B. 2 low cards and 2 high cards (1 non-overlapping, 1 overlapping)
= C(4,1)*C(4,1)*C(4,1)*C(3,1)*C(19,2)
= 32,832

Case 5C. 3 low cards without a pair and 1 high card (3 non-overlapping)
= C(4,3)*C(4,1)*C(4,1)*C(4,1)*C(19,1)
= 4,864

Case 5D. 3 low cards without a pair and 1 high card (2 non-overlapping, 1 overlapping)
= C(4,2)*C(4,1)*C(4,1)*C(4,1)*C(3,1)*C(19,1)
= 21,888

Case 5E. 3 low cards without a pair and 1 high card (1 non-overlapping, 2 overlapping)
= C(4,1)*C(4,1)*C(4,2)*C(3,1)*C(3,1)*C(19,1)
= 16,416

Case 5F. 3 low cards with a pair and 1 high card (2 non-overlapping)
= C(4,2)*C(2,1)*C(4,2)*C(4,1)*C(19,1)
= 5,472

Case 5G. 3 low cards with a pair and 1 high card (1 non-overlapping with pair, 1 single overlapping)
= C(4,1)*C(4,2)*C(4,1)*C(3,1)*C(19,1)
= 5,472

Case 5H. 3 low cards with a pair and 1 high card (1 single non-overlapping, 1 overlapping pair)
= C(4,1)*C(4,1)*C(4,1)*C(3,2)*C(19,1)
= 3,648

Case 5I. 4 low cards without a pair (4 non-overlapping)
= C(4,4)*C(4,1)*C(4,1)*C(4,1)*C(4,1)
= 256

Case 5J. 4 low cards without a pair (3 non-overlapping, 1 overlapping)
= C(4,3)*C(4,1)*C(4,1)*C(4,1)*C(4,1)*C(3,1)
= 3,072

Case 5K. 4 low cards without a pair (2 non-overlapping, 2 overlapping)
= C(4,2)*C(4,1)*C(4,1)*C(4,2)*C(3,1)*C(3,1)
= 5,184

Case 5L. 4 low cards without a pair (1 non-overlapping, 3 overlapping)
= C(4,1)*C(4,1)*C(4,3)*C(3,1)*C(3,1)*C(3,1)
= 1,728

Case 5M. 4 low cards with a pair (3 non-overlapping)
= C(4,3)*C(3,1)*C(4,2)*C(4,1)*C(4,1)
= 1,152

Case 5N. 4 low cards with a pair (2 non-overlapping w/pair, 1 single overlapping)
= C(4,2)*C(2,1)*C(4,2)*C(4,1)*C(4,1)*C(3,1)
= 3,456

Case 5O. 4 low cards with a pair (2 non-overlapping, 1 overlapping pair)
= C(4,2)*C(4,1)*C(4,1)*C(4,1)*C(3,2)
= 1,152

Case 5P. 4 low cards with a pair (1 non-overlapping pair, 2 single overlapping)
= C(4,1)*C(4,2)*C(4,2)*C(3,1)*C(3,1)
= 1,296

Case 5Q. 4 low cards with a pair (1 single non-overlapping, 2 overlapping w/pair)
= C(4,1)*C(4,1)*C(4,2)*C(2,1)*C(3,2)*C(3,1)
= 1,728

Case 5R. 4 low cards with 2 pairs (2 non-overlapping)
= C(4,2)*C(4,2)*C(4,2)
= 216

Case 5S. 4 low cards with 2 pairs (1 non-overlapping, 1 overlapping)
= C(4,1)*C(4,2)*C(4,1)*C(3,2)
= 288

Case 5T. 4 low cards with trips (2 non-overlapping)
= C(4,2)*C(2,1)*C(4,3)*C(4,1)
= 192

Case 5U. 4 low cards with trips (1 non-overlapping trips, 1 single overlapping)
= C(4,1)*C(4,3)*C(4,1)*C(3,1)
= 192

Case 5V. 4 low cards with trips (1 non-overlapping single, 1 overlapping trips)
= C(4,1)*C(4,1)*C(4,1)*C(3,3)
= 64

