The odds of both holecards 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:

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)

I would like the answer in %
thank you

[MortalWombatDotCom]

i just composed a very nice post on how to calculate this, including the usual disclaimer that gaming mouse would come along in a couple of hours and explain the easy way, and right as i was finishing up the formatting bits (in fact, when i was asking a formatting question) i went and hit CTRL-W which in my IDE highlights the current word but in my browser closes the window.

anyway, if you are facing k opponents and were dealt, say, two clubs, the chance that at least one opponent was also dealt two clubs is

.k . . 39! * 11! * 2^i *(50 - 2k)! * (k choose i)
SUM . . ----------------------------------
i=0 . . . .(39 - 2k + i)! * (11 - i)! * 50!

ignore the '.'s, they are just there to make the quotient look nicer. you can pull the constants out, of course, but this way made more sense if you read the original post to see the derivation.

i know you wanted %. if you can't plug in the numbers yourself and nobody else steps up, i'll probably be un-grumpy enough to do it tomorrow. stupid browser. for k=1 it's .045 or 4.5% that an opponent has clubs, and for k=2 it's .088 or 8.8% that at least one opponent has clubs.

i probably made a mistake which would have been easy for someone to point out in the original post but without the derivation they probably won't bother.

[MortalWombatDotCom]

typo.

that sum is the probability that nobody else has two clubs. to get the chance that at least one person does subtract that sum from 1.

[ QUOTE ]
i just composed a very nice post on how to calculate this, including the usual disclaimer that gaming mouse would come along in a couple of hours and explain the easy way, and right as i was finishing up the formatting bits (in fact, when i was asking a formatting question) i went and hit CTRL-W which in my IDE highlights the current word but in my browser closes the window.

anyway, if you are facing k opponents and were dealt, say, two clubs, the chance that at least one opponent was also dealt two clubs is

.k . . 39! * 11! * 2^i *(50 - 2k)! * (k choose i)
SUM . . ----------------------------------
i=0 . . . .(39 - 2k + i)! * (11 - i)! * 50!

ignore the '.'s, they are just there to make the quotient look nicer. you can pull the constants out, of course, but this way made more sense if you read the original post to see the derivation.

i know you wanted %. if you can't plug in the numbers yourself and nobody else steps up, i'll probably be un-grumpy enough to do it tomorrow. stupid browser. for k=1 it's .045 or 4.5% that an opponent has clubs, and for k=2 it's .088 or 8.8% that at least one opponent has clubs.

i probably made a mistake which would have been easy for someone to point out in the original post but without the derivation they probably won't bother.


[/ QUOTE ]

[MortalWombatDotCom]

i was looking over my post and realized that the quotient could actually be simplified alot. in fact, gaming mouse will probably explain why jumping to the simplified version is the easy way [img]/images/graemlins/wink.gif[/img]

anyway, its the same sum, but
39! * 11! / 50! is the same as 1/(50 choose 11)
and
*(50 - 2k)! / (39 - 2k + i)! * (11 - i)!
is the same as
((50 - 2k) choose (11 - i)

so you get

2^i * (k choose i) * ((50 - 2k) choose (11 - i)) / (50 choose 11)

programming this into my programmable calculator, i can then come up with #opp: %chance as follows

1: 4.5%
2: 8.8%
3: 13%
4: 17.1 %
5: 21%
6: 24.8%
7: 28.5%
8: 32.1%
9: 35.5%

since i did the numbers for 1 and 2 opps by hand the long way before coming up with the summation, and got the same answer both ways, i'm pretty confident in this formula.

hope this answered your question.

[gaming_mouse]

[ QUOTE ]
gaming mouse will probably explain why jumping to the simplified version is the easy way

[/ QUOTE ]

You can get a good approximation by assuming independence among your opponent's hands, since they are loosely dependent.

ncr(11,2)=55 (number of possible flush hands left in your suit)
ncr(50,2)=1225 (total possible hands left)

The approximation is very close, except at 7 or more opponents, and even then it is reasonably close.

1 - (1170/1225)=.0449
1 - (1170/1225)^2=.0878
1 - (1170/1225)^3=.1288
1 - (1170/1225)^4=.1679
1 - (1170/1225)^5=.2053
1 - (1170/1225)^6=.2410
1 - (1170/1225)^7=.2750
1 - (1170/1225)^8=.3076
1 - (1170/1225)^9=.3387

Of course, Mortal, you never know how close the approximation is unless you have the exact answers to compare it to -- so your long answers justifies my shortcut. There may be some sort of bounding theorem on the errors in situations like this, but I don't it if there is. Perhaps BruzeZ does...

gm

[Dahlberg]

Thank you both for your great answers.

[gamble4pro]

If you refer to the moment right before flop (no board cards to take into account):
The probability for one opponent (a specific one) to hold the same suit is C(11,2)/C(50,2) = 4.489%.
If you want the probability for at least one opponent to hold such suit, the formula is much complicated and depends on n=number of your opponents.