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On June 11 2011 01:31 Dlok wrote:Show nested quote +On June 11 2011 00:33 Logo wrote:On June 11 2011 00:27 Dlok wrote:On June 10 2011 23:53 Logo wrote:On June 10 2011 23:41 Dlok wrote:
to those who say 1/5: flip two coins, then tell me one of the results in any way that does not exclude the possibility of a pair. you would give me 1/3 odds on a pair, while the odds clearly havent changed and really are 1/2. now we do this 100 times and I make alot of money. If we do the test 100s of times the result will in fact be 1/3rd and I would gladly take you up on such odds at something less than 1/2 but greater than 1/3rd as it'd be free money. You could even write a very simple program to test it. Not understanding probability (which is very counter-intuitive) doesn't change reality. I'll also just leave this here: http://www.ted.com/talks/peter_donnelly_shows_how_stats_fool_juries.html we run it 100 times I recommend you stay away from Vegas. Do you even realize what you'd be betting on? If I win every time we get heads + heads then you'd win with heads + tails (or tails + heads) and every time we get tails + tails we'd call it a draw. You'd be a fool to give someone odds on heads + heads at > 1/3 when you're not winning on two of the 4 possible outcomes. Haha these two sentences from wikipedia are also ambiguous. It's unclear that the many people who argued strongly for both sides knew whether it was a paradox or not. Since that link was already posted at the beginning of the thread, I assume people here knew it's was paradox when arguing, so why simply not admit that the wording has it's importance and that it's pointless to argue forever, since both sides are right depending on how you interpret the wording ? The point is the paradox causes intense reactions by those who don't accept the answer as true while those who understand the right answer are... well right. I don't know why you'd think that people wouldn't argue over a paradox that's known to create controversy and argument. Every time you say Heads i say it will be pair of heads, if you say tailes i say pair of tails, now we know 50 out of a hundred are likely to be pairs so i will do fine. If i say i take only pair of heads i will win 1/4 of said 100 but if im allowed to withdraw when atleast one is not heads i raise my odds to 1/3. This however was not how the problem was stated, and I cant se how it could be interpited that way.
Because that is how the problem was stated? Many people are arguing about the wording of the problem, so you can refer to those posts.
The problem as you say it is:
"What are the odds of getting a pair knowing one result"
when the question is...
"What are the odds of getting a pair if one result is heads"
and they have different answers.
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On June 11 2011 03:03 FairForever wrote:Show nested quote +On June 11 2011 02:59 L3g3nd_ wrote:
i read the solution, then voted 1/3. youre making the mistake of counting ZT and TZ as two seperate cases, when they are actually exactly the same, given that ordering doesnt matter. It technically doesn't matter but ZT (or ZT/TZ) is twice as likely as ZZ, so it does matter in a sense. Alternatively you could look at it as ZZ ZT x2 ZP x2
ZT x2 is the exact same as ZT and TZ
Its like saying instead of 4+4+4+4 do 3*4
Let me do some math,
So i think one issue is the T/P, but im not sure. anyway ill use Z for zerg and N for not zerg.
Z/N (1/3*2/3) = 2/9
N/Z (1/3*2/3 = 2/9
Z/Z (1/3*1/3) = 1/9
N/N (2/3*2/3) = 4/9
Well balls now i have an answer of 1/9 and that wasnt even an option. what i do wrong
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On June 11 2011 03:14 L3g3nd_ wrote:Show nested quote +On June 11 2011 03:03 FairForever wrote:On June 11 2011 02:59 L3g3nd_ wrote:
i read the solution, then voted 1/3. youre making the mistake of counting ZT and TZ as two seperate cases, when they are actually exactly the same, given that ordering doesnt matter. It technically doesn't matter but ZT (or ZT/TZ) is twice as likely as ZZ, so it does matter in a sense. Alternatively you could look at it as ZZ ZT x2 ZP x2 ZT x2 is the exact same as ZT and TZ Its like saying instead of 4+4+4+4 do 3*4 Let me do some math, So i think one issue is the T/P, but im not sure. anyway ill use Z for zerg and N for not zerg. Z/N (1/3*2/3) = 2/9 N/Z (1/3*2/3 = 2/9 Z/Z (1/3*1/3) = 1/9 N/N (2/3*2/3) = 4/9 Well balls now i have an answer of 1/9 and that wasnt even an option. what i do wrong
You aren't getting rid of all the possibilities.
