ok I posted my first blog, asked you to challenge me with some odds questions, got two challenges and answered them (one poorly at first) and got two messages asking for more.... so here it is, this time I am asking the questions, and answering them, the reason is we need to start EASY.....
Question (1) what are my chances of flopping a set with any NON pocket pair?
the math: your hole cards are 8c 9s (any two cards will do, we could even use X Y as your cards , and i started with this, but it is too confusing)
50 cards you have not seen (52 cards in the deck, less 2 you have in the hole )
3 8's left in the deck ( some people have a problem with this cuz "Johnny might have pocket 8's"... I won't argue, but that's why we say the deck is 50 cards and not less)
3 9's left
Now we will calculate the odds of flopping exactly a set (not quads, not a fullhouse, not two pair EXACTLY a set)
First we will divide the problem in half and only look to hit a set of 8's
we use combinitorics. Combinations are permeatitions without order, and here without replacement....
read as : ( 3 choose 2) times (3 choose 0) times ( 44 choose 1) divided by ( 50 choose 3)
I am having a hard time with the notation, ( 3 choose 2 ) is supposed to look like a fraction in parenthases without the horizontal bar but, I can't use paint, and I am too impatient to wait to find something compatable ( my impatients also affects my poker game but anywho...)
Now go to Wikipedia and read ONE and ONLY ONE heading here: Combination without Repitition
http://en.wikipedia.org/wiki/Combinatorics#Combination_without_repetition
the r! is called factorial and can be calculated as follows.... 1! = 1, 2! = 2*1=2, 3! = 3*2*1=6 4! = 4*3*2*1 and so on
that's what my notation is supposed to look like, and how it is calculated. If you are lucky enough to have a calculator with the button nCr you can just type ( for 3 choose 2) ... 3 nCr 2
anyway, before we look at the calculation, lets look at the setup...
the first part ( of three) of the numerator (before the division) is ( 3 choose 2 ), we are trying to hit exactly 2 of the 3 remaining 8's
the second part of the numerator is ( 3 choose 0 ), we must miss all 9's
the third part ( 44 choose 1), signifies that we must hit one card that is left that is not an 8 or 9 (44 cards left in the deck that are not 8 or 9)
the numerator signifies that we are drawing 3 cards from 50 remaining in the deck
notice that if you add the 3 numbers before the choose in the numerator, 3, 3, and 44, we get 50, which is the same number before the choose in the denominator (after the div) which is 50...
likewise if we add the numbers after the "choose" 2, 0,1 we get 3
anyway... the calculation of (3 choose 2) is 3 , three ways to hit two more eights
(3 choose 0) is 1 , there is only one way to hit no nines
( 44 choose 1) is 44, there are 44 ways to hit the remaining 44 cards
( 50 choose 3) is 19600, all the ways the flop could come
so ... 3 times 1 times 44 = 132
so 132/19600 is our answer to hit exactly a set of 8's or .006734693878,
or about 0.67%
the odds of hitting the nines is the same, and they are mutually exclusive events ( they can't happen at the same time ), so you can add em (or just multiply by two)
and THE CHANCES OF HITTING A SET OF 8's or 9's with 8 9 as hole cards is about [EXACTLY but rounded ] 1.347%
now 8 and 9 are arbitrary cards, ACTUALLY this works for any two differently ranked hole cards ( not a pocket pair )
So what... once in a blue moon, or once in 74.2 hands
Is this helpful to anyone?????????
© Copyright 2008
27 comments
I think I might go with flop a set with a pocket pair. flop a full house, and then move into flushes, then straights, but flops only, cuz we all know how to do the 2% thing from there on out right?
Also some stuff with best hands preflop based on the number of players at the table (just the high card aspect)...
Or maybe this will be the end of the road... does anyone REALLY wanna see this crap?
Anything is better than another bad beat blog lol. I think a lot of people join Railbirds to learn more about poker. Any blog that has helpful info is welcomed.
