EPTBrainteaser for TeamLiquid! - Page 5
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rbx270j
Canada540 Posts
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Hamster1800
United States175 Posts
On June 10 2011 11:25 SUSUGAM wrote: Fixed... but nice try. =/ "2nd flip" means you're asking for 1 flip's probability. That's indeed what he was doing, and the answer is not 1/2. | ||
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Kyuukyuu
Canada6263 Posts
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SUSUGAM
United States177 Posts
On June 10 2011 11:28 Hamster1800 wrote: That's indeed what he was doing, and the answer is not 1/2. I disagree, he said 'on the 2nd flip' which excludes all other information, because he's asking for the probability of a single, specific, 50/50 flip. Had he said 'both times', or something similar, it would have been fine. | ||
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Hamster1800
United States175 Posts
On June 10 2011 11:31 SUSUGAM wrote: I disagree, he said 'on the 2nd flip' which excludes all other information, because he's asking for the probability of a single, specific, 50/50 flip. Had he said 'both times', or something similar, it would have been fine. Then do it yourself. Get a coin, flip it in pairs 100 times (so 200 flips). Then cross out all pairs where neither of the ones were heads and count up how many are left and how many had the second one as heads. | ||
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teamsolid
Canada3668 Posts
On June 10 2011 11:29 Kyuukyuu wrote: Since "I played Z once" is a given it has no effect on the probability of the other one; I'm with the "you worded it wrongly" camp You should look this up. http://en.wikipedia.org/wiki/Conditional_probability The question is actually worded fine. It just sounds ambiguous, because the question is in the poll, while the given part is written earlier. If it was all put in one sentence, there'd be 0 ambiguity though. | ||
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radscorpion9
Canada2252 Posts
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nitdkim
1264 Posts
On June 10 2011 11:31 SUSUGAM wrote: I disagree, he said 'on the 2nd flip' which excludes all other information, because he's asking for the probability of a single, specific, 50/50 flip. Had he said 'both times', or something similar, it would have been fine. I meant to say what I've said. If you think about it, you will be confused for a little while. | ||
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freeloader625
United States180 Posts
1) Facts Presented 2) Question asked Then you proceed to answer the question with the facts presented. If you choose to ignore the facts then you get it wrong. The OP's question On June 10 2011 10:11 theDreamStick wrote: CLARIFICATION: I'm talking about -my- race For those of you that are bored, 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? Notice what I underlined. He did not ask for an independent event. His final question however, was simply summarized and shortened. | ||
Whitewing
United States7483 Posts
The question isn't "What are the odds of randomly getting zerg in a match." The question is "What are the odds of getting zerg twice in two matches, given that I got zerg at least once." | ||
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oxidized
United States324 Posts
On June 10 2011 11:22 nitdkim wrote: I have a coin. I flipped it twice. It was heads at least once. What is the probability that it was heads on the 2nd flip? (the real teaser :p) Good teaser. Now, ignoring all those comments about you miswriting the question, here is what we have. There are initially 4 scenarios possible: HT TH HH TT But you gave us the information that heads came up at least once. That rules out TT, leaving HT, TH, and HH. That gives you a 2/3 probability that it was heads on the second flip.  | ||
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oxidized
United States324 Posts
On June 10 2011 11:45 Whitewing wrote: Not really a brainteaser, but it looks like most people here have no idea what conditional probability is >_<. The question isn't "What are the odds of randomly getting zerg in a match." The question is "What are the odds of getting zerg twice in two matches, given that I got zerg at least once." Yep. It seems a lot of people are missing the whole conditional part of the problem statement. I didn't find any ambiguity in it at all (unless it has been edited, which I don't think it was). And as someone else said before, the poll was only the last part of the full question, which may have been the cause for some confusion. | ||
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lachy89
Australia264 Posts
Assuming the Second game was zerg there is a 1/3 chance of the other being zerg. You said that at least one game you played as zerg. Regardless if you played the first or second game as zerg the chance of the other being zerg is still 1/3. | ||
Warpath
Canada1242 Posts
It's either he gets Zerg, or he doesn't. | ||
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Steel
Japan2283 Posts
