I'm kind of new to PL Omaha. I was just wondering if anyone knew the probability of seeing two straight flushes in 5 hands, on a 10 handed table?

And if you can instruct me on how to determine those odds, I'd be even more appreciative.
Thanks in advance
Regards

This type of question comes up quite often around here. There are several different but related possible interpretations. Here are a few:

(1) What is the prob of seeing 2 SF in exactly 5 hands?

(2) What is the prob of seeing 1 SF in 100 (?) and then another SF in the following 4 hands?

(3) What is the prob of seeing 2 SF in 5 consecutive hands over a lifetime of 100,000 (?) hands?

Etc.

You can see that the answers to these questions will be quite different.

Possibly it makes sense to start with the fundamental issue in your post. What is the prob of someone at a 10-handed Omaha table of getting a straight flush on a single hand? You can search for previous threads here on 2+2 dealing with this question.

If OP or anyone wants to delve into this, please go for it!

I think OP is asking 1: what are the combined chances of the 10 players to make a straight flush twice in 5 hands.

This would be 1- chance of a single player not making a straight flush twice within 5 hands^10?

edit: no that way wouldnt work right? that would be chance that 1 player out of 10 makes a straight flush twice within 5 hands.
so im lost.

is it additive? meaning would the chance of the players combined be the chance of one player times 10?

sorry folks, I may have not been clear enuff?

"What is the prob of seeing 2 SF in exactly 5 hands?" Is what I'm asking. I will try to figure out how to search the forum for this.
Thanks for the replies

My analogy is that yesterday I played NLHE and on back-to-back hands I was dealt pocket tens. Wow, that's gotta be pretty rare, right?

Well, I would have been as "surprised" if it was any pocket pair that I got dealt back-to-back. So that makes it less rare.

And I played, say, 1,000 hands yesterday. So there were many possible couplets of hands in which the duplicate pairs could have been dealt. Which also makes the phenomenon less rare.

This forum has many purposes. One is to help people calculate probabilities they seek. Another is to to help people think "better" about probabilities.

My reply above was related to this second rubric.

whoisnext has a good point. Your question can be interpreted in different ways. And each way will give a different result.

I think the point he's really trying to drive home is that in reality seeing a straight flush twice in five hands is not as unlikely as it might seem because the more hands you play the more likely you are to see strings of straight flushes.

I'm very curious how to solve this in a vacuum. For example: I'm going to deal 5 hands to 10 people and never deal another hand ever again. What are the chances that I deal 2 straight flushes?

Was I making any sense in post #3?

Not quite. More than one player can get 2 SF's in 5 hands, and what you describe would overcount those cases. It would be a good approximation of the chance of at least one player getting 2 SF's, but I thought you were trying to calculate the chance of 2 SF's occurring period? E.g. if two different villains each only get one, that should count. It can even happen in the same hand.

You are correct. I see what you mean. I guess that's where it gets confusing.

-scaling from the chance of one to ten players without over counting.
-accounting for the fact that two players can each get a straight flush at once.
-finding the chance of this happening twice between all ten players instead of twice for one player.

It seems like the question would be much easier if it were "how often will I deal AT LEAST ONE straight flush in five hands between ten people?" This way we could treat it similar to a poker hand. Like asking "if i shove, how often will I get called with n players behind if each is calling x"
1 - (1-x)^n
right?

In this case x would be chances of one player making a straight flush once in five hands. n would be the number of players (10).

I want to do that for OP's problem but like I said in post #3 that won't work because OP's question is asking how often it will happen TWICE within five times between ten players instead of asking how often it will happen at least once in five times between ten players.

I'm just trying to explain my thought process here so you can point in the right direction and point out where my understanding might be incorrect.

Hopefully it will light the way for OP too

Thanks for helping out.

As people above have stated, it is important to determine which elements are "dependent" and which elements are "independent".

As mentioned above, one player having (or not having) a straight flush is NOT independent of another player at the same table having (or not having) a straight flush on the same deal. Of course, they share the board cards which make any player more or less likely to make a straight flush. And they can "block" each other as well. So let's come back to that issue.

