Math Geek Help Needed - Poker Odds for Monotone Flop? (1 Viewer)

PokerDogDoc

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Attention all math and stat geeks!
I need your help!

I play a friendly home game on PokerStars and notice way more monotone (all 1 suit) flops than I'd expect or have previously calculated.
I've often counted 19-22 of them over a 3.5 hour game consisting of a couple of hundred hands.
I've also seen many more flopped flush-over-flush hands that I'd expect (i.e. 2 players both having suited hole cards that flop a flush).
;
So I wanted to re-run the odds of this and double-check them with a math/stat/probability minded PCF'r as this can really affect strategies when holding these cards and seeing these flops.

Please see below and let me know your thoughts... and please explain the math if different.

My math logic for a monotone flop:
-4 suits available, choose 1 suit
-13 cards of that specific suit, choose 3 of those cards
-divide that by the total possible flop combinations... 52 cards, choose 3 cards

Stats/math equation:
C(4,1)*C(13,3) / C(52,3)
= 4*286 / 22,100
= 1,144 / 22,100
= 0.052 or 5.2% or roughly 1 out of 19 flops (odds of 18:1) will be monotone for any suit

So for my PokerStars play... when I see 20 monotone flops over 200 hands (10%) that's double the number expected...
... yes, I know that is a relatively small number of hands, but I think that probability way exceeds any standard deviation model.

NOTE: I feel pretty good about the above calculation... but read on!

I also wanted to calculate the odds of a monotone flop matching my suited hole cards... i.e. I have 2 hearts, so what are the odds that the flop will be all hearts?

Math logic:
-only 1 suit available that matches mine
-I know I have 2 cards of that suit, leaving only 11 cards available to be used for the flop
-of those remaining 11 cards, choose 3 for the flop
-divide that by the total remaining flops available... but I think that is choosing 3 cards out of only the 50 remaining cards (since I know 2 of them aren't available b/c they are my specific hole cards)

Equation -- 50 cards for flop:
C(11,3) / C(50,3)
=165 / 19,600
= 0.0084 , or 0.84% , or roughly 1 out of 119 flops (odds of 118:1) will be a monotone flop matching my suited hole cards (i.e. flop a flush)

And the odds of 2 players flopping a flush are the odds of each individual doing it, multiplied together... yes?
= 0.0084 * 0.0084
= 0.000064 , or 0.0064% , or roughly 1 out of 15,625 times (15,624:1) when 1 person flops a flush will a 2nd player also have flopped a flush

I'm definitely seeing way more than that rate of flopped flush-over-flush hands on PS!

Thoughts?
Is my math off?
Is the PS algorithm not truly random and does this to promote crazy play?
 
Last edited:
I am not super into stats, but I am pretty decent at math.
The way I would do this for the first example is (1)*(12/51)*(11/50)=5.2% => exactly what you said.
Example 2 (suited hole cards):
(11/50)*(10/49)*(9/48)=0.84% => exactly what you said.
Example 3 (2 players flopping a flush):
This is where things get tricky. In example 2, we assume the player was dealt a suited hand. If we factor in the likelihood of being dealt suited cards (1*(12/52)=23.5%), there is only a 0.19% (23.5% * 0.84%) chance of flopping a flush on any given hand. If we make the assumption that 2 players were both dealt a suited hand (same suit), then the likelihood they flop a flush will be (9/48)*(8/47)*(7/46)= 0.49%. If we factor in the likelihood of 2 players being dealt suited cards (same suit) to begin with (1*(12/51)*(11/50)*(10/49)= 1.05%) then the odds of 2 players being dealt the same suited cards and flopping a flush would be 0.0051%(1.05%*0.49%) which is very close to your answer.
I am not sure if I’m right, but our answers are very close.
With all of that being said, 200 hands is WAY too small of a sample size to mean anything. I remember seeing a while back that PokerStars uses some kind of lasers to randomize and their RNG is allegedly very good at being random.
 
So for my PokerStars play... when I see 20 monotone flops over 200 hands (10%) that's double the number expected...
... yes, I know that is a relatively small number of hands, but I think that probability way exceeds any standard deviation model.
I don't agree, you'll need to show me that bell curve. And did you count 19-22, or did it feel like that? Same question with hands.

I'm not meaning to be argumentative, but when we're talking about statistics and standard deviations it matters. We're already working with a very small sample size that I suspect isn't actually counted, so no I wouldn't put weight in this.
 
"And the odds of 2 players flopping a flush are the odds of each individual doing it, multiplied together... yes?
= 0.0084 * 0.0084"

That is only correct in a Heads Up Situation
 
I completely agree that all of this math only stands when calculating heads up and the likelihood of flush over flush must go up with more players, but do you know by how much? Is it as simple as 2 opponents doubles the likelihood and 3 triples it, etc.?
 
I completely agree that all of this math only stands when calculating heads up and the likelihood of flush over flush must go up with more players, but do you know by how much? Is it as simple as 2 opponents doubles the likelihood and 3 triples it, etc.?
Actualy I think it is not even correct for heads up.

One thing that I have learned during stochastic class is that the human brain is not made for stochastic. Realy!
 
"And the odds of 2 players flopping a flush are the odds of each individual doing it, multiplied together... yes?
= 0.0084 * 0.0084"

That is only correct in a Heads Up Situation

Actualy I think it is not even correct for heads up.