I guess, but i would be getting alot more interest if i was calling someone a shemale, unfortunatly
Great post...
I personally play the math of the game rather than the player.Its nice to see someone writing about it....
Bee.
Actually after seeing thousands upon thousands of poker hands... 1 in 74 to flop a set seems right on the dot. Seems like everytime I play I will (on average) flop a set somewhere around 1/74.
Great blog thanks for the info.
That's a common misconception Ironman, glad you pointed that out... but it is the wrong line of thaught... since you have no knowlegde of those cards, you can make no assumptions about what they are ( or are not )... therefor you must assume that they are "unknown" and consider them to to be a possible turn or river....
I do understand what your "hang-up" is, and all i can say is that if you continue with that line of logic, then there are no odds besides 100% and 0% [ do or do not ], but since that is NOT useful, we make no assumptions about those cards that we know not
I try to post a link to an article i read that explains this pretty good... if i can find it
Smiley, I calculated this differently and came up with the exact same number, but I do not understand one of your terms.
(3 choose 0) to show that there are no nines. You say there is one way to have no nines. I don't understand this. Any card other than a 9 is not a 9. I think you cover the no 9s in the next term which is (44 choose 1) and the (3 choose 0) is superfluous. Can you explain this a little.
ok i see your point Howie.... here is why I put the (3 choose 0)
If you add the top numbers of the numerator, they add to 50... this is the same as the denomerator... and to me that is important.. and it should be important to everyone really
when you are doing this combinatorics "crap", you have to make sure your book keeping is correct, and by HAVING the (3 choose 0) it shows that it is infact correct
your point about the 1 way to choose no nines is Valid I guess... but the fact remains that (3 choose 0 ) is 1 [ by definition I believe ]... and that is why i said there is one way to have no nines
In short it is ment to be superfluous for didactic reasons
OK, I understand why the term (3 choose 0) is there, but I can't quite grasp why this represents the combinations of no nines.
I checked and (3 choose 0) is 1 because 0! is 1 by definition.
Since we are both very precise individuals, I'll be a bit of a dickhead and point out that technically a "set" is a pocket pair and a matching card on board. You have calculated the odds of trips, not a set.
Smiley, I find it amusing that we approach odds calculations differently. The way I solve this problem is with this formula:
(3/50)*(2/49)*(44/48)*3*2
The first 2 terms are the probability of getting 2 eights as the first 2 cards of the flop. the third term is the probability of not getting an eight or a nine as the third card. The *3 is the number of combinations of 88x on the flop [(3 choose 2)]. The *2 adds the probability of getting nines instead of eights.
This is the way that I see" probability.
I was not aware of any distinction between "set" and "trips"
As far as I understood it, "a set", "trips", and "three of a kind" were all equilavent
Maybe my understanding is wrong... but I doubt that many make any distinction between any of them :-P
That way works too, but I like my way better
the numbers in the combinatorics... show more or less what is going on... your formula looks more or less like a bunch of jumbled numbers
I assure you that even though the notation is different, I put exactly the same formula :-P
Wow Johnny has pockets 8's....this whole time I thought he had 2 thumbs......lol Joking
Hey smiley your pretty smart my brain doesnt work that fast durin the game im always tryin to figure what everyone is betting and should I call Fold or raise
ok I concede that a "set" ad "trips" are different
I doubt many people know that though
http://www.playwinningpoker.com/poker/terms/set.html
It seems that "set" is used both ways. Sklansky defines set as three-of-a-kind. Gary Carson defines set as a pp and one one board. Lou Krieger (in the book I have) doess not have a glossary, but uses set to mean with a pp. Mike Caro straddles the issue:
set
1. (n) In hold 'em and stud, three of a kind. To flop a set in hold 'em means that (most often) one started with a pair and one of those cards was among the flop (the first three community cards). Less often it means a pair was among the flop and the player had another card of that rank in the hole.
Bottom line, you were not wrong. Well, I said I was being a dickhead.