Most flops are.....

Rainbow = All three different suits
Two tone = Two of the same suit
Monotone = Three of the same suit

Which happens the most often?

1, rainbow...2 two tone...3.monotone

way i see it is you have 3 cards coming out and 4 suits, it wouldnt make sense for it to be MORE likely that one of those 4 suits having 2 or 3 cards come out...pretty blazed so not sure im making sense but yea im sticking with rainbow

feel like the right answer is "it depends",

well if we think about it like drawing 3 balls out of a hat containing 52 balls and theres 4 different color balls we can pretty easily determine that monotone will happen least often based on random probability. i feel like all different suits or colors is not as unlikely as all the same color but not as likely as a mix of say color x and color y. im sure theres some sort of equation to figure this out but im pretty bad at setting up equation stuff. we could also run an experiment over a large M (meaning the # of times you run the exp) by just selecting 3 balls at random from the hat..we could also just do this with a deck of cads lol

Originally Posted by thelorax

im sure theres some sort of equation to figure this out but im pretty bad at setting up equation stuff. we could also run an experiment over a large M (meaning the # of times you run the exp)

there certainly is and you certainly could. probably faster to just make the equation though

[16:24] <pdk1010> feel like theres alot of variables that go into it though too
[16:24] <philly> not really :-/
[16:24] <pdk1010> like how many players get dealt cards
[16:24] <philly> you still dont know those players cards
[16:24] <spoonitnow> that doesn't matter at all
[16:24] <philly> theres still 50 cards you dont know
[16:24] <philly> before the flop[
[16:26] *** speedAFK is now known as speedcake
[16:26] <pdk1010> strictly from a math standpoint i think it would matter
[16:26] <pdk1010> could be wrong though
[16:26] <spoonitnow> it doesn't change the answer
[16:27] <bhaley66_work> as far as your concerned, those cards are in the deck
[16:28] <bhaley66_work> you have no idea what they are, so why does it matter if they have them or the deck?

its prolly an insignificant amount but a mathematician that studied probabilities would chime in about how many cards were dealt preflop as well......but im not a mathematician that studies probabilities so who knows.

Originally Posted by JKDS

there certainly is and you certainly could. probably faster to just make the equation though

do you know how to go about setting the equation up? im kinda curious/wanna play around with it.

Originally Posted by thelorax

do you know how to go about setting the equation up? im kinda curious/wanna play around with it.

I'll show it after this thread gets some action.

Originally Posted by spoonitnow

I'll show it after this thread gets some action.

ok fair enough, ill try to think about it some more in the meantime.

Okay so real quick, what's the chance if we just deal a flop without seeing any hole cards that it comes monotone?

When the first card comes, it doesn't matter what the suit is, so 52 out of 52 cards can make a monotone flop. The chance of this happening is 52/52.

When the second card comes, it has to match the same suit as the first card. That's 12 possible cards out of the 51 cards remaining. The chance of this happening is 12/51.

When the third card comes, it has to match the same suit again. That's 11 possible cards out of the 50 cards remaining. The chance of this happening is 11/50.

So to get the chance of all three happening, we can multiply the three together. So we do (52/52) * (12/51) * (11/50) = 0.0518, or about 5.18%.

ohh ok so for rainbow can we go (52/52)x [(51-12)/51)]x [(50-11)/50)] wich gives us like 61%? wait but then my initial theory of 2tone being the most common would be wrong

Originally Posted by thelorax

ohh ok so for rainbow can we go (52/52)x [(51-12)/51)]x [(50-11)/50)] wich gives us like 61%? wait but then my initial theory of 2tone being the most common would be wrong

Nope. Right idea but you messed it up. Think about your equation harder.

ohhh eff we need to make the last part 50-25/50 or something for rainbow huh?

