EPTBrainteaser for TeamLiquid! - Page 11
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zobz
Canada2175 Posts
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Eknoid4
United States902 Posts
Now now, child, this is cheating. | ||
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Clog
United States950 Posts
On June 10 2011 16:33 TheOne85 wrote: The actual question asked was "I played as Zerg at least once. What is the probability that my other game was as Zerg as well?" Brain teaser messed up the OP so badly that he posted it in a retarded manner. No it's not. At least it wasn't when I came in. Reread the OP, I copied and pasted it | ||
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Shaok
297 Posts
On June 10 2011 16:35 Clog wrote: No it's not. At least it wasn't when I came in. Reread the OP, I copied and pasted it Ok it was the original problem spoilered. Who wrote that amazing piece of English? I am also going with this dude: On June 10 2011 16:34 garbobjee wrote: When you said, ZP, ZZ, ZT, PZ, and TZ, you stated the number of possible combinations incorrectly. The order of the games doesn't actually matter, the only thing that does matter is how many times you played as a certain race. So now, ZP and PZ should be one outcome, and ZT and TZ should also be one outcome. With those outcomes combined, there should only be three outcomes: ZZ, ZP, and ZT, so the chance of playing as zerg twice given that you played as zerg at least once, is 1/3. | ||
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SolC361
United States184 Posts
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Clog
United States950 Posts
On June 10 2011 16:38 TheOne85 wrote: Ok it was the original problem spoilered. Who wrote that amazing piece of English? I am also going with this dude: I understand the reasoning behind it, but I'd still argue it's incorrect. The order of the game does matter when considering its independent probability of happening. If you play two games as random, you are twice as likely to get 1 Zerg 1 Protoss than you are getting Zerg both games (which can be shown by splitting the Zerg + Protoss result into two based on order, and given them equal probabilities). It's the same reason that after flipping a coin twice you have a 50% chance to get 1 Heads 1 Tails, and then 25% each for both Heads and both Tails. And this, as well, could be split up based on order saying that its just 25% for all the following: HH, HT, TH, TT Basically, if you do something to identify a specific game or result (such as using the words "the other"), then the order does not matter, and the SC2 example answer would in fact be 1/3. But if you do not (such as the OP currently has in the non-spoiler chunk of his post), it does, for purposes as shown in the coin flipping scenario, and the SC2 example answer would be 1/5. Further reading: http://en.wikipedia.org/wiki/Boy_or_Girl_paradox | ||
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freeloader625
United States180 Posts
Independent Events: The answer is 1/3 if they are independent events. This is the case in which the games are independent: + Show Spoiler + "I play two starcraft games as random. One of them was Zerg. What is the probability that my other game was Zerg?" Total Possible Outcomes:
PT PZ TP TT TZ ZP ZT ZZ We look at game two and see 3/9 or 1/3 Simple. Conditional events The answer is 1/5 if they are not independent events. + Show Spoiler + Total Possible Outcomes:
PT PZ TP TT TZ ZP ZT ZZ However "given that at least one of my games was Zerg, what is the probability that both of my games are zerg." So now we must eliminate to meet our conditions (hence conditional probability) We are now left with: Total Outcomes that meet the conditions stated:
TZ ZP ZT ZZ Now we simply count which meet condition one (one game played was zerg) and condition two (the other game is zerg AS WELL). One out of Five = 1/5. Yes the wording was ambiguous at first, which led to confusion. But the "as well," makes it conditional. Now the OP changed so there is no ambiguity at all simply asking straight out that both games are played as zerg after one of them was already played as zerg. This is just a purely simple proof. + Show Spoiler + EDIT: To garbobjee On June 10 2011 16:34 garbobjee wrote: When you said, ZP, ZZ, ZT, PZ, and TZ, you stated the number of possible combinations incorrectly. The order of the games doesn't actually matter, the only thing that does matter is how many times you played as a certain race. So now, ZP and PZ should be one outcome, and ZT and TZ should also be one outcome. With those outcomes combined, there should only be three outcomes: ZZ, ZP, and ZT, so the chance of playing as zerg twice given that you played as zerg at least once, is 1/3. WRONG. Look at above all possible outcomes to see why order of games does matter. | ||