Let's suppose we derive the probability of at least one player having a straight flush on a random deal of a ten-player Omaha table (either by exact formula, estimation, simulation, etc.). Call this P10.

Then it is clear that the prob of seeing straight flushes in two different deals in a string of five deals at that table is exactly given by the Binomial formula since deals ARE independent.

Based upon P10, you could simply use any Binomial calculator to find the prob of seeing 0, 1, 2, 3, 4, or 5 straight flushes in a 5-deal set at that table.

-----

Here are rough estimates of these key figures based upon simulations.

Prob of 1 player getting a straight flush in Omaha is around once in 1,125 deals.

Prob of seeing at least one player having a straight flush on a 10-player Omaha table is around once in 113 deals.

Using the 113 figure in a binomial calculator shows:

Prob of seeing at least two different deals with straight flushes in a string of five deals at a 10-player Omaha table is around once in 1300 five-deal sets.

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 would instead figure out the chance of at least one player making a SF in one hand, and then let n = #hands (5). Then that formula would give the exact answer because the different hands are independent.

Not really.
P(at least one) = 1 - P(none)
P(at least 2) = above - P(exactly one)

Let q1 = P(none in 1 hand)
Then q5 = P(none in 5 hands) = (q1)^5
and P(exactly 1 in 5 hands) = 5(p1)(q1)^4
where p1 = P(exactly 1 in 1 hand)

The least difficult way to get p1 might be to break it down into the different types of boards. That is, take the sum of [P(board type x) * P(exactly 1 SF | 10 players using board x)]

P(exactly 1 SF | board X) isn't too hard. Often it's only possible for 1 player to have a SF, so you can just take one player's probability and multiply by 10. Sometimes 2 people can have one, in which case you use a simple application of PIE and subtract 2*P(sf over sf).

I don't know if there's a clever direct way to calculate q1, so I'd probably do 1 - P(at least 1). The calculation of P(at least 1) is identical to P(exactly 1) except we only subtract P(sf over sf) once instead of twice.

Phew a lot to take in. I think this is slowly starting to make sense. Yay progress!

Very helpful pointing out the importance of distinguishing between dependence and independence.

I wanted to use this "binomial calculation" on the probability of ONE player making a SF (1/1125 according to sim) instead of using it on P10 (1/113) like you've shown. Thinking I could then expand on that probability to find out how likely it is for ten players. But that clearly won't work because of the dependence issue.

What in the world would that formula give?

I've hit a snag on the way you find P10. It appears you went from the probability of one player making a straight flush in a single hand to the probability of one in ten players making a straight flush in a single hand (P10) by adding the probabilities together? I thought you couldn't do that for the same reason; they are dependent. So, can't add them together.

FYI

I think I understand the overarching principle here. Fuzzy on the details with all the p's and c's and pie's. I don't really know any of the fancy symbols. I know + - / * ^ < > = and when a letter is being used to represent a variable. Does P() mean probability of ()?

You're saying that with 72 you have reverse blockers so it increases the combos of AA left out there. And for each player that does not have AA the chances of the next person having AA increases slightly?




So that would give us the chances of one player making a straight flush in 5 hands right?

Ohhhhhh so it just happens to be about 10x that of one player?

You don't arrive there by adding them together?

Yes, because the error of saying 10x is small. The over-counted cases have tiny probabilities.

Yes P(something) means probability of something.
P(A | B) means "probability of A, given that B is true"
C(x,y) = the number of ways to pick a subset of y objects from a set of x objects.

Yes, which is why 1-q^n is only an estimate (albeit a rather good one in that case). Specifically, it's the lower bound, while the additive estimate is the upper bound. And I forgot to mention I was talking about hold'em, though I'm sure you gathered that.

Yep, the chance of at least one player making a SF in 5 hands or fewer.

cita, you are asking a lot of really good questions (here and in the other thread). Welcome to the Probability Forum.

You will find many posters here who are extremely generous with their time in helping other posters by answering specific questions or by being generally helpful and informative. heehaww is among that group of great Probability Forum posters.

