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 09-26-2016, 05:45 PM #26 whosnext Pooh-Bah     Join Date: Mar 2009 Location: California Posts: 4,853 re: Straight flush odds in PLO (with poker prob primer) Regarding programming simulations, I am essentially self-taught. My programming abilities are quite modest but modern computers are so fast that even poor and inefficient programmers like me can run some neat simulations. Many of the major contributors to this Forum are good programmers and several have even programmed their own poker simulators over the years. I think all of the following posters have experience in poker programming of one sort or the other (I apologize to anybody I have overlooked): statmanhal Rusty Brooks heehaww kapw7 suited fours R Gibert David Lyons Tom Cowley NewOldGuy BaseMetal2 Bottom line: if you have any desires or proclivities, I recommend fiddling around with poker simulators as they are a lot of fun. Last edited by whosnext; 09-27-2016 at 12:09 PM.
09-26-2016, 09:43 PM   #27
whosnext
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Join Date: Mar 2009
Location: California
Posts: 4,853
re: Straight flush odds in PLO (with poker prob primer)

Here are the tallies of the 100 Million deal simulation that just concluded:

Players having SF Tally
Zero
99,116,071
One
879,235
Two
4,694
Three
0
Four
0
Five
0
Six
0
Seven
0
Eight
0
Nine
0
Ten
0
GRAND TOTAL
100,000,000

I will repeat what I said in the post above:

From the table we see the degree of "overlap" (two players having a straight flush on the same deal) is small but not insignificant. This causes P10 to be slightly but not insignificantly lower than 10*P1.

09-26-2016, 11:39 PM   #28
suited fours
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Join Date: Nov 2012
Location: East Coast USA
Posts: 4,150
re: Straight flush odds in PLO (with poker prob primer)

Quote:
 Originally Posted by whosnext Regarding programming simulations, I am essentially self-taught. My programming abilities are quite modest but modern computers are so fast that even poor and inefficient programmers like me can run some neat simulations. Many of the major contributors to this Forum are good programmers and several have even programmed their own poker simulators over the years. I think all of the following posters have experience in poker programming of one sort or the other (I apologize to anybody I have overlooked): statmanhal Rusty Brooks heehaww kapw7 suited fours R Gibert David Lyons Tom Crowley NewOldGuy BaseMetal2 Bottom line: if you have any desires or proclivities, I recommend fiddling around with poker simulators as they are a lot of fun.
My math degree and associated computer classes were in the 80s. But with that foundation and the relative breeze that is Python, learning to do basic simulations on my own was straightforward. (I started learning Python in my spare time last month.) Agree this stuff is great fun.

09-26-2016, 11:53 PM   #29
suited fours
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Join Date: Nov 2012
Location: East Coast USA
Posts: 4,150
re: Straight flush odds in PLO (with poker prob primer)

Quote:
 Originally Posted by whosnext For a single player: In 100,000,000 deals of Omaha, a single player got 88,872 straight flushes*, which works out to around one in every 1,125 deals.
I ran 10M as a check and got 8,814 btw. If one is industrious enough to come up with the direct calculation, this wiki page is where I'd start:

https://en.wikipedia.org/wiki/Poker_...straight_flush

On a separate note, if anyone is looking for a poker probability challenge to work on:

https://projecteuler.net/problem=369

I had some fun and it took me a fair amount of time to crack that one.

09-27-2016, 01:27 AM   #30
citamgine
journeyman

Join Date: Aug 2011
Posts: 277
re: Straight flush odds in PLO (with poker prob primer)

Quote:
 Originally Posted by whosnext Hope this helps.
It helps tremendously. Excellent post. I'm sure there are many other people who stumble into this forum who would also find this extremely helpful.

I have one more hang up and the need for a sanity check.

Is the number below made bold a typo? Is it meant to be 1219?

Quote:
 Originally Posted by heehaww This is an approximation. Suppose I establish a reputation of only shoving with KK+, and so my villains will only call with AA. If I suddenly decide to shove with 72o utg ten-handed, then 1-(1119/1225)^9 < P(at least one caller) < 9*6/1225 The exact chance is 54/1225 - C(9,2)/C(50,4), via Principle of Inclusion-Exclusion (PIE).
I see that we're subtracting C(9,2)/C(50,4) to get rid of over counting by taking out one of the "Ands" but I'm not really sure how exactly we find that.
The C(9,2) must be the two cards from nine players. The 50 in C(50,4) must be the total cards remaining after we remove our 72. Where does the 4 in C(50,4) come from?

Again, can't say thanks enough to everyone who has posted in here. I felt like I was looking at hieroglyphics at first. Now it's all clear.
Seriously, it means a lot.

09-27-2016, 02:29 AM   #31
whosnext
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re: Straight flush odds in PLO (with poker prob primer)

Quote:
 Originally Posted by citamgine It helps tremendously. Excellent post. I'm sure there are many other people who stumble into this forum who would also find this extremely helpful. I have one more hang up and the need for a sanity check. Is the number below made bold a typo? Is it meant to be 1219? I see that we're subtracting C(9,2)/C(50,4) to get rid of over counting by taking out one of the "Ands" but I'm not really sure how exactly we find that. The C(9,2) must be the two cards from nine players. The 50 in C(50,4) must be the total cards remaining after we remove our 72. Where does the 4 in C(50,4) come from? Again, can't say thanks enough to everyone who has posted in here. I felt like I was looking at hieroglyphics at first. Now it's all clear. Seriously, it means a lot.
I will gladly defer to heehaww on this but here are my thoughts on your questions.

(1) Yes, the 1119 is supposed to be 1219. For each villain, there are a total of C(50,2) = 1225 possible holdings in HoldEm when I hold 72o.