Total for Case 5 = 126,984 [confirmed by computer brute-force]


Case 6: Board = 4 Distinct Low Cards [2,1,1,1]

Case 6A. 2 low cards and 2 high cards (2 non-overlapping ranks from low ranks on board)
= C(4,2)*C(4,1)*C(4,1)*C(20,2)
= 18,240

Case 6B. 2 low cards and 2 high cards (1 non-overlapping, 1 overlapping not board pair)
= C(4,1)*C(4,1)*C(3,1)*C(3,1)*C(20,2)
= 27,360

Case 6C. 2 low cards and 2 high cards (1 non-overlapping, 1 overlapping board pair)
= C(4,1)*C(4,1)*C(1,1)*C(2,1)*C(20,2)
= 6,080

Case 6D. 3 low cards without a pair and 1 high card (3 non-overlapping)
= C(4,3)*C(4,1)*C(4,1)*C(4,1)*C(20,1)
= 5,120

Case 6E. 3 low cards without a pair and 1 high card (2 non-overlapping, 1 overlapping not board pair)
= C(4,2)*C(4,1)*C(4,1)*C(3,1)*C(3,1)*C(20,1)
= 17,280

Case 6F. 3 low cards without a pair and 1 high card (2 non-overlapping, 1 overlapping board pair)
= C(4,2)*C(4,1)*C(4,1)*C(1,1)*C(2,1)*C(20,1)
= 3,840

Case 6G. 3 low cards without a pair and 1 high card (1 non-overlapping, 2 overlapping not board pair)
= C(4,1)*C(4,1)*C(3,2)*C(3,1)*C(3,1)*C(20,1)
= 8,640

Case 6H. 3 low cards without a pair and 1 high card (1 non-overlapping, 2 overlapping with board pair)
= C(4,1)*C(4,1)*C(3,1)*C(3,1)*C(1,1)*C(2,1)*C(20,1)
= 5,760

Case 6I. 3 low cards with a pair and 1 high card (2 non-overlapping)
= C(4,2)*C(2,1)*C(4,2)*C(4,1)*C(20,1)
= 5,760

Case 6J. 3 low cards with a pair and 1 high card (1 non-overlapping w/pair, 1 single overlapping not board pair)
= C(4,1)*C(4,2)*C(3,1)*C(3,1)*C(20,1)
= 4,320

Case 6K. 3 low cards with a pair and 1 high card (1 non-overlapping w/pair, 1 single overlapping board pair)
= C(4,1)*C(4,2)*C(1,1)*C(2,1)*C(20,1)
= 960

Case 6L. 3 low cards with a pair and 1 high card (1 single non-overlapping, 1 paired overlapping not board pair)
= C(4,1)*C(4,1)*C(3,1)*C(3,2)*C(20,1)
= 2,880

Case 6M. 3 low cards with a pair and 1 high card (1 single non-overlapping, 1 paired overlapping board pair)
= C(4,1)*C(4,1)*C(1,1)*C(2,2)*C(20,1)
= 320

Case 6N. 4 low cards without a pair (4 non-overlapping)
= C(4,4)*C(4,1)*C(4,1)*C(4,1)*C(4,1)
= 256

Case 6O. 4 low cards without a pair (3 non-overlapping, 1 overlapping not board pair)
= C(4,3)*C(4,1)*C(4,1)*C(4,1)*C(3,1)*C(3,1)
= 2,304

Case 6P. 4 low cards without a pair (3 non-overlapping, 1 overlapping board pair)
= C(4,3)*C(4,1)*C(4,1)*C(4,1)*C(1,1)*C(2,1)
= 512

Case 6Q. 4 low cards without a pair (2 non-overlapping, 2 overlapping not board pair)
= C(4,2)*C(4,1)*C(4,1)*C(3,2)*C(3,1)*C(3,1)
= 2,592

Case 6R. 4 low cards without a pair (2 non-overlapping, 2 overlapping w/board pair)
= C(4,2)*C(4,1)*C(4,1)*C(3,1)*C(3,1)*C(1,1)*C(2,1)
= 1,728

Case 6S. 4 low cards without a pair (1 non-overlapping, 3 overlapping no board pair)
= C(4,1)*C(4,1)*C(3,3)*C(3,1)*C(3,1)*C(3,1)
= 432