1/9 is correct on the odds of getting ZvZ. However we know that N/N did not happen so those gets thrown out completely because it's impossible for that to be the case since we were told one of the games was ZvZ. However this does not change the relative odds of the other situations (Z/N is still x2 as likely as Z/Z)
So then you have
Z/N 2/X
N/Z 2/X
Z/Z 1/X
and the sum of the probabilities have to equal 1 so we know it's 2/5 2/5 1/5.
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On June 11 2011 03:17 Logo wrote:Show nested quote +On June 11 2011 03:14 L3g3nd_ wrote:On June 11 2011 03:03 FairForever wrote:On June 11 2011 02:59 L3g3nd_ wrote:
i read the solution, then voted 1/3. youre making the mistake of counting ZT and TZ as two seperate cases, when they are actually exactly the same, given that ordering doesnt matter. It technically doesn't matter but ZT (or ZT/TZ) is twice as likely as ZZ, so it does matter in a sense. Alternatively you could look at it as ZZ ZT x2 ZP x2 ZT x2 is the exact same as ZT and TZ Its like saying instead of 4+4+4+4 do 3*4 Let me do some math, So i think one issue is the T/P, but im not sure. anyway ill use Z for zerg and N for not zerg. Z/N (1/3*2/3) = 2/9 N/Z (1/3*2/3 = 2/9 Z/Z (1/3*1/3) = 1/9 N/N (2/3*2/3) = 4/9 Well balls now i have an answer of 1/9 and that wasnt even an option. what i do wrong You aren't getting rid of all the possibilities. 1/9 is correct on the odds of getting ZvZ. However we know that N/N did not happen so those gets thrown out completely because it's impossible for that to be the case since we were told one of the games was ZvZ. However this does not change the relative odds of the other situations (Z/N is still x2 as likely as Z/Z) So then you have Z/N 2/X N/Z 2/X Z/Z 1/X and the sum of the probabilities have to equal 1 so we know it's 2/5 2/5 1/5.
ah, thatll be why its called conditional probability!
Well, while i still think its 1/3 when it comes to words,
Math doesnt lie, so i understand why its 1/5, mathematically at least
Thanks to those people pointing out where us people who dont get it are going wrong! for a lot of us its awhile since we did math at school, and like for me i havent used it since i left school
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incase you guys didn't know, on ladder you don't have a 33.3% chance of getting each race. Long detailed explanation that I don't feel like getting into and explaining, but it is not completely random.
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On June 11 2011 03:02 L3g3nd_ wrote:Show nested quote +On June 11 2011 02:11 foxmeep wrote:
Here's an easier way to understand the problem. Say I flip two coins, we can all agree that the possible outcomes are:
H-T 1/4
H-H 1/4
T-T 1/4
T-H 1/4
Now I tell you that if, and only if, a heads comes up, if the other coin is tails, I win. If the other coin is heads, you win. Simply making this proposition does not change the probability of each result coming up.
H-T I win
H-H I lose
T-T Draw
T-H I win
We already know each of the outcomes has an equal chance of occuring, so in 2/4 results, I win, 1/4 you win. Therefore the odds are 2:1, or 2/3 in my favour, and 1/3 in yours. NOT 1/2. no, because you are making ordering a necessaity in this. the OP's problem does not take ordering into account.
explain how thanks.
