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timogl wrote:well, since the two are independent, aren't the odds the same for either of you as when you play alone?
My opponent is working with Monsanto cronies, Rupert Murdoch yes-men and corrupt labor unions.
DoomYoshi wrote:The odds for either of you to have a set is 2/3. The odds for you both to have a set is 1/9.
timogl wrote:(the 3! thing, 3*3*3, that is true for when you have three numbers, you say, how many 3 number combinations can you make. when you have, say, 1, 2 and 3, yes, 113 is different from 311. but these are colors. when you change their position in the order (red red blue and blue red red for intance, it doesn't change their value. they either match or they don't. there are 10 unique cominations, i listed them. four of them are matches. the one in three part i won't disupte, it makes perfect sense. as far as three people drawing cards, if you say the probability that one of them will draw a set is higher because there are higher because there are more of them, ok, if you say so.)
timogl wrote:(the 3! thing, 3*3*3, that is true for when you have three numbers, you say, how many 3 number combinations can you make. when you have, say, 1, 2 and 3, yes, 113 is different from 311. but these are colors. when you change their position in the order (red red blue and blue red red for intance, it doesn't change their value. they either match or they don't. there are 10 unique cominations, i listed them. four of them are matches. the one in three part i won't disupte, it makes perfect sense. as far as three people drawing cards, if you say the probability that one of them will draw a set is higher because there are higher because there are more of them, ok, if you say so.)
timogl wrote:i don't get the controversy. one in three is fine (if the red, green and blue are present in equal numbers in the pool you draw from, that is, the chance of drawing, on any draw, green, red and blue are the same. if there are more red than green, and more green than blue, then the chances are different. the thing there with the 27 combinations of 1, 2 and 3 is perfectly valid. but there are only 10 ways the blue, green and red can fall. and four ways they can match. i listed them. as i said, red red blue is the same as blue red red.)
timogl wrote:don't get me wrong. i am not trying to argue. one in three, ok (provided the red, blue and green are available in equal numbers). you can write 27 different combinations or red, blue and green. but only the unique ones are relevant. all that stuff with dice is ok, i see your point, but it doesn't apply. we are talking about red, blue and green. there are only 10 unique combinations. not 27. red red blue and blue red red are the same combination. green red red and red red green are not unique. they are the same. you are trying to match three cards. the order of it does not matter. as you said, no matter what the first two colors are, the chance of matching with the third card is one in three. it does not change because of the order.
1. the first two can be red red, for instance. you draw blue. you have red red blue.
2. the first two can be red blue. you draw red. you have red blue red (since you want the order of how you write the combinations to have some significance).
the two card sets in (1) and (2) are the same.......thing. they are not different.
Swifte wrote:since they are independent events, they multiply, they do not add. the math goes like this:
in 2v2, if odds of getting a set in 3 cards is 1/3:
Neither has a set: (2/3)*(2/3) = 4/9
Exactly 1 has a set: (2/3)*(1/3)*2 = 4/9
Both have a set: (1/3)*(1/3) = 1/9
Total = 9/9 (100%)
in 3v3:
none have a set: (2/3)*(2/3)*(2/3) = 8/27
exactly 1 has a set: 3*(1/3)*(2/3)*(2/3) = 12/27
exactly 2 have a set: 3*(1/3)*(1/3)*(2/3) = 6/27
all 3 have a set: (1/3)*(1/3)*(1/3) = 1/27
Total = 27/27
The Gambler's fallacy, also known as the Monte Carlo fallacy (because its most famous example happened in a Monte Carlo Casino in 1913),[1][2] and also referred to as the fallacy of the maturity of chances, is the belief that if deviations from expected behaviour are observed in repeated independent trials of some random process, future deviations in the opposite direction are then more likely.
SirSebstar wrote:this is the anwserSwifte wrote:since they are independent events, they multiply, they do not add. the math goes like this:
in 2v2, if odds of getting a set in 3 cards is 1/3:
Neither has a set: (2/3)*(2/3) = 4/9
Exactly 1 has a set: (2/3)*(1/3)*2 = 4/9
Both have a set: (1/3)*(1/3) = 1/9
Total = 9/9 (100%)
in 3v3:
none have a set: (2/3)*(2/3)*(2/3) = 8/27
exactly 1 has a set: 3*(1/3)*(2/3)*(2/3) = 12/27
exactly 2 have a set: 3*(1/3)*(1/3)*(2/3) = 6/27
all 3 have a set: (1/3)*(1/3)*(1/3) = 1/27
Total = 27/27
Re: probability question
Postby HardAttack on Thu Dec 20, 2012 12:39 pm
cool answer swifte,
answer is 1-(4/9) = 5/9 it is.
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