Poker math ?

[ManyBeers]

....how many 3-rank combinations are possible? That is, when the 5 board community cards are dealt how many 3 of a kind combinations are possible? For instance Board:2c,3h,2s,Jd,2h=3 deuces.
I am not looking for permutations just combinations. In other words a board of 2c,3h,2sJd,2h is equivalent to a board of 2h,3h,2c,Jd,2s. The order of the ranks arrival on the board does not matter. I think anyone who can solve this knows what i mean. Thanks in advance.

 

[Mr Pickles]

C

 

[ManyBeers]

Originally posted by: MrLee
C

What?

 

[txrandom]

( 13 choose 1) * (4 choose 3) * (12 choose 2) * (4 choose 1) * (4 choose 1).

This doesn't include 4 of a kinds or full houses.

Edit: This also assumes that all 52 cards have an even chance to be drawn. Obviously cards that are already dealt to the players will have a lower chance of showing up.

 

[ManyBeers]

Originally posted by: txrandom
( 13 choose 1) * (4 choose 3) * (12 choose 2) * (4 choose 1) * (4 choose 1).

This doesn't include 4 of a kinds or full houses.

Edit: This also assumes that all 52 cards have an even chance to be drawn. Obviously cards that are already dealt to the players will have a lower chance of showing up.

Look i am using peoples responses on here to double check my self, so if you don't mind would you please give an answer and not a formula.

 

[txrandom]

Originally posted by: ManyBeers

Originally posted by: txrandom
( 13 choose 1) * (4 choose 3) * (12 choose 2) * (4 choose 1) * (4 choose 1).

This doesn't include 4 of a kinds or full houses.

Edit: This also assumes that all 52 cards have an even chance to be drawn. Obviously cards that are already dealt to the players will have a lower chance of showing up.

Look i am using peoples responses on here to double check my self, so if you don't mind would you please give an answer and not a formula.

Look i am just trying to help you, but now I won't.

 

[Schfifty Five]

Originally posted by: txrandom

Originally posted by: ManyBeers

Originally posted by: txrandom
( 13 choose 1) * (4 choose 3) * (12 choose 2) * (4 choose 1) * (4 choose 1).

This doesn't include 4 of a kinds or full houses.

Edit: This also assumes that all 52 cards have an even chance to be drawn. Obviously cards that are already dealt to the players will have a lower chance of showing up.

Look i am using peoples responses on here to double check my self, so if you don't mind would you please give an answer and not a formula.

Look i am just trying to help you, but now I won't.

LOL awesome! Txrandom for elite!

 

[ManyBeers]

Originally posted by: txrandom

Originally posted by: ManyBeers

Originally posted by: txrandom
( 13 choose 1) * (4 choose 3) * (12 choose 2) * (4 choose 1) * (4 choose 1).

This doesn't include 4 of a kinds or full houses.

Edit: This also assumes that all 52 cards have an even chance to be drawn. Obviously cards that are already dealt to the players will have a lower chance of showing up.

Look i am using peoples responses on here to double check my self, so if you don't mind would you please give an answer and not a formula.

Look i am just trying to help you, but now I won't.

Well, i don't understand why you just didn't answer the question if you knew the answer.

 

[dbk]

Do your own homework, LOSER

 

[Schfifty Five]

Originally posted by: ManyBeers

Originally posted by: txrandom

Originally posted by: ManyBeers

Originally posted by: txrandom
( 13 choose 1) * (4 choose 3) * (12 choose 2) * (4 choose 1) * (4 choose 1).

This doesn't include 4 of a kinds or full houses.

Edit: This also assumes that all 52 cards have an even chance to be drawn. Obviously cards that are already dealt to the players will have a lower chance of showing up.

Look i am using peoples responses on here to double check my self, so if you don't mind would you please give an answer and not a formula.

Look i am just trying to help you, but now I won't.

Well, i don't understand why you just didn't answer the question if you knew the answer.

maybe b/c the way you asked him made it sound like you were demanding it rather than kindly asking.

Furthermore, if you bothered to put even a little effort, you could have done the calculation yourself given what txrandom gave you, but you're jsut one lazy person.

 

[Capt Caveman]

Aren't you a poker pro? Shouldn't you have this already memorized?

 

[ManyBeers]

Originally posted by: dbk
Do your own homework, LOSER

I've already done it and want some confirmation on my answer.I'm 54 and this has nothing to do with homework. Loser.

 

[ViviTheMage]

google this kind of stuff, I am sure everything poker related has been tried.

 

[ManyBeers]

Originally posted by: Capt Caveman
Aren't you a poker pro? Shouldn't you have this already memorized?

No this is just a self improvement math problem i am trying to do.

 

[ManyBeers]

Originally posted by: Schfifty Five

Originally posted by: ManyBeers

Originally posted by: txrandom

Originally posted by: ManyBeers

Originally posted by: txrandom
( 13 choose 1) * (4 choose 3) * (12 choose 2) * (4 choose 1) * (4 choose 1).

