Quote:
Originally Posted by whosnext
Sorry for the confusion on notation.
C(9,2) is the number of ways you can choose 2 items out of 9 items without replacement. C(x,y) is called the Combination function and C(9,2) is sometimes referred to as "9 choose 2".
The formula of C(x,y) is straightforward and is taught in most probability classes in school. You can look it up on the internet if you don't know what it is. C(x,y) is programmed into most modern calculators and computer languages.
Okay, now that that is taken care of. Do we have a fundamental difference in understanding the situation?
Hero (Kc Tc) has 2 clubs. Villain (As Js) has 0 clubs. Flop (Ac 8h 5c) has 2 clubs.
There are 2+2+3=7 cards "known" to us at this point, 4 clubs and 3 non-clubs. The original deck of 52 cards has, of course, 13 clubs and 39 non-clubs. Thus, the turn and river will come from a "stub" which consists of 9 clubs and 36 non-clubs (9+36=45).
You do not "remove" the burn card from the equations since the burn card is unknown. Since you do not know the value of the burn card, it is properly considered part of the "stub" (the set of possible card values from which the turn and river emanate).
From what you have written in your most recent post, you seem to have questions about basic probability concepts rather than poker "theory" questions. There is a Poker Probability forum on 2+2 which may be a better place to continue this conversation.
https://forumserver.twoplustwo.com/25/probability/
Yeah if i scratch my head i can kind of remember this from school, still a bit confused so on point one you say exactly two clubs? So is C(9,2) the probability of a club coming on the turn and river? So its the sum of the 2 out of the 9 clubs coming on the turn and river, the combinations of the 9 clubs that can hit turn and river
Kc10c
Ac 5c 8h
so (2c,3c)(2c,4c)(2c,6c)(2c,7c)(2c,8c)2c9c,2cJc,2cQc, 3c4c,3c6c,3c7c,3c8c,3c9c,3cJc,3cQc,4c6c,4c7c,4c8c, 4c9c,4cJc,4cQc,6c7c,6c8c,6c9c,6cJc,6cQc,7c8c,7c9c, 7cJc,7cQc,8c9c,8cJc,8cQc,9cJc,9cQc,JcQc = 36
So we calculate the sum of combinations for 2 clubs coming because we need to calcualte every single outcome? and then divide by the comination of every other remaning card?