Under the scenario, each will only call if they have pocket aces. There are clearly C(4,2) = 6 holdings of pocket aces.

So using the 1-[(1-p)^n] formula, the prob of at least one villain will call you assuming the villain's calls are independent is given by:

= 1 - [(1 - 6/1225)^9]

= 1 - [(1219/1225)^9]

This will, generally speaking, be a lower bound for the true probability.

(2) The "additive" formula will always give an upper bound as described in my primer post of earlier today.

Since we have 9 villains, and assuming mutually exclusive events , the prob that any of them call (the OR event) is the sum of the 9 individual probs:

= 9 * (6/1225)

= 54/1225

(3) The true probability can be found in various ways. Heehaww is an expert practitioner of the Principle of Inclusion-Exclusion. For now, let's just walk it through from the primer post above.

Of course, since there are only four aces in the deck, at most two villains can ever call. That simplifies our life a fair amount.

True prob = Prob from Additive Formula - Prob 2 Villains Call

since if 2 villains call, we would be double-counting them each calling.

Prob that 2 villains call:

= Prob(2 villains each have pocket aces)

= C(9,2) * Prob(2 specific players have pocket aces) since there are C(9,2) possible duos of villains that could have pocket aces

= C(9,2) *[C(4,2)*C(2,2)]/[C(50,4)*C(4,2)*C(2,2)]

where the numerator is the number of pocket aces duos that are possible (two of the four aces go to one villain and 2 of the remaining 2 aces goes to the other villain), and the denominator reflects that the duo villains will be dealt a total of 4 cards of the 50 unknown cards (excluding the two that you have, of course), and one villain gets 2 of the 4 and the other villain gets 2 of the remaining 2.

Clearly the C(4,2)*C(2,2) cancels from numerator and denominator and we are left with the True Prob:

= 54/1225 - [C(9,2)/C(50,4)]

You may want to verify that this true answer is in between the lower bound and upper bound given above.

Again, hope this helps.

09-27-2016, 03:33 AM   #32
kapw7
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Join Date: Jan 2005
Posts: 3,156
re: Straight flush odds in PLO (with poker prob primer)

Quote:
 Originally Posted by whosnext In 100,000,000 deals of Omaha, a single player got 88,872 straight flushes*, which works out to around one in every 1,125 deals.
There is a very detailed article by Brian Alspach calculating this number which is in agreement with your results (good job!). See article

09-27-2016, 04:10 AM   #33
whosnext
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Join Date: Mar 2009
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Posts: 4,853
re: Straight flush odds in PLO (with poker prob primer)

Quote:
 Originally Posted by kapw7 There is a very detailed article by Brian Alspach calculating this number which is in agreement with your results (good job!). See article
Wow, nice find. Alspach wrote some great poker-related articles over the years.

I honestly can say that I never saw this article before since I can remember that folks in the Probability Forum attempted to derive the exact number of straight flushes in Omaha a year or so ago (maybe longer).

IIRC there were several different approaches and answers. I don't think we ever did zero in on one answer.

It turned out that Alspach had found and published the answer more than 10 years prior!

If I have some time and energy tomorrow, I may try to revisit that thread in light of Alspach's approach and answer.

Thanks again.

09-27-2016, 11:17 AM   #34
heehaww
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re: Straight flush odds in PLO (with poker prob primer)

Quote:
 Originally Posted by citamgine For example: we can find out exactly how likely we are to win a coin flip if we flip a coin twice... Adding the probability together wont work here. I feel like adding would work, for instance, in calculating[....]If I deal in ten players then the chances of dealing the A would be the chance of dealing A to one player * 10? I feel this is correct but I can't articulate why this is.
It would work in that A example, and the difference between that and the coin example is that overlap is impossible in the Ace example. Only one player can get the A (the joint probability "collapses" to 0 like whosnext said), whereas the chance of winning both coinflips is 25%. While adding doesn't work for the coin problem, you get the right answer once you subtract away the 25% overlap.

It's also possible to use multiplication in the card example. But instead of raising P(not)^10, you'd have to multiply a string of conditional probabilities. It would be 1 - (51/52)(50/51)(49/50)...
50/51 is P(2nd player misses | 1st player missed), and so on.

Quote:
 There is only 16 combos of AK. Whats going on here?
The others covered this well, but I'll add one unimportant thing just to keep you thinking: when we count "hand combos", we're really counting suit permutations of a given rank combo. AK is the rank combo and 16 is the number of suit perms it can have. They're not suit combos because we distinguish between AK and AK, otherwise we'd only count 10. (But for some rank combos like AA, suit perms are the same as suit combos since there is no "1st ace" and "2nd ace".)

A "hand" can be thought of in two ways:
a) A hand is a card combo, and a card is a rank matched with a suit.
b) A hand is not really a combo, but a rank combo matched with a suit permutation, or vice versa.

Quote:
 Originally Posted by whosnext = C(9,2) *[C(4,2)*C(2,2)]/[C(50,4)*C(4,2)*C(2,2)] where the numerator is the number of pocket aces duos that are possible (two of the four aces go to one villain and 2 of the remaining 2 aces goes to the other villain), and the denominator reflects that the duo villains will be dealt a total of 4 cards of the 50 unknown cards (excluding the two that you have, of course), and one villain gets 2 of the 4 and the other villain gets 2 of the remaining 2. Clearly the C(4,2)*C(2,2) cancels from numerator and denominator
This is all true, but cancellation isn't necessary. We just need to deal the 2 players 4 aces in no particular way, so we needn't bother counting who gets which 2 cards. There's only C(4,4) = 1 combo of 4 aces, out of C(50,4) possible combos of 4 cards to give those players.

@Cita, where you would* need to think about how the cards are grouped is e.g. if we were talking about dealing two players AK. In a heads-up match, what's the chance both players are dealt AK?

