The odds of holecards both being suited

[Dahlberg]

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

[BottlesOf]

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.

[stinkypete]

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

[/ QUOTE ]

that is correct.

[wegs the wegs]

For them to have the same suit as you:

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

So roughly 4.5% of the time

[kem]

[ 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.

[Cazz]

[ 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.

[sdw]

[ 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%

[Cazz]

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