This doesn't include 4 of a kinds or full houses.

Edit: This also assumes that all 52 cards have an even chance to be drawn. Obviously cards that are already dealt to the players will have a lower chance of showing up.

Look i am using peoples responses on here to double check my self, so if you don't mind would you please give an answer and not a formula.

Look i am just trying to help you, but now I won't.

Well, i don't understand why you just didn't answer the question if you knew the answer.

maybe b/c the way you asked him made it sound like you were demanding it rather than kindly asking.

Furthermore, if you bothered to put even a little effort, you could have done the calculation yourself given what txrandom gave you, but you're jsut one lazy person.

You think i am asking on here because i would understand his formula?

 

[purbeast0]

i really hope you are not 54 cause you sound like you are 16.

 

[ManyBeers]

Originally posted by: purbeast0
i really hope you are not 54 cause you sound like you are 16.

Rather than insulting me why not act like an adult and offer something constructive to the thread.

 

[Epic Fail]

Originally posted by: ManyBeers

Originally posted by: Schfifty Five

Originally posted by: ManyBeers

Originally posted by: txrandom

Originally posted by: ManyBeers

Originally posted by: txrandom
( 13 choose 1) * (4 choose 3) * (12 choose 2) * (4 choose 1) * (4 choose 1).

This doesn't include 4 of a kinds or full houses.

Edit: This also assumes that all 52 cards have an even chance to be drawn. Obviously cards that are already dealt to the players will have a lower chance of showing up.

Look i am using peoples responses on here to double check my self, so if you don't mind would you please give an answer and not a formula.

Look i am just trying to help you, but now I won't.

Well, i don't understand why you just didn't answer the question if you knew the answer.

maybe b/c the way you asked him made it sound like you were demanding it rather than kindly asking.

Furthermore, if you bothered to put even a little effort, you could have done the calculation yourself given what txrandom gave you, but you're jsut one lazy person.

You think i am asking on here because i would understand his formula?

How did you solve it if you don't understand his formula? Post your method of solving first and give us your answer.

 

[purbeast0]

Originally posted by: ManyBeers

Originally posted by: purbeast0
i really hope you are not 54 cause you sound like you are 16.

Rather than insulting me why not act like an adult and offer something constructive to the thread.

i could say the same for you.

 

[txrandom]

http://en.wikipedia.org/wiki/Poker_probability

 

[ManyBeers]

Originally posted by: purbeast0

Originally posted by: ManyBeers

Originally posted by: purbeast0
i really hope you are not 54 cause you sound like you are 16.

Rather than insulting me why not act like an adult and offer something constructive to the thread.

i could say the same for you.

How have i insulted you?

 

[Mr Pickles]

Originally posted by: ManyBeers

Originally posted by: MrLee
C

What?

Most of the time when I don't know the answer I just pick C. I can't leave the question blank. My teacher tells me that to leave an answer blank gives you a 100% chance to get it wrong.

 

[sash1]

Originally posted by: MrLee

Originally posted by: ManyBeers

Originally posted by: MrLee
C

What?

Most of the time when I don't know the answer I just pick C.

good choice. i always picked B. my Bio teached always seemed to like B

 

[ManyBeers]

Originally posted by: MrLee

Originally posted by: ManyBeers

Originally posted by: MrLee
C

What?

Most of the time when I don't know the answer I just pick C. I can't leave the question blank. My teacher tells me that to leave an answer blank gives you a 100% chance to get it wrong.

1,381

 

[Schfifty Five]

Originally posted by: ManyBeers

Originally posted by: Schfifty Five

Originally posted by: ManyBeers

Originally posted by: txrandom

Originally posted by: ManyBeers

Originally posted by: txrandom
( 13 choose 1) * (4 choose 3) * (12 choose 2) * (4 choose 1) * (4 choose 1).

This doesn't include 4 of a kinds or full houses.

Edit: This also assumes that all 52 cards have an even chance to be drawn. Obviously cards that are already dealt to the players will have a lower chance of showing up.

Look i am using peoples responses on here to double check my self, so if you don't mind would you please give an answer and not a formula.

Look i am just trying to help you, but now I won't.

Well, i don't understand why you just didn't answer the question if you knew the answer.

maybe b/c the way you asked him made it sound like you were demanding it rather than kindly asking.

Furthermore, if you bothered to put even a little effort, you could have done the calculation yourself given what txrandom gave you, but you're jsut one lazy person.

You think i am asking on here because i would understand his formula?

It's like if someone tells you "the answer is 5*5", and you reply with "C'mon, just give me the answer". Are you that lazy to solve 5*5?

Now granted, I don't know the answer to your original question and I also do not know if Txrandom's equation is the correct answer, however, your laziness is what makes you sound stupid and a child.

Look up how to do (X choose Y) type problems (permutation/combination formulas) and you will be able to sovle TXrandom's answer in a jiffy.

lol 54 years old.