*Only when using a single numerator and denominator. The alternative would be to multiply two probabilities, which overall would do the same thing, but wouldn't require you to see the inner mechanics of the card grouping.

 09-27-2016, 06:44 PM #35 whosnext Pooh-Bah     Join Date: Mar 2009 Location: California Posts: 4,853 re: Straight flush odds in PLO (with poker prob primer) In case anyone is interested, I read Alspach's article today. Here I will present his finding in a bit more detail than he provides (nothing of what follows is original to me). Query: What is the probability that a single player makes a straight flush on a random deal of Omaha? Of course we all know that in Omaha a player must use exactly three cards from the five-card board and exactly two cards from his four-card hand. (Note: a royal flush is included in everything that follows.) Clearly we need to know two things. First, how many total deals of Omaha to a single player are possible? Second, in how many of these deals does the player make a straight flush (technically, one or more straight flushes)? The total number of deals of Omaha to a single player is clearly C(52,5)*C(47,4) = 463,563,500,400. This can also be written as C(52,9)*C(9,5). Below we will use the fact that there are C(9,5) = 126 partitions of the nine cards into five cards for the board and four cards for the player's hole cards. Now let's tally how many of these deals give the player a straight flush. For the moment just consider all nine cards as a unit and divide the deals into six bins: (1) the nine cards form a nine-card straight flush; (2) the nine cards form exactly an eight-card straight flush; (3) the nine cards form exactly a seven-card straight flush; (4) the nine cards form exactly a six-card straight flush; (5) the nine cards form exactly a five-card straight flush; and (6) the nine cards do not form a five-card straight flush. Clearly these cases are mutually exclusive and collectively exhaustive. Alspach tackles each case in turn. Case 1: The nine cards form a nine-card straight flush There are six possibilities for the ranks: A-6, K-5, Q-4, J-3, T-2, 9-A. There are four possibilities for the suits. It turns out that the player will make a straight flush in all 126 partitions of the nine cards (no matter which cards are on the board and which cards are in his hand). Alspach gives reasoning behind this and it is straightforward to verify. So, there are 6*4*126 = 3,024 deals of this type in which the player makes a straight flush. Case 2: The nine cards form exactly an eight-card straight flush Here we need to subdivide the case into two subcases: Subcase 2A: Ace is one of the eight cards In this subcase, there are two possible rank configs of the eight cards: A-7, 8-A. And the ninth card can be any of the remaining 52-8-1 = 43 cards. So there are 2*4*43 = 344 configs in this subcase. Subcase 2B: Ace is not one of the eight cards Then there are five rank configs: K-6, Q-5, J-4, T-3, 9-2. And the ninth card can be any of the remaining 52-8-2 = 42 cards. So there are 5*4*42 = 840 configs in this subcase. Adding the two subcases, there are a total of 344+840 = 1,184 configs in this case. It turns out that in 120 of the 126 partitions does the player make a straight flush in this case. I imagine that there is a clever way to derive this number but it is straightforward to notice that the following 6 partitions (the player's hole cards are shown below) will not make a straight flush: 1456 2456 2457 3456 3457 3458 where I have coded the eight cards forming the straight flush 1-8. So, there are 1,184*120 = 142,080 deals of this type that give the player a straight flush. Case 3: The nine cards form exactly a seven-card straight flush Here we need to subdivide the case into two subcases: Subcase 3A: Ace is one of the seven cards In this subcase, there are two possible rank configs of the seven cards: A-8, 7-A. And the other two cards can be any of the remaining 52-7-1 = 44 cards. So there are 2*4*C(44,2) = 7,568 configs in this subcase. Subcase 3B: Ace is not one of the seven cards Then there are six rank configs: K-7, Q-6, J-5, T-4, 9-3, 8-2. And the other two cards can be any of the remaining 52-7-2 = 43 cards. So there are 6*4*C(43,2) = 21,672 configs in this subcase. Adding the two subcases, there are a total of 7,568+21,672 = 29,240 configs in this case. It turns out that in 108 of the 126 partitions does the player make a straight flush in this case. I imagine that there is a clever way to derive this number but it is straightforward to notice that the following 18 partitions (the player's hole cards are shown below) will not make a straight flush: 1345 1346 1356 1456 16QK 17QK 2345 2346 2347 2356 2357 2456 2457 27QK 3456 3457 345Q 345K where I have coded the seven