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?
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?
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