One thing that I have learned during stochastic class is that the human brain is not made for stochastic. Realy!
I think his 0.0084^2 is wrong as I explained in a long winded previous reply. But assuming we get to the right number, how do the odds change by adding extra opponents?
 
It depends on your starting assumptions.

If we don't assume our hole cards, the odds of a flopped flush are (odds of monotone hand)*(odds of matching monotone board).
(12/51)*(11/50*10/49*9/48) = .00198

Odds two players heads up flop a flush are (odds of monotone hand)*(odds of matching monotone hand)*(odds of matching monotone board).
(12/51)*(11/50*10/49)*(9/48*8/47*7/46) = .000051

If you assume beforehand that you have a suited hand the odds are a little different. Then the odds you flop a flush are
(11/50*10/49*9/48) = .0084

If we assume both players have suited hands the odds of a flush are
(9/48*8/47*7/46) = .00485

If we only assume our hand is suited, but don't assume our opponents hand then the odds we both flop a flush are
(11/50*10/49)*(9/48*8/47*7/46) = .000218

These are all just variations on 'what are the odds the X random cards from the deck are the same suit', where we can change what X is by assuming what the already dealt cards are.
 
I am not sure if I’m right, but our answers are very close.
With all of that being said, 200 hands is WAY too small of a sample size to mean anything. I remember seeing a while back that PokerStars uses some kind of lasers to randomize and their RNG is allegedly very good at being random.
SE for binomial with p 5.2% with sample size of 200 is ~5% so you’re about 1 std dev away, not that weird
- I agree... super small sample so statistical significance is really hard to apply here... and thanks for the SD calculation.


I am not super into stats, but I am pretty decent at math.
The way I would do this for the first example is (1)*(12/51)*(11/50)=5.2% => exactly what you said.
-Your longhand version of the equation is mathematically quivalent to the "shorthand" COMBINATIONS mathematical function "C"... they are the same thing, so we got the same answer.


I don't agree, you'll need to show me that bell curve. And did you count 19-22, or did it feel like that? Same question with hands.

I'm not meaning to be argumentative, but when we're talking about statistics and standard deviations it matters. We're already working with a very small sample size that I suspect isn't actually counted, so no I wouldn't put weight in this.
-I have actually begun counting the monotone flops once I began noticing it. The number of hands dealt is also listed in the stats section on PS.
-And I agree that it's hard to apply a confidence level to a small number of hands... kinda like flipping a coin is rarely 50-50 unless you do it a gazillion times.


all this flopped flush over flush calcs…we are talking about poker where ranges aren’t random

SRP CO v BB have a lot of suited combos

EP v MP 3! where MP has AQhh, EP basically never has XXhh.
-Whether you're in the hand or not doesn't change the fact that you were dealt 2 same-suited hole cards and the flop came monotone.
Now, whether or not you're in the hand is a different story and is based on ranges, player, etc., as you pointed out.


"And the odds of 2 players flopping a flush are the odds of each individual doing it, multiplied together... yes?
= 0.0084 * 0.0084"

That is only correct in a Heads Up Situation
-I believe your point is that (for example) 9 players remove 18 cards from the deck leaving 34 cards for the flop, whereas 2 players only remove 4 cards, leaving 48 cards for the flop.... so in heads up there is less of a chance that cards of the desired suit are in somebody else's hand and not available.

If so, I would respectfully disagree on that. Although it is inherently understood that the number of hands dealt impacts the number and type of cards available for the flop, that isn't typically how odds are calculated.

But we don't really calculate the odds that way when playing live unless we KNOW ALL the cards in play. That can be only be done when we watch it on TV with the use of RFID readers... the audience knows exactly which cards are still availabe for the flop vs. in someone else's hand. At the live table we stick to the old "8-4-2" formula for calculating outs and odds for flop, turn, and river and understand that it's close enough.


If we only assume our hand is suited, but don't assume our opponents hand then the odds we both flop a flush are
(11/50*10/49)*(9/48*8/47*7/46) = .000218
-I see what you did here and where I went wrong.. I forgot to reduce the number of combinations for the Villain to have 2 of the same suited cards and number of cards left in the stub.

So, knowing that my hole cards are suited (thus only 11 other cards of the same suit that he could possibly have), my logic and Combination shorthand that the Villain has BOTH 2 hole cards matching my suit AND would thus flush a flop with me would be...

Combinations that Villain's suited hole cards of the same suit as me (2 cards of my suit vs. any 2 cards from the entire 50 card stub):
C(11,2) / C(50,2)
= 55/1225
= 0.045

Combinations of flopping a monotone flop the same suit as our cards using the 9 remaining suit cards (3 cards of our suit vs. the entire 48 card stub remaining):
C(9,3) / C(48,3)
= 84/17,296
= 0.00486

Combinations of doing BOTH:
= 0.045 * 0.00485
= 0.000218 ... same number you got... so 0.022% of the time (4545:1 odds) we'd both flop a flush when I flop a flush


Thanks for helping me out on this everyone!
 
By the way, if anyone wants to use an online Combinations Calculator, here's the link:

Combinations Calculator

You'll see the format is C(n,r)... where "n" is the total number of objects possible and "r" is how many you are choosing of those objects.
 

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