cards forming the straight flush 1-7 and the other two cards K and Q for convenience. So, there are 29,240*108 = 3,157,920 deals of this type that give the player a straight flush. Case 4: The nine cards form exactly a six-card straight flush Here we need to subdivide the case into two subcases: Subcase 4A: Ace is one of the six cards In this subcase, there are two possible rank configs of the six cards: A-9, 6-A. And the other three cards can be any of the remaining 52-6-1 = 45 cards. So there are 2*4*C(45,3) = 113,520 configs in this subcase. Subcase 4B: Ace is not one of the six cards Then there are seven rank configs: K-8, Q-7, J-6, T-5, 9-4, 8-3, 7-2. And the other three cards can be any of the remaining 52-6-2 = 44 cards. So there are 7*4*C(44,3) = 370,832 configs in this subcase. Adding the two subcases, there are a total of 113,520+370,832 = 484,352 configs in this case. It turns out that in 90 of the 126 partitions does the player make a straight flush in this case. I imagine that there is a clever way to derive this number but it is straightforward to notice that the following 36 partitions (the player's hole cards are shown below) will not make a straight flush: 1234 1235 1236 1245 1246 1256 1345 1346 1356 1456 16JQ 16JK 16QK 1JQK 2345 2346 234J 234Q 234K 2356 235J 235Q 235K 2456 245J 245Q 245K 2JQK 3456 345J 345Q 345K 3JQK 4JQK 5JQK 6JQK where I have coded the six cards forming the straight flush 1-6 and the other three cards K, Q, and J for convenience. So, there are 484,352*90 = 43,591,680 deals of this type that give the player a straight flush. Case 5: The nine cards form exactly a five-card straight flush Here we need to subdivide the case into two subcases: Subcase 5A: Ace is one of the five cards In this subcase, there are two possible rank configs of the five cards: A-T, 5-A. And the other four cards can be any of the remaining 52-5-1 = 46 cards. So there are 2*4*C(46,4) = 1,305,480 configs in this subcase. Subcase 5B: Ace is not one of the five cards Then there are eight rank configs: K-9, Q-8, J-7, T-6, 9-5, 8-4, 7-3, 6-2. And the other four cards can be any of the remaining 52-5-2 = 45 cards. So there are 8*4*C(45,4) = 4,767,840 configs in this subcase. Adding the two subcases, there are a total of 1,305,480+4,767,840 = 6,073,320 configs in this case. It turns out that in 60 of the 126 partitions does the player make a straight flush in this case. I imagine that there is a clever way to derive this number but it is straightforward to notice that the following 66 partitions (the player's hole cards are shown below) will not make a straight flush: 1234 1235 123T 123J 123Q 123K 1245 124T 124J 124Q 124K 125T 125J 125Q 125K 1345 134T 134J 134Q 134K 135T 135J 135Q 135K 145T 145J 145Q 145K 1TJQ 1TJK 1TQK 1JQK 2345 234T 234J 234Q 234K 235T 235J 235Q 235K 245T 245J 245Q 245K 2TJQ 2TJK 2TQK 2JQK 345T 345J 345Q 345K 3TJQ 3TJK 3TQK 3JQK 4TJQ 4TJK 4TQK 4JQK 5TJQ 5TJK 5TQK 5JQK TJQK where I have coded the five cards forming the straight flush 1-5 and the other four cards K, Q, J, and T for convenience. So, there are 6,073,30*60 = 364,399,200 deals of this type that give the player a straight flush. Final Answer: Combining all the Cases The tally of all deals on which the player has a straight flush is simply the sum of the cases above. Clearly, it is impossible for the player to make a straight flush in case 6 in which no five cards of the nine make a straight flush. This sum is 411,293,904. Above we saw that the total number of Omaha deals to a single player is 463,563,500,400. Thus, the probability that a single player makes a straight flush on a random Omaha deal is: 0.000887244 or approximately once in every 1,125 deals. ---------- Addendum, from the above it is clear that we can write the tally as follows: = Sum [S goes from 5 to 9] 4 * [2*C(51-S,9-S) + (13-S)*C(50-S,9-S)] * A(S) where A(S) is the number of the 126 partitions of nine-cards of which exactly S cards form a straight flush into five board cards and four hole cards which give a straight flush under Omaha rules. We saw above that: A(9) = 126 A(8) = 120 A(7) = 108 A(6) = 90 A(5) = 60 Last edited by whosnext; 09-27-2016 at 06:54 PM.
 09-27-2016, 06:57 PM #36 suited fours Pooh-Bah     Join Date: Nov 2012 Location: East Coast USA Posts: 4,150 re: Straight flush odds in PLO (with poker prob primer) Very nice! A neat example of finding a better way to calculate something. Counting the potential boards and going from there seems much messier. Although a better path for counting probablity of 2 SF's amongst x players?
09-28-2016, 01:01 PM   #37
heehaww
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Join Date: Aug 2011
Location: Tacooos!!!!
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re: Straight flush odds in PLO (with poker prob primer)