edit: nvm 
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^if you read a few posts above, i had some revelations 
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On June 11 2011 03:03 FairForever wrote:Show nested quote +On June 11 2011 02:59 L3g3nd_ wrote:
i read the solution, then voted 1/3. youre making the mistake of counting ZT and TZ as two seperate cases, when they are actually exactly the same, given that ordering doesnt matter. It technically doesn't matter but ZT (or ZT/TZ) is twice as likely as ZZ, so it does matter in a sense. Alternatively you could look at it as ZZ ZT x2 ZP x2
19 pages and the first thing that actually makes sense to me. I didn't realize the probability of ZT is twice the probability of ZZ. This helps, thanks
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Probability of getting zerg two consecutive times is 1/9..Given that probability to get zerg is 1/3
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On June 11 2011 03:43 Kikimiki wrote:
Probability of getting zerg two consecutive times is 1/9..Given that probability to get zerg is 1/3
the reason this is wrong is on this page, read a little
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On June 11 2011 03:32 Liveon wrote:Show nested quote +On June 11 2011 03:03 FairForever wrote:On June 11 2011 02:59 L3g3nd_ wrote:
i read the solution, then voted 1/3. youre making the mistake of counting ZT and TZ as two seperate cases, when they are actually exactly the same, given that ordering doesnt matter. It technically doesn't matter but ZT (or ZT/TZ) is twice as likely as ZZ, so it does matter in a sense. Alternatively you could look at it as ZZ ZT x2 ZP x2 19 pages and the first thing that actually makes sense to me. I didn't realize the probability of ZT is twice the probability of ZZ. This helps, thanks
Explain me why the chance to get ZT is twice of ZZ, as one plays zerg and one plays random.
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On June 11 2011 04:08 Robje wrote:Show nested quote +On June 11 2011 03:32 Liveon wrote:On June 11 2011 03:03 FairForever wrote:On June 11 2011 02:59 L3g3nd_ wrote:
i read the solution, then voted 1/3. youre making the mistake of counting ZT and TZ as two seperate cases, when they are actually exactly the same, given that ordering doesnt matter. It technically doesn't matter but ZT (or ZT/TZ) is twice as likely as ZZ, so it does matter in a sense. Alternatively you could look at it as ZZ ZT x2 ZP x2 19 pages and the first thing that actually makes sense to me. I didn't realize the probability of ZT is twice the probability of ZZ. This helps, thanks Explain me why the chance to get ZT is twice of ZZ, as one plays zerg and one plays random.
please reread the OP
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Guys the answer is 100% because everytime I choose random I "randomly" get zerg 
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On June 11 2011 03:47 L3g3nd_ wrote:Show nested quote +On June 11 2011 03:43 Kikimiki wrote:
Probability of getting zerg two consecutive times is 1/9..Given that probability to get zerg is 1/3 the reason this is wrong is on this page, read a little
OK ill explain it to you,why you are wrong. This is right if the question was "whats the probability of getting zerg 2 consecutive times ?"
This situation could be described using joint probability rules....
If "A" was used to denote getting zerg at game 1, P(A)= 1/3
If "B" was used to denote getting zerg at game 2, P(B)= 1/3
The probability of occurrence of both of A and B is denoted ![[image loading]](http://upload.wikimedia.org/math/1/c/5/1c5070056d7351f8090e9436778a2cf5.png) (for independant variables or probabilities, which means the result of event A (getting zerg in game 1) in the 1st try doesn't alter event B)
Which is also equal to P(A) x P(B) = 1/9
If the question was "Whats the probability of occurrence of event be given that A had occured ?"
That would be denoted by: P(B|A) which is also equal to P(B)= 1/3, for independent event..
Thats because getting zerg in the 1st try doesn't change your chances of getting zerg the 2nd try..
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On June 11 2011 04:29 Kikimiki wrote:Show nested quote +On June 11 2011 03:47 L3g3nd_ wrote:On June 11 2011 03:43 Kikimiki wrote:
Probability of getting zerg two consecutive times is 1/9..Given that probability to get zerg is 1/3 the reason this is wrong is on this page, read a little OK ill explain it to you,why you are wrong. This is right if the question was "whats the probability of getting zerg 2 consecutive times ?" This situation could be described using joint probability rules.... If "A" was used to denote getting zerg at game 1, P(A)= 1/3 If "B" was used to denote getting zerg at game 2, P(B)= 1/3 The probability of occurrence of both of A and B is denoted (for independant variables or probabilities, which means the result of event A (getting zerg in game 1) in the 1st try doesn't alter event B) Which is also equal to P(A) x P(B) = 1/9If the question was "Whats the probability of occurrence of event be given that A had occured ?" That would be denoted by: P(B|A) which is also equal to P(B)= 1/3, for independent event.. Thats because getting zerg in the 1st try doesn't change your chances of getting zerg the 2nd try..
