EPTand 2/6 if you do not know, ZT, ZP, (Z revealed)Z, PZ,TZ, Z(Z revealed)
Brainteaser for TeamLiquid! - Page 13
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Dlok
Sweden7 Posts
and 2/6 if you do not know, ZT, ZP, (Z revealed)Z, PZ,TZ, Z(Z revealed) | ||
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Llama
United Kingdom69 Posts
![[image loading]](http://i.imgur.com/68oPu.png) As probability goes this one is pretty basic. Here's a write-up of a slightly more complex variant of this problem: http://gregegan.customer.netspace.net.au/ESSAYS/TUESDAY/Tuesday.html | ||
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Dimagus
United States1004 Posts
On June 10 2011 20:14 Dlok wrote: Answer is 1/3 that the other is zerg if you know witch one is being revealed and 2/6 if you do not know, ZT, ZP, (Z revealed)Z, PZ,TZ, Z(Z revealed) This is a common error when dealing with set operations and the "at least" function. For instance, if you wanted all the two digit numbers that have at least one 5 in it, you would search for: *5 = (15, 25, 35, 45, 55, 65, 75, 85, 95) 5* = (50, 51, 52, 53, 54, 55, 56, 57, 58 , 59) and then if the sets aren't merged correctly you end up with two 55s. For a three digit number, three 555s, etc etc. | ||
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Cyber_Cheese
Australia3615 Posts
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niteReloaded
Croatia5282 Posts
I answered 1/3 before reading carefully. | ||
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Euronyme
Sweden3804 Posts
The poll question is: "Probability that my other game was Zerg?" There are 3 options. T Z and P. You didn't ask about the specific order of games. Reword the poll question and it'd be fixed. | ||
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aRRoSC2
Denmark241 Posts
For any problem of this kind, use this formula: Chance = 1 / (races^(matches-1) + (races-1)*(races^(matches-1) - (races-1)^(matches-1))) Of course, the word "matches" can be replaced with "attempt" or any other fitting word, as can "races" with "choices" etc. Examples: In the OP we have 3 races and 2 matches. Chance = 1 / (3^1 + 2*(3^1-2^1)) = 1 / (3 + 2*1) = 1 / 5 This shows that it at least works on the question in the OP. Let's try with 4 races, such as Human, Orc, Undead and Night Elf. You play random for 4 matches and at least one of them you play Undead. What are the odds you played Undead in all 4 games? Chance = 1 / (4^3 + 3*(4^3 - 3^3)) = 1 / (64 + 3*(64 - 27) = 1 / (64 + 3*37) = 1 / 175 Had we not known you played Undead in at least one game, there would have been 4^4 = 256 possibilities, however, there are only 175 possibilities where Undead has been played (or in other words 81 where Undead has not been played), so the chances of playing Undead 4 times in a row KNOWING it has been played at least once is 1/175. | ||
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Tektos
Australia1321 Posts
http://people.hofstra.edu/stefan_waner/realworld/tutorialsf3/frames6_5C.html Independent Events The events A and B are independent if any one of the following three equivalent conditions hold. P(AB) = P(A)P(B) P(A|B) = P(A) - - - - - B has no effect on A P(B|A) = P(B) - - - - - A has no effect on B Intuitively, two events are independent if the occurrence of one has no effect on the probability of the other. If two events E and F are not independent, then they are dependent. | ||
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DarkPlasmaBall
United States45320 Posts
Chance of the other being Zerg is just one out of those 5, as 2 of those 5 show a Protoss and 2 show a Terran. Only 1 gives you a double Zerg. Therefore, 1/5. | ||
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Llama
United Kingdom69 Posts
On June 10 2011 21:13 Tektos wrote: Everybody saying 1/5 is assuming this is a dependent event problem when it isn't. Can you explain how the diagram I drew is wrong? | ||
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aRRoSC2
Denmark241 Posts
On June 10 2011 21:13 Tektos wrote: I can't believe people can still think it is 1/5: http://people.hofstra.edu/stefan_waner/realworld/tutorialsf3/frames6_5C.html Independent Events The events A and B are independent if any one of the following three equivalent conditions hold. P(AB) = P(A)P(B) P(A|B) = P(A) - - - - - B has no effect on A P(B|A) = P(B) - - - - - A has no effect on B Intuitively, two events are independent if the occurrence of one has no effect on the probability of the other. If two events E and F are not independent, then they are dependent. Everybody saying 1/5 is assuming this is a dependent event problem when it isn't. I can't believe how bad you are at reading. | ||
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Tektos
Australia1321 Posts
On June 10 2011 21:21 Llama wrote: Can you explain how the diagram I drew is wrong? We're answering different questions. The poll says probability that the other game is zerg (being 1/3), the diagram shows the answer to ZZ given at least 1 Z (1/5). I jumped back into the thread after a few hours break and didn't see OP had edited everything to ask a completely different question to what he was asking earlier in the day. We're both right just answering different questions. The OP sucks. | ||
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Llama
United Kingdom69 Posts
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Pyo
United States738 Posts
On June 10 2011 21:13 Tektos wrote: I can't believe people can still think it is 1/5: http://people.hofstra.edu/stefan_waner/realworld/tutorialsf3/frames6_5C.html Independent Events The events A and B are independent if any one of the following three equivalent conditions hold. P(AB) = P(A)P(B) P(A|B) = P(A) - - - - - B has no effect on A P(B|A) = P(B) - - - - - A has no effect on B Intuitively, two events are independent if the occurrence of one has no effect on the probability of the other. If two events E and F are not independent, then they are dependent. They are dependent given the original statement of the problem. He says that one of the two games he played zerg. That mean that if game 1 was protoss, game 2 HAD to be zerg. That is dependence. | ||
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aRRoSC2
Denmark241 Posts
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Tektos
Australia1321 Posts
"Probability of the other game being zerg given at least one game had zerg" are different questions. I was answering the first, you're answering the second. Or, there is the other question: "I played as Zerg at least once. What is the probability that my other game was as Zerg as well?" Played zerg at least once, so the possible outcomes are: ZZ ZT ZP TZ PZ Then "other game was zerg" if we take it as game 1 is the "other" game then the probability is 3/5. If you take it as game 2 is the "other" game then it is also 3/5. All different questions, similar wording = confusion between everyone in the thread. | ||
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foxmeep
Australia2337 Posts
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aRRoSC2
Denmark241 Posts
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Llama
United Kingdom69 Posts
On June 10 2011 21:33 Tektos wrote: "Probability of the other game being zerg" and "Probability of the other game being zerg given at least one game had zerg" are different questions. I was answering the first, you're answering the second. The first question was never actually asked though. Both wordings imply the same thing. | ||
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Tektos
Australia1321 Posts
On June 10 2011 21:37 Llama wrote: The first question was never actually asked though. Both wordings imply the same thing. If you look at ONLY the poll without reading the rest of OP then yes that is the question. If you read the thread it should be 1/5, if you read the old question it should be 3/5. See my previous post's edit. | ||
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