Quote:
 Originally Posted by whosnext It turns out that in __ of the 126 partitions does the player make a straight flush in this case. I imagine that there is a clever way to derive this number
This approach can be used.

 09-28-2016, 04:58 PM #38 whosnext Pooh-Bah     Join Date: Mar 2009 Location: California Posts: 4,853 re: Straight flush odds in PLO (with poker prob primer) Motivated by heehaww's post, here is one more slice of the onion in case there is anybody still interested. From above, remember the key to Alspach's approach was these A(S) numbers, where A(S) is the number of the 126 partitions of nine-cards where exactly S cards form a straight flush into five board cards and four hole cards that gives a straight flush under Omaha rules. We saw above that: A(9) = 126 A(8) = 120 A(7) = 108 A(6) = 90 A(5) = 60 Here goes my attempt. A(5): Exactly five cards of the nine form a straight flush Clearly, there are C(5,3)*C(4,2) boards = C(5,2)*C(4,2) hands = 10*6 = 60 possible in this case in which the five cards are divided 3 to the board and 2 to the hand. Which is the only way that a straight flush would be possible under Omaha rules. Obviously, no matter which cards are put where, in all 60 of these cases would a straight flush be made by the player. So A(5) = 60 A(6): Exactly six cards of the nine form a straight flush There are two subcases to consider. Subcase 6A: 4 of the 6 are on board, and other 2 are in hand There are C(6,4)*C(3,1) boards = C(6,2)*C(3,2) hands = 15*3 = 45 with this config. Consider the hands. Which of these hands would not allow a straight flush? Clearly, if the max card minus the min card were 5 or higher. Here is a tally of such hands (SF cards only): 16 So there are [C(6,2)-1]*C(3,2) = (15-1)*3 = 14*3 = 42 configs that allow a SF in this subcase. Subcase 6B: 3 of the 6 are on board, and other 3 are in hand There are C(6,3)*C(3,2) boards = C(6,3)*C(3,1) hands = 20*3 = 60 with this config. Consider the boards. Which of these boards would not allow a straight flush? Clearly, if the max card minus the min card were 5 or higher. Here is a tally of such boards (SF cards only): 126 136 146 156 So there are [C(6,3)-4]*C(3,2) = (20-4)*3 = 16*3 = 48 configs that allow a SF in this subcase. Adding the two subcases, we see that there are 42+48 = 90 configs in this case. A(7): Exactly seven cards of the nine form a straight flush There are three subcases to consider. Subcase 7A: 5 of the 7 are on board, and other 2 are in hand There are C(7,5)*C(2,0) boards = C(7,2)*C(2,2) hands = 21*1 = 21 with this config. Consider the hands. Which of these hands would not allow a straight flush? Clearly, if the max card minus the min card were 5 or higher. Here is a tally of such hands (SF cards only): 16 17 27 So there are [C(7,2)-3]*C(2,2) = (21-3)*1 = 18*1 = 18 configs that allow a SF in this subcase. Subcase 7B: 4 of the 7 are on board, and other 3 are in hand There are C(7,4)*C(2,1) boards = C(7,3)*C(2,1) hands = 35*2 = 70 with this config. Consider the boards. Which of these boards would not allow a straight flush? Clearly, if in each triple the max card minus the min card were 5 or higher. Here is a tally of such boards (SF cards only): 1267 So there are [C(7,4)-1]*C(2,1) = (35-1)*2 = 34*2 = 68 configs that allow a SF in this subcase. Subcase 7C: 3 of the 7 are on board, and other 4 are in hand There are C(7,3)*C(2,2) boards = C(7,4)*C(2,0) hands = 35*1 = 35 with this config. Consider the boards. Which of these boards would not allow a straight flush? Clearly, if the max card minus the min card were 5 or higher. Here is a tally of such boards (SF cards only): 126 127 136 137 146 147 156 157 167 237 247 257 267 So there are [C(7,3)-13]*C(2,2) = (35-13)*1 = 22*1 = 22 configs that allow a SF in this subcase. Adding the three subcases, we see that there are 18+68+22 = 108 configs in this case. A(8): Exactly eight cards of the nine form a straight flush There are two subcases to consider. Subcase 8A: 5 of the 8 are on board, and other 3 are in hand There are C(8,5)*C(1,0) boards = C(8,3)*C(1,1) hands = 56*1 = 56 with this config. Consider the boards. Which of these boards would not allow a straight flush? Clearly, if in each triple the max card minus the min card were 5 or higher. By inspection, there are no such boards. So there are 56 configs that allow a SF in this subcase. Subcase 8B: 4 of the 8 are on board, and other 4 are in hand There are C(8,4)*C(1,1) boards = C(8,4)*C(1,0) hands = 70*1 = 70 with this config. Consider the boards. Which of these boards would not allow a straight flush? Clearly, if in each triple the max card minus the min card were 5 or higher. Here is a tally of such boards (SF cards only): 1267 1268 1278 1368 1378 2378 So there are [C(8,4)-6]*C(1,1) = (70-6)*1 = 64*1 = 64 configs that allow a SF in this subcase. Adding the two subcases, we see that there are 56+64 = 120 configs in this case. A(9): Nine cards of the nine form a straight flush Clearly, 5 of the 9 are on board, and other 4 are in hand There are C(9,5)*C(0,0) boards = C(9,4)*C(0,0) hands = 126*1 = 126 with this config. Consider the boards. Which of these boards would not allow a straight flush? Clearly, if in each triple the max card minus the min card were 5 or higher. By inspection, there are no such boards.* So there are 126 configs that allow a SF in this case. * I am reminded of the old joke about two mathematicians discussing a math problem. One of them says that some mathematical statement is obvious. The other says just a moment and dashes to the nearest blackboard and begins writing furiously. After fifteen minutes he returns. Ah, yes, it is obvious. Last edited by whosnext; 10-01-2016 at 04:50 AM. Reason: clarification
09-30-2016, 11:57 PM   #39
citamgine
journeyman

Join Date: Aug 2011
Posts: 277
re: Straight flush odds in PLO (with poker prob primer)

Please pardon my hiatus. There's a lot of well explained information here to absorb. I've read back over the thread a few times now and I think I'm pretty clear on the concepts discussed. Hopefully OP got as much out of this thread as I feel I have!

Also, the straight flush solution is absolutely amazing. How anyone can come up with this stuff is way beyond me.

Quote:
 Originally Posted by whosnext Again, hope this helps.
Again, you da man (or woman). Another great explanation.

Quote:
 Originally Posted by heehaww ...
Thanks, heehaww. Your post really helped clear some things up.

Quote:
 Originally Posted by suited fours On a separate note, if anyone is looking for a poker probability challenge to work on: https://projecteuler.net/problem=369 I had some fun and it took me a fair amount of time to crack that one.
I utilized the newfound probability magic, bestowed upon me by the wizards ITT, in an attempt to solve that badugi problem but quickly ran into trouble.

I feel like I have all the knowledge required to work it out now. Maybe I'm just missing where/how to apply it. I'm going to hide my thoughts on it in case anyone else wants to solve it first. Not that seeing my sorry attempt to solve it would provide anything useful. Still....