If you would like to refresh your own understanding of probability concepts in a systematic manner, there are several great places to do so online.

As I said above, welcome.

citamgine
"It seems like the question would be much easier if it were "how often will I deal AT LEAST ONE straight flush in five hands between ten people?"


That's probably closer to what I should have said, sorry, 1st time in a Prob thread, didn't think of all the scenarios.

Maybe, "how often will I deal a straight flush to ONLY ONE player in ONLY five hands dealt between ten people?" Additionally, it was 2 different players in each instance, so I'm guessing those odds would differ from the SF's being dealt to the SAME player, making it 2 separate equations.

Furthermore, the equations you guys are giving are so above my ahead for some reason, maybe it's because I don't know what the letters/variables stand for? I'm kinda like citamgine here, I know symbols lol.

Wanna say thanks again for all the input, didn't mean to be so much trouble

Feel like I need to brush up on my Algebra for Dummies book

I should have said "Welcome to the Probability Forum" to Betta as well!

If you have questions on any of this, including on any answers given above, do not hesitate to ask.

Seriously, I cannot emphasize this enough, all questions are welcome. No matter at what "level".

If a member does not know what C(52,5) means, for example, they should certainly ask.

Most probability concepts are rather basic and I truly believe that they can be explained and understood by all 2+2 poker enthusiasts.

Welcome!

Last night I ran new simulations on this topic. As per usual, I hope that my simulation code is correct and that the random number generator that I am using is "accurate". Of course these simulations assume that each player goes to showdown in each deal.

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.

If we are interested in sets of five deals, the binomial formula tells us that a single player will get one or more straight flushes around once every 225 sets of five deals, and two or more straight flushes once every 126,836 sets of five deals.

For a 10-player table:

In 100,000,000 deals of Omaha, 881,282 deals "saw" a straight flush (by any player or players at the table) at a 10-player table, which works out to around one in every 113.5 deals.

If we are interested in sets of five deals, the binomial formula tells us that one or more straight flushes will be seen around once every 23 sets of five deals, and two or more straight flushes will be seen around once every 1,311 sets of five deals.

(Note: all of these simulations are including royal flushes.)

Egads, I now realize that I was not keeping track of the number of deals in which multiple players at the 10-player table made a straight flush with the same board. Of course, this can be inferred from the tallies posted above, but I want to explicitly keep track of this.

As heehaww mentioned above, P10 will be lower than 10*P1 due to the possibility of two players making a straight flush on the same deal. If the board is KhQh7s6s5s, then one player could hold 9s8s and another player could hold 4s3s. (Note: It is not possible for three people to have a straight flush on the same Omaha deal, right?)

So I just kicked off another simulation to report this explicitly. Here are the results of 1 Million Omaha deals to a 10-player table. I will update the results when the simulation ultimately reaches 100 Million.

Players having SF	Tally
Zero	991,185
One	8,774
Two	41
Three	0
Four	0
Five	0
Six	0
Seven	0
Eight	0
Nine	0
Ten	0
GRAND TOTAL	1,000,000

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.

P(whosnext and heehaww being THE NUTS)

hope I did that right.

Huge thanks for helping out so much. Definitely appreciate you guys taking the time out to give such comprehensive explanations. The fog is lifting!


Yeah I believe that is true. There is less of a chance for the straight flushes being dealt to a specific player, rather than to any of the ten players. Take what I say with a grain of salt, but it seems like with probabilities the more specific you get the less likely the chances get. So, it would be two different equations. Although it looks like you can build on the same information to solve both. Same ingredients different recipes.



Hmmm ahhhh this is what you were showing here?

Does this concept apply only to dependent probabilities?

I'm a little confused about when its appropriate to add probabilities instead of raising their inverse to the nth then subtracting that from 1.

For example: we can find out exactly how likely we are to win a coin flip if we flip a coin twice by doing the ol 1-.5^2. This would be the same as two people flipping a coin once each right? Both examples are indys.
Adding the probability together wont work here.

I feel like adding would work, for instance, in calculating the combined chances of anyone being dealt a specific card:
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.