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freeloader625
United States180 Posts
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Rammblin Man
Canada19 Posts
1 = zerg, 2 = toss, 3 = terran Now, generate a random number twice to create a set of data. Next, do this 9999 more times. Done? Good. Now, decide which question you want to ask. If you want to look only at the series that have at least one zerg in them, then eliminate all the data sets that don't have a 1 in them. Next, count how many of these sets have two zergs. The answer will approximately be equal to the chances of having two zergs in that situation. If you want to look at all of the series, choose one set at random. Does it have a 1 in it? If so, record the other number in that set. Next, do this 9999 more times. The number of times that you record a double zerg will be approximately equal to the probability of finding a second zerg in a set if you randomly choose one that already contains a zerg. There you go, there should be two different answers, both derived empirically. No more arguing should be allowed until everybody has done this. | ||
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oBlade
United States5908 Posts
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freeloader625
United States180 Posts
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Geo.Rion
7377 Posts
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Cambium
United States16368 Posts
Let's say IMNestea (always Z) is playing against, say, TLO, and TLO is playing random. They're going to play exactly two games (for simplicity). Your friend spoils the fun by saying "OMG Nestea's ZvZ is absolutely sick!" (implying he saw a ZvZ game). This leaves a lot up for interpretation. Your friend could have said that to you after the first game, with the 2nd game yet to be played, in this case, the answer is 1/3. What you really meant was Let's say IMNestea (always Z) is playing against, say, TLO, and TLO is playing random. They played exactly two games (for simplicity). Your friend spoils the fun by saying "OMG Nestea's ZvZ is absolutely sick!" (implying he saw at least one ZvZ game). This is a simple application of conditional probability, and it's hardly a brainteaser. Those who say 1/5 are correct: P(A | B) = P(A \cap B) / P(B) (\cap is the intersection sign) In this case: P(A) = twice Z P(B) = at least Z once P(A \cap B): Since P(A) is a subset of P(B), P(A \cap B) = P(A) = 1/3 * 1/3 = 1/9 P(B) = P(at least Z once) = 1 - P(no Z) = 1 - (2/3)^2 = 5/9 P(A | B) = 1/9 / 5/9 = 1/5 | ||
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Sneakyz
Sweden2361 Posts
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fy12345
Canada151 Posts
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semantics
10040 Posts
http://www.teamliquid.net/blogs/viewblog.php?topic_id=155158 + Show Spoiler [quite long somethinggraphic] + ![[image loading]](http://www.teamliquid.net/staff/disciple/light_copy.jpg) On June 10 2011 16:32 Eknoid4 wrote: No, I'm serious. This doesn't deserve a thread. Ask a friend who is decent at math. Don't start a 10+ page argument. I don't have to try to make anybody i nthis thread look stupid. People arguing with basic math concepts when the OP has more to do with reading comprehension (which, in the context of math, you also should have learned in middle school) are making themselves look stupid enough. | ||
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han_han
United States205 Posts
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Cambium
United States16368 Posts
On June 10 2011 16:59 Geo.Rion wrote: Monthy python paradox reloaded? Monty Python is very different. Monty Python is a "paradox" because the state space was never fully defined. | ||
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greendestiny
Bosnia-Herzegovina114 Posts
Here's my explanation: In two matches, a player can get match 1 - Z P T match 2 - Z P T ZZ ZP ZT PZ PP PT TZ TP TT (9 combinations) Since we know he was Zerg once, we rule out all possibilities without Z. ZZ ZP ZT PZ The answer is: 1/5. | ||
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Ivs
Australia139 Posts
On June 10 2011 17:04 Cambium wrote: Monty Python is very different. Monty Python is a "paradox" because the state space was never fully defined. You mean the Monty Hall paradox? Zerg tears  Monty Hall | ||
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