Spoiler:

 10-01-2016, 12:45 AM #40 whosnext Pooh-Bah     Join Date: Mar 2009 Location: California Posts: 4,853 re: Straight flush odds in PLO (with poker prob primer) How many 4-card badugis are there? Spoiler: What is a badugi? 4 different ranks in 4 different suits. Let's do the ranks first. How many ways are there to make 4 different ranks? Clearly it is C(13,4) = 715. Okay, given that we have four different ranks, say KQJT, how many different ways are there to have a badugi with KQJT. In this case the order of the cards (suits) matters. SHDC is a different badugi than HSDC, etc. So instead of using the "Choose" or "Combination" operator, we need the parallel operator which considers different orders to count. This is the permutation operator. P(4,4) = 4! = 4*3*2*1 = 24 So, combining there are 715*24 = 17,160 four-card badugis. P.S. Determining how many N-card hands have a 4-card badugi inside seems challenging (for larger N certainly). Many/most of the Euler problems "require" computer programming solutions, so maybe that is the direction they want you to go in.
 10-01-2016, 02:25 AM #41 suited fours Pooh-Bah     Join Date: Nov 2012 Location: East Coast USA Posts: 4,150 re: Straight flush odds in PLO (with poker prob primer) The question has been on the reasonably popular site for 4-1/2 years and only 327 people have entered a correct answer. Some people can answer it quickly, but for most of us mortals, it is quite a challenging question.
10-01-2016, 04:42 AM   #42
Buzz
gramps

Join Date: Sep 2002
Location: Los Angeles, California
Posts: 18,256
re: Straight flush odds in PLO (with poker prob primer)

Quote:
 Originally Posted by whosnext How many 4-card badugis are there?
52*36*22*10/1/2/3/4=17,160

Buzz

10-01-2016, 08:43 AM   #43
heehaww
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re: Straight flush odds in PLO (with poker prob primer)

Quote:
 Originally Posted by citamgine Would it work to do: C(4,4)*C(13,1)*C(12,1)*C(11,1)*C(10,1)
Yes, and imo that's the best possible approach for the 4-card version of that problem.

Notice how whosnext picked a combo of ranks and then a permutation of suits, while you did the opposite and got the same answer. In this case, your way was better because C(n,n)=1, so the answer is just (13 permute 4) aka "13 P 4". Granted, if you do it the other way and say C(13,4)*4!, you should instantly know that that's the same as (13 P 4) because a combo of 4 objects that counts all 4! arrangements is the definition of a permutation of 4 objects.

C(13,1)*...*C(10,1) = 13*12*11*10 = (13 P 4)

Last edited by heehaww; 10-01-2016 at 08:49 AM.

 10-03-2016, 03:16 AM #44 citamgine journeyman   Join Date: Aug 2011 Posts: 277 re: Straight flush odds in PLO (with poker prob primer) Beautiful, guys! I'd like to continue learning some stuff for fun. I'm particularly interested in learning how to build the sims. Maybe some day I could actually solve the whole badugi problem . I've searched around a bit, but I don't really even know what these types of sims are called and most of the stuff I have been able to find is wayy above my head. Any recommended places to start? Thanks a million!!! Over and out.
 10-03-2016, 08:33 AM #45 suited fours Pooh-Bah     Join Date: Nov 2012 Location: East Coast USA Posts: 4,150 re: Straight flush odds in PLO (with poker prob primer) If you're saying you'd like to learn to write your own poker simulations but you've never written a program, there are as many approaches as there are people. I'd suggest the book "python crash course" which has been out for about a year. That steps you through the basics right from downloading python and an editor to some pretty advanced stuff. I'd read and understand that book up through and including the sections on - dictionaries - lists - loops - if statements. Then Google "python random module" and go from there, or that book may cover random too. Anyway, you'll need basic understanding of a programming language similar to above to be able to ask the right questions and to enjoy writing up a simulator. It isn't hard, but you've gotta learn to walk first.
09-01-2017, 04:52 AM   #46
whosnext
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Location: California
Posts: 4,853
Re: Straight flush odds in PLO (with poker prob primer)

Re-opening this thread for breaking news ....

I made the "mistake" of going over to the Wizard of Odds website over the past weekend. I can't remember what I was looking for, but I came across a table of the distribution of (single player) Omaha High hands.

Of course, I remembered this thread so I looked at the straight flush figure (including royal flushes) and was perplexed to see that the Wizard came up with a different number than Brian Alspach's number which was reported in some detail above.

Since these calculations are pretty straightforward, I decided to derive the correct answer on my own from first principles. Hopefully, I will identify what is the correct figure. Anyway, here goes.

Case 1: Royal Flush

In Omaha, as everyone knows, you must use exactly two cards from your four card hand and exactly three cards from the five card board. So there are C(5,2) = 10 ways to divide the five royal flush cards of each suit into two cards in hand and three cards on board. (Clearly you can never make more than one royal flush in any one deal, so there is no issue of double counting.)

For each division, there are clearly 47 other cards that need to be placed two in the hand and two more on the board. Building this up we have:

= C(47,2)*C(45,2) = 1,070,190 for each division

* 10 divisions = 10,701,900

* 4 suits = 42,807,600.

Case 2. King High Straight Flush

In what follows, I find it easiest to simply tally the exclusions that need to be subtracted from the total possible given a correct alignment. For example, suppose we have KQ of a suit in hand and the JT9 of the same suit on board. We need to fill in the other two cards for the hand and two additional cards for the board.

As in the royal flush case, the total possible (before any exclusions) is clearly seen to be C(47,2)*C(45,2) = 1,070,190 (per alignment and per suit).

Now why isn't this what we want? Because in some of the cases in which we fill in the "other" cards, the player will actually make a royal flush (in addition to the King high straight flush). And since we already counted all the royal flushes in case one above, we need to make sure to subtract them out here or we will be double counting them.

It is easy to see that the player will make a royal flush whenever the Ace of the suit appears with the 9 of the suit (either in the hand or on the board).

Thus we need to exclude:

= C(1,1)*C(46,1)*C(45,2) = 45,540 for each alignment combo

Subtracting from the maximum amount gives:

= 1,070,190 - 45,540 = 1,024,650

* 10 combos = 10,246,500

* 4 suits = 40,986,000.

