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Re: probability question

Postby HardAttack on Fri Dec 21, 2012 8:26 am

[quote="timogl"]if the
Last edited by HardAttack on Fri Dec 21, 2012 9:15 am, edited 1 time in total.
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Re: probability question

Postby timogl on Fri Dec 21, 2012 9:05 am

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.)
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Re: probability question

Postby Swifte on Fri Dec 21, 2012 9:28 am

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.)


The math for the 27 combinations support the 1/3 odds that you say you accept.

What the 27 combinations shows you is that the odds of getting the 10 ultimate combinations are different, so it is incorrect to assign them equal probability (i.e. saying it is 4 out of 10).

The permutations show that:
1. there is only 1 way to get red red red (So this will happen 1 out of 27 times)
2. there is only 1 way to get blue blue blue (So this will happen 1 out of 27 times)
3. there is only 1 way to get green green green (So this will happen 1 out of 27 times)
4. there are 6 ways to get a set with 1 blue 1 red 1 green (So this will happen 6 out of 27 times)
5. there are 3 ways of ending up with 2 red, 1 blue (So this will happen 3 out of 27 times)
6. there are 3 ways of ending up with 2 red, 1 green (So this will happen 3 out of 27 times)
7. there are 3 ways of ending up with 2 blue, 1 green (So this will happen 3 out of 27 times)
8. there are 3 ways of ending up with 2 blue, 1 red (So this will happen 3 out of 27 times)
9. there are 3 ways of ending up with 2 green, 1 red (So this will happen 3 out of 27 times)
10. there are 3 ways of ending up with 2 green, 1 blue (So this will happen 3 out of 27 times)

so of the 10 combinations you listed, the odds of each are:
1: 1/27
2: 1/27
3: 1/27
4: 6/27
5: 3/27
6: 3/27
7: 3/27
8: 3/27
9: 3/27
10: 3/27

Combinations 1-4 result in a 3 card set - the sum of those probabilities is 9/27, or 1/3.

That's as clear as I can make it. boils down to, just because the order doesn't matter, doesn't mean the odds of each combinations of the same. Just like the odds of rolling a sum of 12 with two dice is much lower than the sum being a 7, because there are way more combinations that create a sum of 7 than a 12.
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Re: probability question

Postby timogl on Fri Dec 21, 2012 10:16 am

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.
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Re: probability question

Postby timogl on Fri Dec 21, 2012 10:18 am

(and i do aplogize. we just may be talking about two entirely different things and i am too dense to understand.)
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Re: probability question

Postby SirSebstar on Fri Dec 21, 2012 11:38 am

its okay, we are all posting here cause we fee the seasonal love, or something. its okay
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Re: probability question

Postby Metsfanmax on Fri Dec 21, 2012 3:24 pm

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.


Yes, but the point is, getting a set with two reds and a blue is more likely to occur than having red/green/blue, which is obvious if you write out the possible combinations. The number of unique combinations doesn't matter when the different unique combinations are not equally probable.

If you had a seven sided die, with 1,1,2,3,4,5,6 then the fact that there are multiple 1's does not change what happens when you roll a 1, but it does change how likely it is to get a 1. Your argument is analogous to saying that for this die, there's a 1/6 chance that you'll roll a 6 because there's only 6 unique numbers.
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Re: probability question

Postby timogl on Fri Dec 21, 2012 3:55 pm

not to belobor it, but the question at the top of this thread was not about drawing a set when you already had two, it was about two players (on a team) that have already drawn three cards each. it is beside the point now what their chances were of getting whatever combination. they either have a match, either or them, both of them, or neither. and thats it. and if one of them has blue blue red it does not matter how he drew it. if one of the drew blue blue red (in that order) and the other drew red blue blue (in that order) each of them now has the same card set.

how any of that relates now, now that they both have three cards (they're not drawing, they have them. its like you played cards, you dealt them both three from a crazy big shoe with blues, reds and greens. the shoe has so many cards what card already went out scarcely affects what a player subsequently draws (if i am understanding right, you are all saying the reds, blues and greens are available in the same levels, if not, then i guess you would be working out all kinds of odds for what it is to draw a blue card when you already have to blues, as opposed to a red or green card). anyway, the cards are dealt, imgine it like bridge, with partners. north and south have three cards each. what are the odds one of them has a match? i am not denying whatever odds you want to say. but if north drew his cards in this order

blue red blue

and south drew his

red blue blue

they have the same set of cards.
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Re: probability question

Postby SirSebstar on Fri Dec 21, 2012 5:54 pm

this is the anwser
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
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Re: probability question

Postby Fewnix on Fri Dec 21, 2012 6:25 pm

a fun discussion that often appears :D

When someone holds two spoils there are only two possibilities they are of the same colour or they are of different colour. Doesn't matter what the colours are or the order in which they are drawn they are either of the same colour or different colour

Any situation two spoils, the odds are 1/3 the third spoils will give a set. If the two are the same colour, 1/3 chance the third will be of the same colour, needed to make a matching set. In the different situation where the two are of two different colours, 1/3 the third spoils will be of the missing third colour needed for a rainbow set.

This is not a certainty it is probability and we have to be careful of the Gamblers Fallacy

http://en.wikipedia.org/wiki/Gambler%27s_fallacy

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.


But to answer the question asked I think the multiplication approach is correct. When you are looking to the future probability, not a certainty, probability, of at least one of the two players holding two spoils, getting a set, maybe just one of them, OR call me maybe baby ,two of them getting a set, the odds are 5/9. the chances of none of them getting a set, 4/.

So if I was looking at two oppos both holding two spoils. and boiling it down to two and only two likely outcomes the chances at least one of them getting a set or noen of them getting a set would say a very slight favouring at leat one of them getting a set the odds 5/9 yes versus 4/9 no.

With escalating spoils i would not just look at two probabilities I would look at multiple probabilities- starting with 1/ 3 the first oppo up #1 gets a set, Then regardless of whether s/he does get a set or not. 1/3 the next oppo up #2 gets a set. Which leads to four difffernt possible outcomes all of which can be big factors in escalating spoils

#! no set #2 no set
#1 no set #2 set
#1 set #2 no set
#1 set #2 set

Agreed

SirSebstar wrote:this is the anwser
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



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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