EPT
I even took Probability a couple semesters ago, lol.EDIT: I probably would have tried to apply Bayes theorem to it anyway, but I'm having a hard time getting it to work out.
Brainteaser for TeamLiquid! - Page 4
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Triscuit
United States722 Posts
I even took Probability a couple semesters ago, lol.EDIT: I probably would have tried to apply Bayes theorem to it anyway, but I'm having a hard time getting it to work out. | ||
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oxidized
United States324 Posts
On June 10 2011 10:39 n.DieJokes wrote: Yep, like I said its a brainteaser (actually its an anecdote I read in a introductory number theory text but its more fun phrased this way  ) so a1+1=a2, a2+1=a3 and so on and so forthHehe, the trick here is in using factorials. At first, I was trying to do 50!, but then 50!+1 is not guaranteed to be a non-prime number. So... you need to start on 50!+2, because that is definitely divisible by 2. But since we start on 2, we need to bump up our initial factorial by 1. So the range will be [51!+2 , 51!+51]. Because 51! is divisible by each number between 2 and 51. so that guarantees that 51!+x is non-prime as long as x is between 2 and 51. Cool problem. | ||
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numLoCK
Canada1416 Posts
A problem that turns many people off math, unfortunately | ||
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ObliviousNA
United States535 Posts
On June 10 2011 10:36 ZeromuS wrote: unfortunately you worded the question incorrectly and the answer you provide is incorrect as a result of the way you asked the question. The chances the "other game was Zerg as well" is 1/3 You used the phrase "as well" in your mind as meaning "in addition to" whereas colloquially "as well" means "similarly to". Therefore the question can be justifiably read as saying that You played Zerg in one game and what are the chances that you similarly played Zerg in the second game. Randomly selected you could play Zerg as one of 3 races a second time in a row since each event (playing a game as Random) always has the same probability (1/3 for each race). Therefore the chances you play Zerg the second game is 1/3 since each game is its own roll of the dice if you will. This is assuming that you meant "as well" as meaning "similar to" referencing the probability to roll Zerg just like you rolled Zerg in the first game. Had you used a different phrase then the 1/5 would be correct. Unfortunately, due to English ambiguity and the colloquial meaning of the phrase "as well" your current brain teaser is flawed. Thank you, you explained that really well. I thought "as well" meant "similarly to". I no longer consider everyone else morons for saying 1/5 :D | ||
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n.DieJokes
United States3443 Posts
On June 10 2011 10:52 oxidized wrote: + Show Spoiler + Hehe, the trick here is in using factorials. At first, I was trying to do 50!, but then 50!+1 is not guaranteed to be a non-prime number. So... you need to start on 50!+2, because that is definitely divisible by 2. But since we start on 2, we need to bump up our initial factorial by 1. So the range will be [51!+2 , 51!+51]. Because 51! is divisible by each number between 2 and 51. so that guarantees that 51!+x is non-prime as long as x is between 2 and 51. Cool problem. ^^ This is correct | ||
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InvalidID
United States1050 Posts
Using Bayes Theorem: P(A|B) = P(B|A)*P(A) / P(B) . P(B|A) = 1 P(A) = 1/3 P(B)=1 . P(A|B) = 0.33 = 1/3 This is because you already stated that you were Zerg in one game. The probability that you were Zerg in one game is 100% because you already knew that. Because you already knew you were Zerg one game the only three possible options are: ZZ ZP or ZT. You asked what the other game was, so the order is irrelevant. | ||
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Disquiet
Australia628 Posts
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Madoga
Netherlands471 Posts
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Isaac
United States810 Posts
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Tektos
Australia1321 Posts
On June 10 2011 10:11 theDreamStick wrote: Poll: Probability that my other game was Zerg? 1/3 (262) 1/5 (249) 1/2 (28) 1/4 (9) 548 total votes Your vote: Probability that my other game was Zerg? "Poll: Probability that my other game was Zerg?" The probability of your OTHER game being Zerg is mutually exclusive to the race of your original game. Hence the option of what race you play in your other game are: Z, P, T. Z is 1 of 3 possible options, hence the probability is 1/3 You worded your question incorrectly. The question you provided an answer for is: "What is the probability that both my games were played as zerg given at least one game was played as zerg" which is indeed 1/5. | ||
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Schneeflocke
Canada89 Posts
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Risen
United States7927 Posts
On June 10 2011 11:11 Tektos wrote: "Poll: Probability that my other game was Zerg?" The probability of your OTHER game being Zerg is mutually exclusive to the race of your original game. Hence the option of what race you play in your other game are: Z, P, T. Z is 1 of 3 possible options, hence the probability is 1/3 You worded your question incorrectly. The question you provided an answer for is: "What is the probability that both my games were played as zerg given at least one game was played as zerg" which is indeed 1/5. This, this, this. I was like how the hell are people confused about this? lol | ||
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W2
United States1177 Posts
You say you played zerg at least once... so it's zt zp zz as possibilities... I took all of them same likelihood so 1/3? Where did i go wrong? | ||
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Risen
United States7927 Posts
On June 10 2011 11:20 W2 wrote: im confused, i thought it was 1/3 as well. You say you played zerg at least once... so it's zt zp zz as possibilities... I took all of them same likelihood so 1/3? Where did i go wrong? You didn't. OP worded his question wrong. The answer is 1/3 | ||
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nitdkim
1264 Posts
(the real teaser :p) | ||
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SUSUGAM
United States177 Posts
Moral: Don't try to adapt 'brainteasers' to practical examples when you suck at language. The real brainteaser is trying to figure out what you are actually asking. | ||
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Zuramed
13 Posts
Find 3 numbers in an arithmetic sequence that when multiplied together yield a prime number. Answer below: + Show Spoiler + -3, -1, 1 | ||
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Reborn8u
United States1761 Posts
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? | ||
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funtimeplayer
United States15 Posts
On June 10 2011 11:11 Tektos wrote: [...] The probability of your OTHER game being Zerg is mutually exclusive to the race of your original game. Hence the option of what race you play in your other game are: Z, P, T. Z is 1 of 3 possible options, hence the probability is 1/3 You worded your question incorrectly. The question you provided an answer for is: "What is the probability that both my games were played as zerg given at least one game was played as zerg" which is indeed 1/5. I agree, well said. | ||
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SUSUGAM
United States177 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 both times? (the real teaser :p) Fixed... but nice try. =/ "2nd flip" means you're asking for 1 flip's probability. | ||
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