Home / Community / Forum / Poker Community / Off-Topic Discussions /

Poker Math Probabilities 1/19/2013

Old
Default
Poker Math Probabilities 1/19/2013 - Thu Jan 19, 2012, 09:36 AM
(#1)
XXChris123's Avatar
Since: Jul 2010
Posts: 1,512
BronzeStar
We have a bag with 9 red marble, 2 blue marbles , and 3 green marbles in it . What is the probability of randomly selecting a non - blue marble from the bag.

total possibilities = 9+2+3 = 14 possibilities
2 blue marbles or 12 non blue marbles

so probability = 12 / 14 divide by 2/2 so probability is 6/7

what is probability of drawing a green marble ?
 
Old
Default
Thu Jan 19, 2012, 10:28 AM
(#2)
mtnestegg's Avatar
Since: Feb 2011
Posts: 1,336
this is too easy chris.. 3/14 probability to draw a green one and since 3 is prime and 14 cant be divided by 3 this ratio cant be reduced. so 3/14 to draw green we can convert to a percentage though by simply dividing the numerator by the denominator.. or 3/ 14 we get .214 (again rounded) or 21.4% chance to draw green, which for poker discussion purposes is close enough to 21% to call it that.
some folks find ratios easier and some find %ages easier. just depends on how individuals brains fire i guess. ( synapsis?)
and just for the non math peeps out there ( / ) means divided by

Last edited by mtnestegg; Thu Jan 19, 2012 at 10:44 AM..
 
Old
Default
Thu Jan 19, 2012, 10:37 AM
(#3)
XXChris123's Avatar
Since: Jul 2010
Posts: 1,512
BronzeStar
they get harder

researching them now

Last edited by XXChris123; Thu Jan 19, 2012 at 11:16 AM..
 
Old
Default
Thu Jan 19, 2012, 10:45 AM
(#4)
Keldraco's Avatar
Since: Jul 2011
Posts: 102
yeah... how about adding 1 or 2 more bags or marbles...
 
Old
Default
Thu Jan 19, 2012, 11:55 AM
(#5)
mtnestegg's Avatar
Since: Feb 2011
Posts: 1,336
Quote:
Originally Posted by Keldraco View Post
yeah... how about adding 1 or 2 more bags or marbles...
ya how bout a bag of 52 marbles in 4 different colors and 13 different sizes..
 
Old
Default
Thu Jan 19, 2012, 12:03 PM
(#6)
XXChris123's Avatar
Since: Jul 2010
Posts: 1,512
BronzeStar
coin flips
P(HH ) whats probability to get heads and then heads again
lets start with one flip and then keep adding one more
P(H) = 1/2 or 50%
coin can be either heads or tails
P(HH) 1st flip = 1/2 P = H , 2nd flip = 1/2P = H
so P (HH) = 1/2 x 1/2 = 1/4
so P ( HHH) = 1/2 x1/2 x 1/2 = 1/8
P(HHHH) = 4 heads in a row is 1/16
P ( HHHHH ) = 1/2 to the 5th power or 1/32
P (HHHHHH) = 1/2 to the 6th power or 1/64

each coin flip is always a P of 1/2 a flipped coin has no memory
so 12 heads in a row has no bearing on outcome of 13th flip as an individual flip but as a cumulative effect it does. :P
a flipped coin has no memory but probability does :P
.
.
this is why some idiot in a large field of a MTT can seem to magically win 14 hands in a row with any 2 cards
Well you have 6000 players out of 6000 players a few are gonna get hot streaks if they choose to play any two cards . It's all a part of the probability matrix or probability tree

Last edited by XXChris123; Thu Jan 19, 2012 at 12:07 PM..
 
Old
Default
Thu Jan 19, 2012, 12:11 PM
(#7)
mtnestegg's Avatar
Since: Feb 2011
Posts: 1,336
...... duplicate
 
Old
Default
Thu Jan 19, 2012, 12:13 PM
(#8)
XXChris123's Avatar
Since: Jul 2010
Posts: 1,512
BronzeStar
Even I have weaknesses in math and this is just one area where things are not crystal clear to me ,
but I'm working on it today , and will for next few days . I feel that you can create a formula for all decisions in poker kinda a unified field theory of poker . Thats my goal , I can see it mentally but cant express in math yet. I can break it up in parts though. Which is part of my goal here. Game theory and probability are great concepts , the more you understand probability the more intuition you will have in making proper decisions. This will be a long post I think .
 
Old
Default
Thu Jan 19, 2012, 12:28 PM
(#9)
XXChris123's Avatar
Since: Jul 2010
Posts: 1,512
BronzeStar
ok probailty of 4 heads in a row we determined earlier was
P ( HHHH ) 1/2 to the 4th power or 1/2 x 1/2 x 1/2 x 1/2 = 1/16
there are 16 possible outcomes and 1/16 of the time we will get 4 heads in a row

what happens when we flip the coin 5 times
thats 1/2 to the 5th power 1/2 x 1/2 = 1/4 x 1/2 = 1/8 x 1/2 = 1/16 x 1/2 = 1/32
so 5 coin flips has 32 possibilities

whats the odds of getting exactly 1 heads only in 5 flips

T T T T H
T T T H T
T T H T T
T H T T T
H T T T T

out of 32 possibilities only 5 of those contain 1 heads
1/32+1/32+1/32+1/32+1/32 = 5/32

now heres a tough one
flip the coin 5 times
whats the probability that we will not get exactly one heads

.
.