Case 6T. 4 low cards without a pair (1 non-overlapping, 3 overlapping w/board pair)
= C(4,1)*C(4,1)*C(3,2)*C(3,1)*C(3,1)*C(1,1)*C(2,1)
= 864

Case 6U. 4 low cards with a pair (3 non-overlapping)
= C(4,3)*C(3,1)*C(4,2)*C(4,1)*C(4,1)
= 1,152

Case 6V. 4 low cards with a pair (2 non-overlapping w/pair, 1 single overlapping no board pair)
= C(4,2)*C(2,1)*C(4,2)*C(4,1)*C(3,1)*C(3,1)
= 2,592

Case 6W. 4 low cards with a pair (2 non-overlapping w/pair, 1 single overlapping board pair)
= C(4,2)*C(2,1)*C(4,2)*C(4,1)*C(1,1)*C(2,1)
= 576

Case 6X. 4 low cards with a pair (2 non-overlapping, 1 pair overlapping no board pair)
= C(4,2)*C(4,1)*C(4,1)*C(3,1)*C(3,2)
= 864

Case 6Y. 4 low cards with a pair (2 non-overlapping, 1 pair overlapping board pair)
= C(4,2)*C(4,1)*C(4,1)*C(1,1)*C(2,2)
= 96

Case 6Z. 4 low cards with a pair (1 non-overlapping w/pair, 2 singles overlapping no board pair)
= C(4,1)*C(4,2)*C(3,2)*C(3,1)*C(3,1)
= 648

Case 6AA. 4 low cards with a pair (1 non-overlapping w/pair, 2 singles overlapping w/board pair)
= C(4,1)*C(4,2)*C(3,1)*C(3,1)*C(1,1)*C(2,1)
= 432

Case 6AB. 4 low cards with a pair (1 non-overlapping single, 2 overlapping w/pair and no board pair)
= C(4,1)*C(4,1)*C(3,2)*C(2,1)*C(3,2)*C(3,1)
= 864

Case 6AC. 4 low cards with a pair (1 non-overlapping single, 2 overlapping w/pair, single=board pair)
= C(4,1)*C(4,1)*C(3,1)*C(3,2)*C(1,1)*C(2,1)
= 288

Case 6AD. 4 low cards with a pair (1 non-overlapping single, 2 overlapping w/pair, pair=board pair)
= C(4,1)*C(4,1)*C(3,1)*C(3,1)*C(1,1)*C(2,2)
= 144

Case 6AE. 4 low cards with 2 pairs (2 non-overlapping)
= C(4,2)*C(4,2)*C(4,2)
= 216

Case 6AF. 4 low cards with 2 pairs (1 non-overlapping, 1 overlapping not board pair)
= C(4,1)*C(4,2)*C(3,1)*C(3,2)
= 216

Case 6AG. 4 low cards with 2 pairs (1 non-overlapping, 1 overlapping board pair)
= C(4,1)*C(4,2)*C(1,1)*C(2,2)
= 24

Case 6AH. 4 low cards with trips (2 non-overlapping)
= C(4,2)*C(2,1)*C(4,3)*C(4,1)
= 192

Case 6AI. 4 low cards with trips (1 non-overlapping trips, 1 overlapping not board pair)
= C(4,1)*C(4,3)*C(3,1)*C(3,1)
= 144

Case 6AJ. 4 low cards with trips (1 non-overlapping trips, 1 overlapping board pair)
= C(4,1)*C(4,3)*C(1,1)*C(2,1)
= 32

Case 6AK. 4 low cards with trips (1 non-overlapping single, 1 trips overlapping not board pair)
= C(4,1)*C(4,1)*C(3,1)*C(3,3)
= 48

TOTAL for Case 6 = 123,776 [confirmed by computer brute-forced]


Case 7: Board = 5 Distinct Low Cards [1,1,1,1,1]

Case 7A. 2 low cards and 2 high cards (2 non-overlapping ranks from board)
= C(3,2)*C(4,1)*C(4,1)*C(20,2)
= 9,120