But A does not equal getting zerg on your first try. It's getting Zerg on one of the two tries. The events are not independent. B is also "Zerg on both of two games" not "Zerg on the 2nd game"
It's easy to illustrate. If you look at one of the two games and it's not zerg the probability of the other game being Zerg is 1 (100%) otherwise our known condition (at least one game is zerg) wouldn't be true.
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The question is ambiguous for the same reason the Boy/Girl Paradox question is ambiguous. Draw an analogy from this quote about the Boy/Girl Paradox:
"While it is certainly true that every possible Mr. Smith has at least one boy - i.e., the condition is necessary - it is not clear that every Mr. Smith with at least one boy is intended. That is, the problem statement does not say that having a boy is a sufficient condition for Mr. Smith to be identified as having a boy this way."
i.e. - if your friend would not always say that he saw a ZvZ, but perhaps some of the times that TLO played zerg in one game, he said instead that Nestea has sick ZvT because he only watched the other game where TLO played Terran... then the conditional probability of your friend telling you that Nestea has a sick ZvZ when TLO played at least once as zerg is less than one. This provides a different answer than if TLO playing one game as zerg is sufficient for your friend to report that Nestea has sick ZvZ.
If your friend could possibly have told you that Nestea has sick ZvT when TLO played zerg in the other game, then the probability that TLO played zerg twice is much higher because the conditional probability of your friend telling you Nestea played a sick ZvZ when TLO rolled ZZ is 1... and 1/2 for all other combinations.
Therefore:
P(XX) = 1/9
P("ZvZ"|ZZ) = 1
P("ZvZ"|ZX OR XZ) = 1/2
P("ZvZ") = 1/9 + 4 * (1/18) = 3/9 = 1/3
P(ZZ|"ZvZ") = P("ZvZ"|ZZ) * P(ZZ) / P("ZvZ") = 1 * 1/9 / 1/3 = 3/9 = 1/3
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From what I gathered on the Wikipedia article previously posted, the problem is only paradoxical because of the ambiguous nature of the statement of the problem. The statistical analysis is correct when viewed from both sides, it is only the view which causes the inconsistency. By changing a few words in the statement the view can change from a statistical analysis of a singular event (e.g. I played 2 games. The first game was played as zerg. What is the probability the next game will be played as zerg? In this case the first and second statement are true, but unimportant, as the problem asks for the probability that the next game will be zerg. We can deduce that only 1 of 3 options fits the parameters of the question, therefore the answer is 1/3.) to an analysis of all possibilities involving at least one zerg (e.g. I played 2 games. One game was played as Zerg. What is the probability of playing zerg in both games? Given these parameters you weed out all non-zerg possibilities and come to the aforementioned 1/5 answer.) as well as a further possibility of analysis involving all possibilities over two games when what race is played either game is unknown (e.g. I played 2 games. What is the probability of playing zerg in both games? This time the statement of one game being played as zerg is removed and all possibilities must be included. Since there are 9 possibilities, and only 1 of them is ZZ, the answer is 1/9.).
For an additional example of the linguistic nature of this type of paradox I will quote a different one from Wikipedia:
For example, consider a situation in which a father and his son are driving down the road. The car crashes into a tree and the father is killed. The boy is rushed to the nearest hospital where he is prepared for emergency surgery. On entering the surgery suite, the surgeon says, "I can't operate on this boy. He's my son."