wich gives us like 39% and if thats right it means my 1st equation was correct for 2tone! which makes sense (i think). but 39+61 doesnt leave ~5% left for monotone so i effed something up :\
EDIT: so maybe my 1st equation isnt right for 2tone

so mono=5.18%
2tone=1 x (51/51) x (24/50) = 48%<---------winner!
rainbow= 1 x (39/51) x (26/50) = 39.8%

if im right about 2 tone every card left in the deck can come out as the second card on the flop and we still have the possibility of having a two tone board on the 3rd card so ----> 1 x (51/51) x (24/50) = 48%


actually this isnt right either i forgot the third part of the equation changes if the same suit comes out for the second card

card A : 52/52
card B: M: 12/51 R: 39/51
card C: M: 11/50 R: 26/50

crap, I'll be back when I'm done @ the tables.. got same

39% for rainbow as previous....
We need to weight card B because it affects our wanted outcomes for card C
ok so if card B is same suit as A which is P (12/51)

then card c must be any of the 39 offsuit cards left of 50

if card B is different suit from A which is P (39/51)
then card c must be any of the 25 (no... 26) cards which are offsuit from both previous.

(12/51)*39/50 = .1835
(39/51)*26/50 = .385

sum of 56.58% for 2 tone.
roughly 39.8% for rainbow
remaining 5.2ish for mono

= really close to 100%.. Windows calculator sucks.. also. I miscounted 1 card. aah, screw it, really close.. miss counting something, u get the idea though
yea ez way is 1-P(r)-P(m) where P(r) is probably of rainbow/mono..

Originally Posted by pdk1010

feel like the right answer is "it depends",

Yes, it depend on
a) which cards are missing from the deck (as in you have a 50 card deck cuz your dog ate 2 of the s .
b) if you have AA on jokerstars the flop will come monotone of a suit you don't have.

Originally Posted by pdk1010

so mono=5.18%
2tone=1 x (51/51) x (24/50) = 48%<---------winner!
rainbow= 1 x (39/51) x (26/50) = 39.8%

if im right about 2 tone every card left in the deck can come out as the second card on the flop and we still have the possibility of having a two tone board on the 3rd card so ----> 1 x (51/51) x (24/50) = 48%

Your numbers have to add up to 100%, otherwise you have an error somewhere. So, you have an error somewhere (two-tone is actually 55.06%).

Spoon already showed how to calculate the monotone flop. Your rainbow calculation is right, too.

Your two-tone value is wrong. The easy way to figure it out is to just do 1 - (5.18% + 39.8%), which assumes that you have accounted for all possible outcomes with "mono, rainbow, 2tone".

More rigorous approach: calculate the probability that the 2nd and 3rd card contain only one of the same suit as the first card, and add it to the probability that the 2nd and 3rd card are the same suit but different from the first card's suit.

It's been a long time since stats class, but it's a lot more fun when calculating stuff for poker than doing homework.

The probablilty of a two tone flop is 0.5506

The first card dealt can be any card, so the chance of dealing this is 52/52.

For a two tone flop the second card has to match the suit of the first. 12 cards do this, so the chance is 12/51.

The third card can't match the suit of the first two or there would be a monotone flop. There are 39 cards that do not match the suit of the first two, so the chance is 39/50.

So the probablility of a two tone flop is 52/52 * 12/51 * 39/50 EXCEPT that this just calculates the probability of a flop where the first two cards have the same suit. (ie SSX) We also need to include SXS and XSS flops too.

So the probablility of a two tone flop is 52/52 * 12/51 * 39/50 *3 = 0.5506 = 55.1%

Originally Posted by NightGizmo

Your numbers have to add up to 100%, otherwise you have an error somewhere. So, you have an error somewhere (two-tone is actually 55.06%).

Spoon already showed how to calculate the monotone flop. Your rainbow calculation is right, too.

Your two-tone value is wrong. The easy way to figure it out is to just do 1 - (5.18% + 39.8%), which assumes that you have accounted for all possible outcomes with "mono, rainbow, 2tone".

More rigorous approach: calculate the probability that the 2nd and 3rd card contain only one of the same suit as the first card, and add it to the probability that the 2nd and 3rd card are the same suit but different from the first card's suit.

It's been a long time since stats class, but it's a lot more fun when calculating stuff for poker than doing homework.