Case 7B. 2 low cards and 2 high cards (1 non-overlapping, 1 overlapping)
= C(3,1)*C(4,1)*C(5,1)*C(3,1)*C(20,2)
= 34,200

Case 7C. 2 low cards and 2 high cards (2 overlapping)
= C(5,2)*C(3,1)*C(3,1)*C(20,2)
= 17,100

Case 7D. 3 low cards without a pair and 1 high card (3 non-overlapping)
= C(3,3)*C(4,1)*C(4,1)*C(4,1)*C(20,1)
= 1,280

Case 7E. 3 low cards without a pair and 1 high card (2 non-overlapping, 1 overlapping)
= C(3,2)*C(4,1)*C(4,1)*C(5,1)*C(3,1)*C(20,1)
= 14,400

Case 7F. 3 low cards without a pair and 1 high card (1 non-overlapping, 2 overlapping)
= C(3,1)*C(4,1)*C(5,2)*C(3,1)*C(3,1)*C(20,1)
= 21,600

Case 7G. 3 low cards without a pair and 1 high card (3 overlapping)
= C(5,3)*C(3,1)*C(3,1)*C(3,1)*C(20,1)
= 5,400

Case 7H. 3 low cards with a pair and 1 high card (2 non-overlapping)
= C(3,2)*C(2,1)*C(4,2)*C(4,1)*C(20,1)
= 2,880

Case 7I. 3 low cards with a pair and 1 high card (1 non-overlapping pair, 1 overlapping single)
= C(3,1)*C(4,2)*C(5,1)*C(3,1)*C(20,1)
= 5,400

Case 7J. 3 low cards with a pair and 1 high card (1 non-overlapping single, 1 overlapping pair)
= C(3,1)*C(4,1)*C(5,1)*C(3,2)*C(20,1)
= 3,600

Case 7K. 3 low cards with a pair and 1 high card (2 overlapping)
= C(5,2)*C(2,1)*C(3,2)*C(3,1)*C(20,1)
= 3,600

Case 7L. 4 low cards without a pair (3 non-overlapping 1 overlapping)
= C(3,3)*C(4,1)*C(4,1)*C(4,1)*C(5,1)*C(3,1)
= 960

Case 7M. 4 low cards without a pair (2 non-overlapping, 2 overlapping)
= C(3,2)*C(4,1)*C(4,1)*C(5,2)*C(3,1)*C(3,1)
= 4,320

Case 7N. 4 low cards without a pair (1 non-overlapping, 3 overlapping)
= C(3,1)*C(4,1)*C(5,3)*C(3,1)*C(3,1)*C(3,1)
= 3,240

Case 7O. 4 low cards without a pair (4 overlapping)
= C(5,4)*C(3,1)*C(3,1)*C(3,1)*C(3,1)
= 405

Case 7P. 4 low cards with a pair (3 non-overlapping)
= C(3,3)*C(3,1)*C(4,2)*C(4,1)*C(4,1)
= 288

Case 7Q. 4 low cards with a pair (2 non-overlapping w/pair, 1 single overlapping)
= C(3,2)*C(2,1)*C(4,2)*C(4,1)*C(5,1)*C(3,1)
= 2,160

Case 7R. 4 low cards with a pair (2 non-overlapping, 1 overlapping pair)
= C(3,2)*C(4,1)*C(4,1)*C(5,1)*C(3,2)
= 720

Case 7S. 4 low cards with a pair (1 non-overlapping w/pair, 2 overlapping)
= C(3,1)*C(4,2)*C(5,2)*C(3,1)*C(3,1)
= 1,620

Case 7T. 4 low cards with a pair (1 non-overlapping, 2 overlapping w/pair)
= C(3,1)*C(4,1)*C(5,2)*C(2,1)*C(3,2)*C(3,1)
= 2,160

Case 7U. 4 low cards with a pair (3 overlapping)
= C(5,3)*C(3,1)*C(3,2)*C(3,1)*C(3,1)
= 810

Case 7V. 4 low cards with 2 pairs (2 non-overlapping)
= C(3,2)*C(4,2)*C(4,2)
= 108