If my understanding of heehaww's explanation is correct then C(52,5) would be the number of different possible flops. wow!

I played around with this a little further for hand combos and ran into a little hiccup.

It works for pairs: Combos of AA= C(4,2)= 6
sets: combos of AA on board of Axx= (3,2) = 3

But doesnt work for unpaired hands.
Combos of AK = C(8,2)=28
There is only 16 combos of AK. Whats going on here?

Thanks again!

And wowww, whosnext. It looks like the chances are way more likely than I expected. Where do you learn to build these sims?

I will reply more later, but ...

C(8,2) would include all pairs of aces or kings in addition to AK.

Using combination notation, the number of AK hands would be:

C(4,1)*C(4,1) = 4*4 = 16

C(8,2) is the number of ways of choosing 2 of 8 cards. That would include AhAs or KhKs for example, which you do not want. The correct expression for AK would be 1 from each rank or C(4,1)*C(4,1) or simply 4*4=16.

BTW, the number of possible flops is not C(52,5) since a flop only has 3 cards and the deck size depends on what you assume is given.

ahah! yes that makes sense. dont know how i didnt c that. get it...c that

Yes i meant boards, not flops. C(52,3) for flops. And tht would not be taking into consideration cards removed for #players dealt in?

Here is a brief primer on some of the things you asked about. Terminology is important in probability to help keep things straight. Terms such as trials, events, and outcomes are important to distinguish.

Let's work through a few simple ideas.

(1) Suppose you have an event that has exactly two mutually exclusive outcomes. Call them "success" and "failure" (or "heads" and "tails" in a coin flipping scenario). Suppose you are interested in a series of N independent trials. The probability of success in each trial is given by p.

Then, you may be interested in finding the probability of getting any (one or more) successes in the N trials.

Prob(any success in N trials)
= 1 - Prob (no successes in N trials) [this is just logic]
= 1 - Prob (all failures in N trials) [logic]
= 1 - {(1-p)^N} [this relies upon independence of trials; see number 3 below]

(2) For two events A and B, the following must hold:

P(A or B) = P(A) + P(B) - P(A and B)

Clearly, if A and B are mutually exclusive (meaning that they both cannot occur simultaneously), then P(A and B) = 0. If that is true, the above formula collapses to P(A or B) = P(A) + P(B).

So it is correct to ADD probabilities of mutually exclusive events to obtain the probability of EITHER event occurring.

(3) If two events are independent, then the probability of them BOTH occurring is the PRODUCT of the individual probabilities.

----------

Many poker examples could be given. Let's do some simple ones.

(1) Prob of being dealt a pair of aces in holdem is 1/221 (same as for any pair). Suppose you want to figure out the prob of getting dealt a pair of aces over the course of a series of 50 deals (meaning one or more pairs of aces). Then, from above, it is simply 1 - [(1 - 1/221)^50] which is around 20%.

(2) Using a standard deck of cards, what is the probability that a card chosen at random is either a spade or a diamond? Of course, a card cannot be both a spade and a diamond, so this is simply prob(Spade) + prob(Diamond) = 1/4 + 1/4 = 1/2.

Now, suppose you want to know the probability of a card being either a spade or a face card. Here the events are not mutually exclusive so we have to take care. P(Spade or Face) = P(Spade) + P(Face) - P(Spade & Face). This is clearly = 13/52 + 12/52 - 3/52 = 22/52.

(3) Suppose you and 8 friends are about to play NLHE. You would like to know the probability of getting dealt pocket aces in the big blind in the very first hand dealt at the table. Clearly these two events (getting dealt pocket aces and being the big blind in the very first deal) are independent.

So to figure out the probability of both events occurring, we simply multiply their probabilities. Prob(getting dealt aces) = 1/221. Prob(being big blind in very first deal) = 1/9. So the prob of getting dealt aces in the big blind in the very first deal at the table is (1/221)*(1/9) = 1/1989.

These are but a few of the "fundamental" relationships that are used over and over in probability settings. It is sometimes a challenge to remember them or to keep them straight. But after awhile it becomes easier.

Hope this helps.