Case 3: Queen High Straight Flush

Case 3A: 98 in hand and QJT on board

Here the player makes a higher straight flush whenever the King is in hand
= C(1,1)*C(46,1)*C(45,2) = 45,540

So that the subtotal of Queen high straight flushes
= 1,070,190 - 45,540 = 1,024,650

Case 3B:
Q8 in hand & JT9 on board
J8 in hand & QT9 on board
T8 in hand & QJ9 on board

Here we need to exclude when King is in hand OR [Ace in hand & King on board]
= C(1,1)*C(46,1)*C(45,2) + C(1,1)*C(45,1)*C(1,1)*C(44,1) = 47,520

So that the subtotal of Queen high straight flushes per combo is
= 1,070,190 - 47,520 = 1,022,670

*3 combos = 3,068,010

Case 3C:
QJ in hand & T98 on board
QT in hand & J98 on board
JT in hand & Q98 on board

Here we need to exclude when King is on board
= C(1,1)*C(46,1)*C(45,2) = 45,540

So that the subtotal of Queen high straight flushes per combo is
= 1,070,190 - 45,540 = 1,024,650

*3 combos = 3,073,950

Case 3D:
Q9 in hand & JT8 on board
J9 in hand & QT8 on board
T9 in hand & QJ8 on board

Here we need to exclude when King is on board OR [Ace on board & King in hand]
= C(1,1)*C(46,1)*C(45,2) + C(1,1)*C(45,1)*C(1,1)*C(44,1) = 47,520

So that the subtotal of Queen high straight flushes per combo is
= 1,070,190 - 47,520 = 1,022,670

*3 combos = 3,068,010

Total for the case = 10,234,620

*4 suits = 40,938,480.

Case 4: Jack High Straight Flush

Case 4A: 87 in hand & JT9 on board

Need to exclude Queen in hand OR [both Ace & King in hand & Queen on board]

= C(1,1)*C(46,1)*C(45,2) + C(2,2)*C(1,1)*C(44,1) = 45,584

So that the subtotal of Jack high straight flushes is
= 1,070,190 - 45,584 = 1,024,606

Case 4B: 97 in hand & JT8 on board

Need to exclude Queen in hand OR [King in hand & Queen on board]

= C(1,1)*C(46,1)*C(45,2) + C(1,1)*C(45,1)*C(1,1)*C(44,1) = 47,520

So that the subtotal of Jack high straight flushes is
= 1,070,190 - 47,520 = 1,022,670

Case 4C:
J7 in hand & T98 on board
T7 in hand & J98 on board

Need to exclude Queen in hand OR [King in hand & Queen on board] OR [Ace in hand & both King & Queen on board]

= C(1,1)*C(46,1)*C(45,2) + C(1,1)*C(45,1)*C(1,1)*C(44,1) + C(1,1)*C(44,1)*C(2,2) = 47,564

So that the subtotal of Jack high straight flushes per combo is
= 1,070,190 - 47,564 = 1,022,626

*2 combos = 2,045,252

Case 4D: JT in hand & 987 on board

Need to exclude Queen on board

= C(1,1)*C(46,1)*C(45,2) = 45,540

So that the subtotal of Jack high straight flushes is
= 1,070,190 - 45,540 = 1,024,650

Case 4E:
J8 in hand & T97 on board
T8 in hand & J97 on board

Need to exclude Queen on board OR [Queen in hand & King on board]

= C(1,1)*C(46,1)*C(45,2) + C(1,1)*C(45,1)*C(1,1)*C(44,1) = 47,520

So that the subtotal of Jack high straight flushes per combo is
= 1,070,190 - 47,520 = 1,022,670

*2 combos = 2,045,340

Case 4F: 98 in hand & JT7 on board

Need to exclude Queen on board OR [Queen in hand & King on board] OR [both King & Queen in hand & Ace on board]

= C(1,1)*C(46,1)*C(45,2) + C(1,1)*C(45,1)*C(1,1)*C(44,1) + C(2,2)*C(1,1)*C(44,1) = 47,564

So that the subtotal of Jack high straight flushes is
= 1,070,190 - 47,564 = 1,022,626

Case 4G:
J9 in hand & T87 on board
T9 in hand & J87 on board

Need to exclude Queen on board OR [Queen in hand & both Ace & King on board]

= C(1,1)*C(46,1)*C(45,2) + C(1,1)*C(44,1)*C(2,2) = 45,584

So that the subtotal of Jack high straight flushes per combo is
= 1,070,190 - 45,584 = 1,024,606

*2 combos = 2,049,212

Total for the case = 10,234,356

* 4 suits = 40,937,424

Case 5: Ten High Straight Flush

Case 5A: T9 in hand & 876 on board

Need to exclude Jack on board

= C(1,1)*C(46,1)*C(45,2) = 45,540

So that the subtotal of Ten high straight flushes is
= 1,070,190 - 45,540 = 1,024,650

Case 5B: T8 in hand & 976 on board

Need to exclude Jack on board OR [Jack in hand & both King & Queen on board]

= C(1,1)*C(46,1)*C(45,2) + C(1,1)*C(44,1)*C(2,2) = 45,584

So that the subtotal of Ten high straight flushes is
= 1,070,190 - 45,584 = 1,024,606

Case 5C: T7 in hand & 986 on board

Need to exclude Jack on board OR [Jack in hand & Queen on board]

= C(1,1)*C(46,1)*C(45,2) + C(1,1)*C(45,1)*C(1,1)*C(44,1) = 47,520

So that the subtotal of Ten high straight flushes is
= 1,070,190 - 47,520 = 1,022,670

Case 5D:
T6 in hand & 987 on board
96 in hand & T87 on board

Need to exclude Jack in hand OR [Jack on board & Queen in hand] OR [King in hand & both Queen & Jack on board]