Yea i realized my math was RAH tarded after i posted that

Originally Posted by Duffryn

So the probablility of a two tone flop is 52/52 * 12/51 * 39/50 *3 = 0.5506 = 55.1%

doing the math like this works but for me it was easier to think about like this


probability ssx

52/52 * 12/51 * 39/50= .183

probability sxs

52/52 * 39/51 * 12/50= .183

probablity sxx

52/52 * 39/51 *12/50= .183


.183+.183+.183= 55%


Exactly the same thing but just wanted to write it out like that cause like i said was easier for me to think about this way

so to answer the original question....


Mono= 5.2%

Rainbow= 39.8%

2tone=55%


so obv 2tone flop is going to be the most likely

52/52 for first card-
43/51 for second
34/50 for third

57 % for rainbow....rainbow wins yaaay

nevermind these numbers are off im an idiot

well it comes rainbow {1*(39/51)*(26/50)}=39.7% of flops come rainbow
two tone= 100%-(% it comes rainbow+% it comes mono)
times it comes mono={1*(12/51)*(11/50)}=5.17%
two tone = 100-(39.7+5.38)
two tone = 55.12% of flops are two tone

edit**fixed monotone flop thx kiwi

Muh GUT sez two tone and I'm stickin' with that y'hear?

Pokar's about FEEL man. FEEEEEELLLLL

...ace high when I hold KK!

I'm glad this thread worked out like I thought it would.

Originally Posted by Imthenewfish

well it comes rainbow {1*(39/51)*(26/50)}=39.7% of flops come rainbow
two tone= 100%-(% it comes rainbow+% it comes mono)
times it comes mono={1*(12/50)*(11/49)}=5.38%
two tone = 100-(39.7+5.38)
two tone = 54.92% of flops are two tone

your mono calculations should be with 51 and 50 cards left not 50 and 49 and you're rounding wrong (39.76XXX should become 39.8 not 39.7), but otherwise I agree with your answers.

I think all this 52 cards talk is screwing up the math. I don't care about any flop I wasn't dealt cards for, and depending on holding suited or unsuited cards, it really effects the numbers. I don't want to do all the math, but Two tone wins everytime. Is there a stage two question spoon?

1 x 39/51 x 26/50 = .4 rainbow
(1 x 11/51 x 40/50)x3 = .51 2-tone
1 x 11/51 x 10/50 = .04 mono

And that adds up to 95%. The other 5% are rigged obviously. Write support imo.
2-tone has to be the most likely tho.


ffffuuuu there are 13 cards of each suit that's where I went wrong in 2 and 3. Can't be bothered to correct it.
Good news everybody! Laws of nature are intact after all.

so i tried to take this one step further and figure out how likely the different boards were if we were hold a suited(x) hand....thanks to kiwimark for spotting my squirrelly math

for monotone suit x flop

x x x 11/50 * 10/49 * 9/48 = .008

=.8%

for board with one suit x

x s s 11/50 * 38/49 * 38/48=.1386

s x s 39/50 * 11/49 * 38/48=.1386

s s x 39/50 * 38/49 * 11/48=.1386


.1386+.1386+.1386=.4158

=41.58%

for board with 2 cards suit x

s s x 11/50 * 10/49 * 39/48=.0365

s x s 11/50 * 39/49 * 10/48=.0365

x s s 39/50 * 11/49 * 10/48=.0365

.0365+.0365+.0365= .1095

=10.95%

Board with no cards suit x

x x x 39/50 * 38/49 * 37/48=.4662

= 46.62 %



.8% + 41.58% + 10.95% + 46.62%= 100.00%

for board with one suit x

x s s 11/50 * 38/49 * 37/48=.1304

s x s 39/50 * 11/49 * 38/48=.1386

s s x 39/50 * 38/49 * 11/48=.1386

I didn't check what you were actually doing, but I highly suspect these should all be the same. Prolly 'cause for the first one you used 38/49 and 37/48 when you were meant to use 39/49 and 38/48.