Last edited by XXChris123; Thu Jan 19, 2012 at 01:29 PM..
 
Old
Default
Thu Jan 19, 2012, 12:41 PM
(#10)
mtnestegg's Avatar
Since: Feb 2011
Posts: 1,336
Quote:
Originally Posted by XXChris123 View Post

out of 32 possibilities only 5 of those contain 1 heads
1/32+1/32+1/32+1/32+1/32 = 5/32

now heres a tough one
flip the coin 5 times
whats the probability that we will not get at least exactly one heads
.
.
at least or exactly. you pick.
 
Old
Default
Thu Jan 19, 2012, 01:26 PM
(#11)
TrumpinJoe's Avatar
Since: Jun 2010
Posts: 4,557
Quote:
Originally Posted by XXChris123 View Post
I feel that you can create a formula for all decisions in poker kinda a unified field theory of poker . Thats my goal , I can see it mentally but cant express in math yet. I can break it up in parts though.
I think its called Expected Value. Good decisions are +EV.

The biggest problem is that poker is a game of incomplete information so you only have estimates, some with large error bars, for a number of parameters.
 
Old
Default
Thu Jan 19, 2012, 01:30 PM
(#12)
XXChris123's Avatar
Since: Jul 2010
Posts: 1,512
BronzeStar
I reworded it try again. Meanwhile lets look at a probability tree for coin flips
1st flip is either H or T
HT = 2 P
HT HT = 4 P
HT HT HT HT = 8 P
HT HT HT HT HT HT HT HT = 16 P
we see a pattern of 2,4,8,16,32,64,128, 256 , and so on
limited by how this text works
there are other patterns in there as well

each branch is this
..........H..... .. ..T......
........H.T.. ...H.T
......H.T..H.T....H.T..H.T....and so on with only 1 event for HHH and one event for TTT

Last edited by XXChris123; Thu Jan 19, 2012 at 01:49 PM..
 
Old
Default
Thu Jan 19, 2012, 03:19 PM
(#13)
r0ck.carver's Avatar
Since: Apr 2011
Posts: 201
Quote:
Originally Posted by XXChris123 View Post
I reworded it try again. Meanwhile lets look at a probability tree for coin flips
1st flip is either H or T
HT = 2 P
HT HT = 4 P
HT HT HT HT = 8 P
HT HT HT HT HT HT HT HT = 16 P
we see a pattern of 2,4,8,16,32,64,128, 256 , and so on
limited by how this text works
there are other patterns in there as well

each branch is this
..........H..... .. ..T......
........H.T.. ...H.T
......H.T..H.T....H.T..H.T....and so on with only 1 event for HHH and one event for TTT
You should have a look at finite math(with business and/or digital applications) for the construction method of truth sets.....I think your coin example would be clearer if you used a spread sheet style application. This outlines the total possible outcomes for set variables in an easily visualized style.

single coin : N flips=X possibilities ..=or
1 flip = 2 outcomes......H..T
2 flips =4 outcomes....HH..HT..TH..TT
3 flips =8 outcomes....HHH..HHT..HTH..THH..TTH..THT..HTT...TT T
...... base 2 math.....the trouble is when you start to include multiple coin flips "pattern specific".... you rapidly begin to deal with an amount of data that is overwhelming even if you're rainman....it drove Cantor insane.......and he's considered one of the finest Mathematical minds ever. ( ...The Chinese have a story about it......give me half your kingdom or a chess board of rice grains doubled on each square and starting with 1 grain....if all the matter in the universe was rice you would get just over half way through the 64 squares on the chess board...)
If you start with a finite set 13 ranks 4 suits....and want to know 5 card combo's using 2,3,4,5,6&7 cards with the original 2 being hole cards (and u can guess the rest lol) then you need to run exclusive and /or sets for all the various combos as dictated by the original 2 cards ....it can be done using numerical sets algebra (Mandelbrot , Pythagoras, Newton and numerous others...) but the work /value is not a good return (especially given that it's already been done)...as there are too many unquantifiable influences during play (as previously pointed out).
Anyhow good luck with the unified theory of poker....but you know if it does exist and you find it and make it public then there is no point in playing any more is there ????? LOLOL
cheers r0ck
 
Old
Default
Thu Jan 19, 2012, 03:38 PM
(#14)
mtnestegg's Avatar
Since: Feb 2011
Posts: 1,336
if in fact there are 5 out of 32 chances to flip exactly 1 heads in 5 flips (which there are), doesn't that mean that the ratio for fliping any other combo but exactly 1 heads would be 27/32 or the remainder of the total possibilities? high school algebra II was as far as i got, too busy gettin stoned i guess cough, cough
 
Old
Default
Thu Jan 19, 2012, 04:23 PM
(#15)
XXChris123's Avatar
Since: Jul 2010
Posts: 1,512
BronzeStar
well done you got it
 
Old
Default
Thu Jan 19, 2012, 04:26 PM
(#16)
XXChris123's Avatar
Since: Jul 2010
Posts: 1,512
BronzeStar
those are great responses thats the type of thing this type of post needs to generate or I hoped it would. Now I have other avenues to research. Thnx guys. I have been doing math for about 6 hours to day so I am burned out for now . Reading up on cantor now though.