Case 7W. 4 low cards with 2 pairs (1 non-overlapping, 1 overlapping)
= C(3,1)*C(4,2)*C(5,1)*C(3,2)
= 270

Case 7X. 4 low cards with 2 pairs (2 overlapping)
= C(5,2)*C(3,2)*C(3,2)
= 90

Case 7Y. 4 low cards with trips (2 non-overlapping)
= C(3,2)*C(2,1)*C(4,3)*C(4,1)
= 96

Case 7Z. 4 low cards with trips (1 non-overlapping trips, 1 single overlapping)
= C(3,1)*C(4,3)*C(5,1)*C(3,1)
= 180

Case 7AA. 4 low cards with trips (1 single non-overlapping, 1 overlapping trips)
= C(3,1)*C(4,1)*C(5,1)*C(3,3)
= 60

Case 7AB. 4 low cards with trips (2 overlapping)
= C(5,2)*C(2,1)*C(3,3)*C(3,1)
= 60

TOTAL for Case 7 = 136,127 [confirmed by computer brute-force]

I'm way too lazy to do all of that by hand. Respect.

I will present the results of the simulations which have now completed in the next few posts.

Here are the results of 1,000,000 deals for 4-player O8. Note to keep things relatively simple, for each half of the pot I have lumped all ties into one bucket, no matter how many or which players tie.

HIGH WINNER	P1 Wins Low	P2 Wins Low	P3 Wins Low	P4 Wins Low	Any Tie for Low	Nobody has a Low
P1 Wins High	30,248	32,534	32,953	32,793	12,279	99,957
P2 Wins High	32,471	30,280	32,450	32,710	12,405	100,726
P3 Wins High	32,238	32,806	30,171	32,287	12,411	99,691
P4 Wins High	32,909	32,978	32,696	29,955	12,329	99,613
Any Tie for High	4,051	4,167	4,238	4,288	6,098	15,268

Let's try to summarize a few high-level results:

(1) Pct of deals for which there is No Low = 41.5255%

(2) Pct of deals for which one player scoops both high and low pots = 12.0654%

(3) Pct of deals for which two different players win high and low pots = 39.1825%

(4) Pct of deals for which both high and low pots exist and there is a tie for one or more halves = 7.2266%

Here are the results of 1,000,000 deals for 5-player O8. Note to keep things relatively simple, for each half of the pot I have lumped all ties into one bucket, no matter how many or which players tie.

HIGH WINNER	P1 Wins Low	P2 Wins Low	P3 Wins Low	P4 Wins Low	P5 Wins Low	Any Tie for Low	Nobody has a Low
P1 Wins High	19,586	19,802	19,815	20,129	20,107	13,149	76,955
P2 Wins High	19,950	19,584	19,980	19,861	20,286	13,085	77,176
P3 Wins High	20,041	19,912	19,781	20,034	19,949	13,249	77,058
P4 Wins High	20,011	20,006	19,845	19,556	20,128	13,077	77,646
P5 Wins High	19,971	19,997	20,166	20,030	20,074	13,120	76,904
Any Tie for High	4,395	4,460	4,540	4,436	4,400	8,650	19,099

Let's try to summarize a few high-level results:

(1) Pct of deals for which there is No Low = 40.4838%

(2) Pct of deals for which one player scoops both high and low pots = 9.8581%

(3) Pct of deals for which two different players win high and low pots = 40.0020%

(4) Pct of deals for which both high and low pots exist and there is a tie for one or more halves = 9.6561%

Here are the results of 1,000,000 deals for 6-player O8. Note to keep things relatively simple, for each half of the pot I have lumped all ties into one bucket, no matter how many or which players tie.