= C(1,1)*C(46,1)*C(45,2) + C(1,1)*C(45,1)*C(1,1)*C(44,1) + C(1,1)*C(44,1)*C(2,2) = 47,564

So that the subtotal of Ten high straight flushes per combo is
= 1,070,190 - 47,564 = 1,022,626

*2 combos = 2,045,252

Case 5E: 98 in hand & T76 on board

Need to exclude Jack on board OR [Jack in hand & both King & Queen on board] OR [both Queen & Jack in hand & both Ace & King on board] OR [both King & Jack in hand & both Ace & Queen on board]

= C(1,1)*C(46,1)*C(45,2) + C(1,1)*C(44,1)*C(2,2) + C(2,2)*C(2,2) + C(2,2)*C(2,2) = 45,586

So that the subtotal of Ten high straight flushes is
= 1,070,190 - 45,586 = 1,024,604

Case 5F: 97 in hand & T86 on board

Need to exclude Jack on board OR [Jack in hand & Queen on board] OR [both Queen & Jack in hand & both Ace & King on board]

= C(1,1)*C(46,1)*C(45,2) + C(1,1)*C(45,1)*C(1,1)*C(44,1) + C(2,2)*C(2,2) = 47,521

So that the subtotal of Ten high straight flushes is
= 1,070,190 - 47,521 = 1,022,669

Case 5G: 87 in hand & T96 on board

Need to exclude Jack on board OR [Jack in hand & Queen on board] OR [both Queen & Jack in hand & King on board]

= C(1,1)*C(46,1)*C(45,2) + C(1,1)*C(45,1)*C(1,1)*C(44,1) + C(2,2)*C(1,1)*C(44,1) = 47,564

So that the subtotal of Ten high straight flushes is
= 1,070,190 - 47,564 = 1,022,626

Case 5H: 86 in hand & T97 on board

Need to exclude Jack in hand OR [Queen in hand & Jack on board] OR [both Ace & King in hand & both Queen & Jack on board]

= C(1,1)*C(46,1)*C(45,2) + C(1,1)*C(45,1)*C(1,1)*C(44,1) + C(2,2)*C(2,2) = 47,521

So that the subtotal of Ten high straight flushes is
= 1,070,190 - 47,521 = 1,022,669

Case 5I: 76 in hand & T98 on board

Need to exclude Jack in hand OR [both King & Queen in hand & Jack on board] OR [both Ace & King in hand & both Queen & Jack on board] OR [both Ace & Queen in hand & both King & Jack on board]

= C(1,1)*C(46,1)*C(45,2) + C(2,2)*C(1,1)*C(44,1) + C(2,2)*C(2,2) + C(2,2)*C(2,2) = 45,586

So that the subtotal of Ten high straight flushes is
= 1,070,190 - 45,586 = 1,024,604

Total for the case = 10,234,350

* 4 suits = 40,937,400

Case 6: Nine High Straight Flush

Same as Ten high straight flush (the ranks can no longer "reach" higher straights)

= 40,937,400

Case 7: Eight High Straight Flush

Same as Ten high straight flush (the ranks can no longer "reach" higher straights)

= 40,937,400

Case 8: Seven High Straight Flush

Same as Ten high straight flush (the ranks can no longer "reach" higher straights)

= 40,937,400

Case 9: Six High Straight Flush

Same as Ten high straight flush (the ranks can no longer "reach" higher straights)

= 40,937,400

Case 10: Five High Straight Flush

Here we need to start with the tally from the Ten high straight flush. However, here we need to exclude a few strange cases.

Consider 54JT in hand and 32AKQ on board (all the same suit). We see that in addition to the 5432A steel wheel straight flush, the player also makes a royal flush (AKQJT). Of course, the player is credited with the royal flush (the higher ranked hand).

It is straightforward to see that the 5432 can be divided up in C(4,2) ways where two of the cards go to the hand and the other two cards go to the board. Similarly, there are C(4,2) ways to divide up KQJT where two of the cards go to the hand and the other two cards go to the board. Clearly, the Ace needs to be on the board as well.

In total, then, we need to exclude:

= 4*C(4,2)*C(4,2) = 144

Making the total of Five high straight flushes

= 40,937,400 - 144
= 40,937,256

Summary

The following table summarizes the results reported above.

Straight FlushNumber of Deals
Ace High (Royal Flush)
42,807,600
King High
40,986,000
Queen High
40,938,480
Jack High
40,937,424
Ten High
40,937,400
Nine High
40,937,400
Eight High
40,937,400
Seven High
40,937,400
Six High
40,937,400
Five High (Steel Wheel)
40,937,256
GRAND TOTAL
411,293,760

You will see that the grand total of all deals containing a straight flush (including a royal flush) by a single player in Omaha is given by 411,293,760.

For perspective, remember that there are a grand total of C(52,9)*C(9,5) = 463,563,500,400 total possible Omaha deals to a single player.

The 411,293,760 is the figure given by the Wizard of Odds. The method employed by Brian Alspach and presented in detail earlier in this thread did not notice that there are 144 deals that actually make two different straight flushes (both a steel wheel and a royal flush). Alspach should have deducted these 144 since his method double counted them as two different cases of five consecutive cards making up a straight flush (AKQJT and 5432A).

I fervently hope that this answers the question once and for all.

 11-16-2017, 03:51 AM #47 whosnext Pooh-Bah     Join Date: Mar 2009 Location: California Posts: 4,853 Re: Straight flush odds in PLO (with poker prob primer) I recently became aware again of a good tutorial thread over in the Beginners Questions forum explaining some basic poker probability concepts. I hope they won't mind if I place a link to the thread below: https://forumserver.twoplustwo.com/3...poker-1164777/

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