Originally Posted by kiwiMark

I didn't check what you were actually doing, but I highly suspect these should all be the same. Prolly 'cause for the first one you used 38/49 and 37/48 when you were meant to use 39/49 and 38/48.

good call fixed it ty sir

in case this is a trick question, my guess is that the plurality of flops are two-toned, but none of them are in the majority

Originally Posted by surviva316

in case this is a trick question, my guess is that the plurality of flops are two-toned, but none of them are in the majority

stfu it's a majority too

Originally Posted by spoonitnow

Rainbow = All three different suits
Two tone = Two of the same suit
Monotone = Three of the same suit

Which happens the most often?

and does it vary depending on whether our hole cards are suited or unsuited?

.... rigged imo

Originally Posted by dneureiter

if card B is different suit from A which is P (39/51)
then card c must be any of the 25 (no... 26) cards which are offsuit from both previous.

(12/51)*39/50 = .1835
(39/51)*26/50 = .385

NO, lol. one of the 24 cards which are the same suit as either of two previous. That's why I was adding to a little over 100%.

Originally Posted by daven

and does it vary depending on whether our hole cards are suited or unsuited?

The %'s change a little but it still breaks down to being pretty close to the same percentages, something like 55/40/5.

Originally Posted by philly and the phanatics

1, rainbow...2 two tone...3.monotone

way i see it is you have 3 cards coming out and 4 suits, it wouldnt make sense for it to be MORE likely that one of those 4 suits having 2 or 3 cards come out...pretty blazed so not sure im making sense but yea im sticking with rainbow

By far its two tone first, then rainbow, then monotone is like 10 times less likely

Agree with the numbers already posted for rainbow and monotone.

I calculated two tone slightly differently from everyone else. The way I thought about it is there are two ways to make a two tone flop: When the second card matches the first card and when the second card does not match the first card.

When the second card matches the first card then we want the third card to be one of the other three suits, so I set the equation up like this:

52/52 * 12/51 * 39/50 = 18.4%

When the second card does not match the first card then the third card has to match either the first card or the second card. There are 12 left of the first suit and 12 left of the second suit so the equation looks like this:

52/52 * 39/51 * 24/50 = 36.7%

These are the two ways to make a two tone flop, so add the probabilities together and you get

Probability Two tone flop = 18.4 + 36.7 = 55.1%

Is this gunna have something to do with people assuming there up against fd's too often/ dont bet/fold enough when the 3rd of a suit falls?

Originally Posted by spoonitnow

Rainbow = All three different suits
Two tone = Two of the same suit
Monotone = Three of the same suit

Which happens the most often?

Was this supposed to be a math problem? I would think that most of you would be able to answer this based on your experience at the tables. I voted that 2-tone was the most common based on what I've seen over thousands of hands. amirite?

Originally Posted by PlayToWin

Was this supposed to be a math problem? I would think that most of you would be able to answer this based on your experience at the tables. I voted that 2-tone was the most common based on what I've seen over thousands of hands. amirite?

It was just to get people to think about it.

Originally Posted by PlayToWin

Was this supposed to be a math problem? I would think that most of you would be able to answer this based on your experience at the tables. I voted that 2-tone was the most common based on what I've seen over thousands of hands. amirite?

Yeah fuk math! thousands of hands is more than enough to develop a gut feeling for pokers!

Originally Posted by d0zer

Yeah fuk math! thousands of hands is more than enough to develop a gut feeling for pokers!

Don't put words in my mouth. I only said that you should be able to answer the question based on observation.

Originally Posted by PlayToWin

Don't put words in my mouth. I only said that you should be able to answer the question based on observation.

He was joking. =P

Originally Posted by spoonitnow

I'm glad this thread worked out like I thought it would.

You asked a question that has only one correct answer, secure in the knowledge that you would receive multiple responses that disagree and cannot possibly all be correct.
Very clever.

By the way, what are the odds of having suited pocket cards?

Originally Posted by Tasha

By the way, what are the odds of having suited pocket cards?

12/51

Whenever I play online its always 2 of the same suit. When I play my 2 suited cared 2 others come out thinking Im going to get a flush & it doesnt come =(

Originally Posted by spoonitnow

12/51

I've never really thought about this but in a full ring game most of the time at least two players are looking at suited hole cards.

