The odds of holecards both being suited
 

Hey everyone.

I like the maths that are in poker but unfortunatelly I am rather new to the game, and i havent gotten all the math I need. so here is my question for you:

What are the odds that the two holecards are in the same suit?

I would appreciate it if you gave me the answer in %

*EDIT*

I see now that the question was lacking an important piece of info...

If I have two suited cards, what are the odds of someone other having it in the same suit to?

(could be valuable info on flush draws, thats my thought)

thank you

Re: The odds of holecards both being suited
 

12/51= .2353= 23.53%

I'm not too good at maths either, though, so I'd wait for someone else to confirm this.

Also, this is a pretty useless stat to know, there are lots of other more relevant things to learn. For further probability questions, you should try that forum.

Re: The odds of holecards both being suited
 

[ QUOTE ]
12/51= .2353= 23.53%

[/ QUOTE ]

that is correct.

Re: The odds of holecards both being suited
 

For them to have the same suit as you:

=(11/50)*(10/49)= .044897

So roughly 4.5% of the time

Re: The odds of holecards both being suited
 

[ QUOTE ]
For them to have the same suit as you:

=(11/50)*(10/49)= .044897

So roughly 4.5% of the time

[/ QUOTE ]

These are the odds that a specific opponent holds two cards of the same suit as you (given you are suited, and have not seen the flop yet).

These are not the odds that _any_ opponent holds two cards of the same suit as you.

Re: The odds of holecards both being suited
 

[ QUOTE ]
If I have two suited cards, what are the odds of someone other having it in the same suit to?


[/ QUOTE ]

Heads up, the chance the other player has 2 cards of the same suit is

p = (11/50) * (10/49) = 4.490 %
or once in every 22.3 attempts.

Versus N opponents, an approximation that at least one player has 2 cards of the same suit is

P(N) = 1 - (1-p)^N

for N = 9 this is ~ 33.9 %

As approximations go, its not too bad.


Since the hands are not completely independent, I believe the second formula is only correct if

You keep your two cards (leaving 50 in the deck)
The other N guys randomly select 2 cards from the remaining 50.
After each opponent selects 2 cards, he puts them back in the deck and you reshuffle for the next guy.

Re: The odds of holecards both being suited
 

[ QUOTE ]
Heads up, the chance the other player has 2 cards of the same suit is

p = (11/50) * (10/49) = 4.490 %
or once in every 22.3 attempts.

Versus N opponents, an approximation that at least one player has 2 cards of the same suit is

P(N) = 1 - (1-p)^N

for N = 9 this is ~ 33.9 %

As approximations go, its not too bad.


Since the hands are not completely independent, I believe the second formula is only correct if

You keep your two cards (leaving 50 in the deck)
The other N guys randomly select 2 cards from the remaining 50.
After each opponent selects 2 cards, he puts them back in the deck and you reshuffle for the next guy.

[/ QUOTE ]

My probability is a little fuzzy, but doesn't P(N) = 1 - (1-p)^N calculate the probability that exactly one opponent has 2 cards of the same suit?

If you wanted one or more players isn't it p * N?


Here is another interesting one:

Heads up, the chance the other player has 2 cards of the suit when 3 of the 5 cards on the board are the suit and your holding 2 cards of the suit is:

p = (8/45) * (7/44) = 2.83%

Re: The odds of holecards both being suited
 

I believe N*p is the average number of opponents that have 2 cards of the same suit as you.

Let's pretend p = 0.20 (some other condition).
6*p = 1.20. This cannot be the percentage of times
that at least one guy has met the condition. It is
more the 100%.

The probability is
1 - (no one meets the condition)

The chance that one player *misses* (doesn't meet the condition) is 1-p.
The change that N players miss is (1-p)^N .
The probably this doesn't happen is
1- (1-p)^N