HIGH WINNER	P1 Wins Low	P2 Wins Low	P3 Wins Low	P4 Wins Low	P5 Wins Low	P6 Wins Low	Any Tie for Low	Nobody has a Low
P1 Wins High	13,664	13,071	13,377	13,185	13,155	13,275	13,771	63,033
P2 Wins High	13,111	13,373	13,178	13,221	13,419	13,201	13,718	63,020
P3 Wins High	13,329	13,465	13,553	13,218	13,205	13,298	13,809	63,412
P4 Wins High	13,425	13,279	13,211	13,490	13,214	13,204	13,707	63,041
P5 Wins High	13,167	13,356	13,305	13,182	13,594	13,305	13,619	62,662
P6 Wins High	13,353	13,114	13,186	13,258	13,396	13,442	13,548	62,967
Any Tie for High	4,397	4,476	4,539	4,532	4,524	4,605	11,110	22,731

Let's try to summarize a few high-level results:

(1) Pct of deals for which there is No Low = 40.0866%

(2) Pct of deals for which one player scoops both high and low pots = 8.1116%

(3) Pct of deals for which two different players win high and low pots = 39.7663%

(4) Pct of deals for which both high and low pots exist and there is a tie for one or more halves = 12.0355%

Note here at N=6 players it becomes more likely, when a low is made, for a person to scoop the pot compared to that same person splitting the pot with any other specific opponent. That is, for the first time the diagonal elements of the above matrix are larger than the off-diagonal elements.

I am guessing this is related to the added value that Aces provide in O8 as the number of players at the table increases the quality of the winning low hand increases.

Here are the results of 1,000,000 deals for 7-player O8. Note to keep things relatively simple, for each half of the pot I have lumped all ties into one bucket, no matter how many or which players tie.

HIGH WINNER	P1 Wins Low	P2 Wins Low	P3 Wins Low	P4 Wins Low	P5 Wins Low	P6 Wins Low	P7 Wins Low	Any Tie for Low	Nobody has a Low
P1 Wins High	9,801	9,218	9,293	9,369	9,162	9,269	9,198	14,252	53,335
P2 Wins High	9,276	9,795	9,188	9,334	9,319	9,290	9,241	13,999	53,400
P3 Wins High	9,175	9,218	9,936	9,414	9,298	9,301	9,235	14,179	53,240
P4 Wins High	9,145	9,298	9,476	9,654	9,071	9,091	9,194	14,319	53,048
P5 Wins High	9,225	9,245	9,284	9,163	10,010	9,259	9,240	14,082	52,865
P6 Wins High	9,440	9,252	9,037	9,373	9,356	9,619	9,218	14,094	53,136
P7 Wins High	9,104	9,235	9,253	9,211	9,139	9,377	9,908	13,983	53,426
Any Tie for High	4,297	4,450	4,389	4,348	4,461	4,461	4,362	14,088	26,579

Let's try to summarize a few high-level results:

(1) Pct of deals for which there is No Low = 39.9029%

(2) Pct of deals for which one player scoops both high and low pots = 6.8723%

(3) Pct of deals for which two different players win high and low pots = 38.8484%

(4) Pct of deals for which both high and low pots exist and there is a tie for one or more halves = 14.3764%

Here are the results of 1,000,000 deals for 8-player O8. Note to keep things relatively simple, for each half of the pot I have lumped all ties into one bucket, no matter how many or which players tie.

HIGH WINNER	P1 Wins Low	P2 Wins Low	P3 Wins Low	P4 Wins Low	P5 Wins Low	P6 Wins Low	P7 Wins Low	P8 Wins Low	Any Tie for Low	Nobody has a Low
P1 Wins High	7,449	6,599	6,645	6,790	6,608	6,749	6,714	6,739	14,342	46,044
P2 Wins High	6,658	7,294	6,855	6,590	6,770	6,857	6,631	6,699	14,298	46,068
P3 Wins High	6,519	6,781	7,375	6,705	6,622	6,834	6,685	6,749	14,473	46,266
P4 Wins High	6,825	6,785	6,670	7,277	6,633	6,758	6,916	6,623	14,410	46,129
P5 Wins High	6,721	6,651	6,724	6,742	7,441	6,546	6,675	6,700	14,176	46,229
P6 Wins High	6,665	6,855	6,568	6,666	6,725	7,350	6,743	6,634	14,061	46,387
P7 Wins High	6,527	6,749	6,626	6,595	6,707	6,681	7,514	6,774	14,396	46,181
P8 Wins High	6,658	6,745	6,495	6,692	6,745	6,532	6,665	7,323	14,451	46,057
Any Tie for High	4,392	4,308	4,407	4,265	4,294	4,216	4,255	4,339	17,379	30,339

Let's try to summarize a few high-level results:

(1) Pct of deals for which there is No Low = 39.9700%

(2) Pct of deals for which one player scoops both high and low pots = 5.9023%

(3) Pct of deals for which two different players win high and low pots = 37.4815%

(4) Pct of deals for which both high and low pots exist and there is a tie for one or more halves = 16.6462%

Here are the results of 1,000,000 deals for 9-player O8. Note to keep things relatively simple, for each half of the pot I have lumped all ties into one bucket, no matter how many or which players tie.