We've seen above that half the time the flop will be two tone.
The odds of flopping a flush draw is about 10%.
Does that mean that about a fifth of the time at least one player is drawing for a flush?

Originally Posted by Tasha

I've never really thought about this but in a full ring game most of the time at least two players are looking at suited hole cards.

We've seen above that half the time the flop will be two tone.
The odds of flopping a flush draw is about 10%.
Does that mean that about a fifth of the time at least one player is drawing for a flush?

No because all players don't see a flop.

Originally Posted by spoonitnow

No because all players don't see a flop.

A-ha! I forgot about that.
However, if three players including yourself are still in the hand then the chances are that one of the other two is drawing for a flush?

Originally Posted by Tasha

A-ha! I forgot about that.
However, if three players including yourself are still in the hand then the chances are that one of the other two is drawing for a flush?

While it would depend on their ranges (and yours) to an extent, if you knew that one player held two hearts, another player held two spades, you held two clubs, and all three of you saw a flop, I'm pretty sure it's still a less than 50% chance that anyone flops a flush draw.

"12/51"

Glaringly obvious.

But I voted rainbow, so I'm stupid too.

If 3 players each have suited cards of different suits, doesn't each player have a 11/50 chance of flopping a fd? Making it 33/46 chance for one of them to flop a flush draw? So about 71%?

Originally Posted by supahaole

If 3 players each have suited cards of different suits, doesn't each player have a 11/50 chance of flopping a fd? Making it 33/46 chance for one of them to flop a flush draw? So about 71%?

No! Think about things before you type dummy. That would be the chance of hitting one card of a suit. Back to the calculator.

If about 55% of flops are two-tone if no hole cards were seen, and with our example 3/4 of the flushes are covered, it should be less than 75% of 55% do you see why?

Originally Posted by spoonitnow

If about 55% of flops are two-tone if no hole cards were seen, and with our example 3/4 of the flushes are covered, it should be less than 75% of 55% do you see why?

Yes. I was over complicating it.

Originally Posted by spoonitnow

If about 55% of flops are two-tone if no hole cards were seen, and with our example 3/4 of the flushes are covered, it should be less than 75% of 55% do you see why?

Help! No I don't, but that's what I was aiming for.

Originally Posted by OngBonga

"12/51"

Glaringly obvious.

Yes it was a bit. I had a feeling it was going to be very straight-forward.

Originally Posted by Tasha

Originally Posted by spoonitnow

Originally Posted by spoonitnow
If about 55% of flops are two-tone if no hole cards were seen, and with our example 3/4 of the flushes are covered, it should be less than 75% of 55% do you see why?

Help! No I don't, but that's what I was aiming for.

With 52 cards unknown, the chance of a two-tone flop is 55%. If we want to know the likelihood a specific suit will flop, it's 25%*55%. The chance that one of 3 specific suits will fall is 75%*55%.

But that's the probability with 52 cards unknown. We are stating that we know that each of the 3 players has 2 cards of their suit, so the chance of a two-tone flop falling for their suit is diminished, because two of their suited cards are taken out of the deck.

Okay, thanks, now I get it. Wish I could get this down once and for all.
How does this affect preflop play? If you have suited hole cards then your chances of flopping a flush draw are about 25%. Does that make it worthwhile calling a preflop bet up to 4 x BB?

Originally Posted by Tasha

How does this affect preflop play? If you have suited hole cards then your chances of flopping a flush draw are about 25%. Does that make it worthwhile calling a preflop bet up to 4 x BB?

Read post 31 by pdk1010 in this thread. It shows the probabilities of the different kinds of boards you can flop with a suited hand (flopped flush, flopped draw, 1 of your suit, none of your suit).

Even if you flop a draw, you're still only 35% or so to make it by the river.

Ultimately, the decision to call a 4xBB bet with a suited hand hoping for a flush is very situational and relies on implied odds. What are the stack sizes? How many people are already in the pot? Will someone behind try to squeeze you out? Will your opponent pay you off if you hit? Will he play too aggressively to be able to get to the river on just a draw? Do you have any steal equity postflop to make it profitable even if you miss hitting a flush draw? etc.