Last edited by XXChris123; Thu Jan 19, 2012 at 04:30 PM..
 
Old
Default
Thu Jan 19, 2012, 05:27 PM
(#17)
XXChris123's Avatar
Since: Jul 2010
Posts: 1,512
BronzeStar
jeez I have read more about poker theory game theory and probability today than I have typically before.
A lot of the math is above my pay grade . sigh
You guys are right it has been done . Once probabilities reach a certain point it is very difficult to get a good answer really in my opinion.
For instance against 5 players there are more than 9 trillion possible hands ack !!!!!

Against 9 players it is 21 octillion

An exhaustive analysis of all of the match ups in Texas Hold 'em of a player against nine opponents requires evaluating each possible board for each distinct starting hand against each possible combination of hands held by nine opponents, which is (more than 21 octillion).

However I do not belive they are moving duplicate hands. In other words there are 6 ways to be dealt AA
{A♣, A♥}, {A♣, A♠}, {A♣, A♦}, {A♥, A♠}, {A♥, A♦} and {A♠, A♦}. There are 52 ways to pick the first card and 51 ways to pick the second card and two ways to order the two cards yielding
(52X51) / 2 = 1326 possible outcomes when being dealt two cards (also ignoring order). This gives a probability of being dealt two aces of 6/1326 = 1/221.
That is also the probability of being dealt any pair.
22 is dealt every 221 hands also . I would imagine then that 13/221 yields probability of getting any pair.

it is always 1 / 221 for any given pair which I think implies that in any group of 221 hands you will get 13 pairs .
 
Old
Default
Thu Jan 19, 2012, 05:39 PM
(#18)
XXChris123's Avatar
Since: Jul 2010
Posts: 1,512
BronzeStar
I have been doing math since 9:00 AM !!!!!!!!
Thats 8 and 1/2 hours. !!!!! Ok I did nap for about 1 or 2 hours . Needed a break.
Theres some thing about the probability formulas that bugs me. most notably is that you have to form a tree to get an answer in some cases. It would seem to since there are repeating patterns in say a flip coined tree then formulas could account for the patterns. That would enable formulas to remove the duplicates of any given event such as 6 ways to get AA. Actually I think they do have such a formula but I can't swear to it.
.....On KhanAcademy's math tree I have completed 151 of 276 math modules. I dont think the remaining 125 modules will answer my questions. Hopefully though it will prepare me for understanding it better if I can achieve those levels. The higher levels contain concepts and formulas that I have not or have rarely been exposed to. I am addicted to http://www.khanacademy.org/ as much as I was addicted to playing poker . Once I complete the tree I will repeat it to insure that I can do it all smoothly. Thats my goal anyways.
....Today I have found tons of new info on poker though . Also on probability game theory etc.

Last edited by XXChris123; Thu Jan 19, 2012 at 05:53 PM..
 
Old
Default
Thu Jan 19, 2012, 06:27 PM
(#19)
TrumpinJoe's Avatar
Since: Jun 2010
Posts: 4,557
Quote:
Originally Posted by XXChris123 View Post
it is always 1 / 221 for any given pair which I think implies that in any group of 221 hands you will get 13 pairs .
Not quite Chris.

The implication is that over a large enough sample of sets of 221 hands you will find 13 pairs on average, but every set will not have 13.

I leave you with some interesting tidbits. You are more likely to be dealt AA than AKs. You have the exact same probability of being dealt AcAdAhAsKc as you do 4c3h9dQsAh. One you remember forever and one barely registers on your mind.

Joe
 
Old
Default
Thu Jan 19, 2012, 06:38 PM
(#20)
XXChris123's Avatar
Since: Jul 2010
Posts: 1,512
BronzeStar
AcAdAhAsKc as you do 4c3h9dQsAh.
same probability exists as long as you are saying being dealt those exact cards


52 cards in deck any exact card = 1/52
 

Getting PokerStars is easy: download and install the PokerStars game software, create your free player account, and validate your email address. Clicking on the download poker button will lead to the installation of compatible poker software on your PC of 51.7 MB, which will enable you to register and play poker on the PokerStars platform. To uninstall PokerStars use the Windows uninstaller: click Start > Control Panel and then select Add or Remove programs > Select PokerStars and click Uninstall or Remove.

Copyright (c) PokerSchoolOnline.com. All rights reserved, Rational Group, Douglas Bay Complex, King Edward Road, Onchan, Isle of Man, IM3 1DZ. You can email us on support@pokerschoolonline.com