

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 ? 




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 




they get harder
researching them now 




yeah... how about adding 1 or 2 more bags or marbles...








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 




...... duplicate





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 . 




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 . . 







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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. 




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 




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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 




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





well done you got it





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.





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 . 




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. 




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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 




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 