The odds that you play zerg will always be 1/3. No matter if you play as zerg 6 times before, random doesn't remember. You still have 1/3 chance of getting zerg. The probability of getting things, 'in a row' id different. Anyways, I'm not a math major (but I will be in a few year =D ), but this thread has been an overall bad experience. On June 10 2011 11:51 Warpath wrote: Guys, the answer is 1/2 It's either he gets Zerg, or he doesn't. Warpath fanclub coming up | ||
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nitdkim
1264 Posts
On June 10 2011 11:51 Warpath wrote: Guys, the answer is 1/2 It's either he gets Zerg, or he doesn't. Chance for me to win the lottery is 1/2? damn, let's all go buy some lotto tickets! I win the lottery or I don't right? 50/50!! | ||
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00Visor
4337 Posts
On June 10 2011 11:50 lachy89 wrote: Assuming the First game was zerg, there is a 1/3 chance of the other being zerg. Assuming the Second game was zerg there is a 1/3 chance of the other being zerg. You said that at least one game you played as zerg. Regardless if you played the first or second game as zerg the chance of the other being zerg is still 1/3. But you can't seperate the problem like this in first + second game. Then you count the case Zerg/Zerg twice. There lies the problem. | ||
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neo_sporin
United States516 Posts
On June 10 2011 11:25 Reborn8u wrote: I have almost no understanding of probability, so help me out with something. If I flip a coin 100 times and it lands heads every time, wouldn't each flip after that still have a 50/50 chance of coming up heads? I'm not understanding how the fact that you got zerg 1 game has any affect on the chances of you getting zerg in the next game. I would be inclined to answer 1/3, being that in any truly random event with 3 possible outcomes the likely hood of each of those outcomes would be 1/3. The post didn't ask "what are the chances of getting zerg in both games if you got zerg at least once." Even if it did ask that, by definition how can 1 random event change the odds of another random event? The question only asks about 1 game, so how is the first game relevant to the odds of the second game? As for the boy girl part above I am not following. You are saying that girl girl is an impossibility of course, but (boy + girl) and (girl + boy) why would you count this possibility twice? It is making both children variables, when only 1 is actually a variable.You just don't know which one is the variable. If you count (boy + girl) twice then wouldn't you have to count (Boy + Boy) twice? I will respond using the boy/girl as it only has 2 variable. The boy/girl and girl/boy are split up for clarity purposes, but really have no effect on the outcome of boy/boy being 1/3 So by using this word choice you have to account for the two very different situations of the boy coming first and the boy coming second. They do not have to statistically count for 2 different values though. IF i told you the FIRST was a boy, then yes there is a 50% as they are exclusive events and do not depend on eachother. But since the question is being asked in such a way that the two events do depend on eachother for a combined probability, you have to look at the entire problem concurrently, not independently. So the choices become 1st child 2nd child boy girl boy girl If you create all possible events in this scenario you could have since having a boy and girl each have 50% chance of occuring. 1st boy 2nd boy 1st boy 2nd girl 1st girl 2nd boy 1st girl 2nd boy So now you see there is a 25% boy boy, 50% boy/girl combination in some fashion. 25% girl girl. However the question as presented has told you that there is at least one boy (but possibly 2) so you take out the girl girl option. You are now left with 25% boy/boy or 50% boy/girl combination. Since it will never be girl/girl there is now a total of 75 times out of 100 that these solutions will come up. 25 out of the 75 are solutions involving boy boy, the other 50 are boy/girl or girl/boy. 25 out of 75 is 1/3. I hope that makes a bit more sense? | ||
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Hamster1800
United States175 Posts
On June 10 2011 11:52 Steel wrote: It really depends how you word it. Nitpick: It doesn't matter how you word it. It matters what the situation is. As it turns out there are lots of wordings that correspond to more than one situation, and this is arguably one of them (although it's better than a lot of things that I've seen). | ||
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oxidized
United States324 Posts
On June 10 2011 11:52 Steel wrote: I know, right?!?!?!? What the hell is it with people getting so worked up about math. Remember the whole 0.999...=1 proofs? God, I don't get why people get so worked up over this stuff. X_Xbut this thread has been an overall bad experience. On June 10 2011 11:51 Warpath wrote: Guys, the answer is 1/2 It's either he gets Zerg, or he doesn't. HAHAHA, nice one! | ||
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