In this situation the reader may first assume that it is an impossibility and completely illogical. The father was killed in the crash, therefore it is impossible for him to be a surgeon at the hospital when his son arrives. While this conclusion is logically correct, it is founded on the assumption that the father is the only possible parent. If it is given that the mother is a surgeon at the hospital, then it can be concluded that the mother is the surgeon in question and as such the statement is entirely possible and logical. It is the specific way in which the information is presented that leads to false assumptions, providing the ambiguity of the problem.
Hopefully this will help shift the focus of the discussion from whether the answer is 1/3 or 1/5 to the actual problem of the inherent ambiguity of the original question and the erroneous assumptions and conclusions made from it. And for anyone who denies that the original question is ambiguous, the words used are specifically designed to create confusion leading to it being "open to more than one interpretation" which is the definition of ambiguous. If it wasn't designed in such a way we wouldn't be discussing it.
[An Aside] I've been up all night and I am terribly tired, so if I messed something up I apologize.
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On June 10 2011 22:52 Tektos wrote:Show nested quote +On June 10 2011 22:45 DarkPlasmaBall wrote:On June 10 2011 22:33 Tektos wrote:On June 10 2011 22:30 DarkPlasmaBall wrote:
Agreed. You're not twice as likely to play Zerg in ZZ as you are in TZ. You just happen to have twice as many Zs. Big difference. I flip a standard coin, what is the probability of heads? I flip a coin that has heads on both sides, what is the probability of heads? If you have twice as many of something you're twice as likely to pick said thing. Two different games are two different coin flips. Not the same coinflip. It's not even close to the same thing. And you're still forgetting about conditional probability. Anyways, I gotta go. Enjoy your day You're utterly confused.If I have two replays and I play the same race in each replay, and you select one of those replays at random, it is twice as likely to be that race than if you have two replays where you play a different race in each replay. The coin flip was to represent which of the two replays in the series gets sent to you. I'm upset that the education system hasn't worked for so many people.
Oh, the irony in this post. You're one of the few people who has been arguing for 1/3 throughout pages and pages of this thread, despite us attempting to explain conditional probability to you.
Me too. Me too.
I guess we math teachers need to do a little better in explaining these problems so that people don't misinterpret the question and end up with a wrong analogy and answer, huh? 
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On June 11 2011 04:51 Logo wrote:Show nested quote +On June 11 2011 04:29 Kikimiki wrote:On June 11 2011 03:47 L3g3nd_ wrote:On June 11 2011 03:43 Kikimiki wrote:
Probability of getting zerg two consecutive times is 1/9..Given that probability to get zerg is 1/3 the reason this is wrong is on this page, read a little OK ill explain it to you,why you are wrong. This is right if the question was "whats the probability of getting zerg 2 consecutive times ?" This situation could be described using joint probability rules.... If "A" was used to denote getting zerg at game 1, P(A)= 1/3 If "B" was used to denote getting zerg at game 2, P(B)= 1/3 The probability of occurrence of both of A and B is denoted (for independant variables or probabilities, which means the result of event A (getting zerg in game 1) in the 1st try doesn't alter event B) Which is also equal to P(A) x P(B) = 1/9If the question was "Whats the probability of occurrence of event be given that A had occured ?" That would be denoted by: P(B|A) which is also equal to P(B)= 1/3, for independent event.. Thats because getting zerg in the 1st try doesn't change your chances of getting zerg the 2nd try.. But B does not equal getting zerg on your first try. It's getting Zerg on one of the two tries. The events are not independent.
The original problem is :
"So I was playing random today, and I played 2 games of Starcraft 2!
I played as Zerg at least once. What is the probability that my other game was as Zerg as well?"
2 Games played.
A = getting zerg in 1st game
B = getting zerg in 2nd game
My translation of these words would be P(B|A), the probability of getting zerg in the 2nd try is independent of the 1st try.....In other words getting zerg the 1st game doesnt change the probability of getting zerg 2nd game, and thus events are independent.
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infinity21
Canada6683 Posts
P(at least 1 Z & 2 Zs)/P(at least 1 Z)=P(2 Zs)/[1-P(no Z)]
=1/9 / [1- (2/3)^2] = 1/5
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EPT