HIGH WINNER	P1 Wins Low	P2 Wins Low	P3 Wins Low	P4 Wins Low	P5 Wins Low	P6 Wins Low	P7 Wins Low	P8 Wins Low	P9 Wins Low	Any Tie for Low	Nobody has a Low
P1 Wins High	5,697	4,916	4,925	5,099	5,016	4,871	4,990	4,963	4,994	14,782	40,076
P2 Wins High	5,096	5,627	4,973	4,997	5,042	5,009	5,031	5,089	4,891	14,518	40,936
P3 Wins High	5,050	5,032	5,705	4,987	4,925	5,080	4,920	4,979	4,939	14,459	40,508
P4 Wins High	5,045	5,071	5,071	5,679	5,072	5,006	5,099	5,055	4,933	14,566	40,364
P5 Wins High	4,964	4,922	5,096	5,060	5,560	5,011	4,906	4,865	5,110	14,587	40,262
P6 Wins High	4,964	4,983	4,978	5,152	4,985	5,610	4,974	5,033	5,075	14,708	40,542
P7 Wins High	5,036	5,002	4,961	4,927	5,017	5,030	5,677	4,910	4,916	14,698	40,727
P8 Wins High	5,020	5,085	5,031	5,118	5,098	4,946	4,899	5,672	5,011	14,530	40,804
P9 Wins High	5,013	5,136	5,011	4,928	5,069	4,902	5,003	4,987	5,505	14,589	40,784
Any Tie for High	4,205	4,159	4,186	4,152	4,235	4,250	4,169	4,127	4,125	20,895	34,025

Let's try to summarize a few high-level results:

(1) Pct of deals for which there is No Low = 39.9028%

(2) Pct of deals for which one player scoops both high and low pots = 5.0732%

(3) Pct of deals for which two different players win high and low pots = 36.0300%

(4) Pct of deals for which both high and low pots exist and there is a tie for one or more halves = 18.9940%

Okay, here is a summary table of the results posted thus far. The first row results are the exact probabilities while the other rows in the table are the results from simulations of 1,000,000 deals with that many players.

Players at Table	No Low Made	Low Made	Scoop Pot	Split Pot	Tied Pot
1	65.1882%	34.8118%	na	na	na
2	50.4024%	49.5976%	21.1035%	26.3360%	2.1581%
3	44.0788%	55.9212%	15.4666%	35.7575%	4.6971%
4	41.5255%	58.4745%	12.0654%	39.1825%	7.2266%
5	40.4838%	59.5162%	9.8581%	40.0020%	9.6561%
6	40.0866%	59.9134%	8.1116%	39.7663%	12.0355%
7	39.9029%	60.0971%	6.8723%	38.8484%	14.3764%
8	39.9700%	60.0300%	5.9023%	37.4815%	16.6462%
9	39.9028%	60.0972%	5.0732%	36.0300%	18.9940%

Miscellaneous comments on these results:

1. All of these results are based upon every player going to showdown on every deal. Of course, that is an extreme assumption and real-world O8 results are likely to be different from the above percentages.

2. We know that a Low is Available (based purely upon a random Board) in 60.0905% of O8 deals. The table suggests that this cap is approached in most full-ring O8 games.

3. Standard sampling theory suggests that the percentages in the table (except the first row) are within 0.1% of the true values.

4. Deals that involve ties (more than one player sharing one or both halves of the pot) could well be considered split pots and lumped in with the "Split Pot" (one player winning the high pot and another player winning the low pot) percentages above.

5. From the previously reported results, the low half of the pot is much more likely to be tied than the high half of the pot.