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Brainteaser for TeamLiquid! - Page 7

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Tektos
Profile Joined November 2010
Australia1321 Posts
June 10 2011 03:45 GMT
#121
On June 10 2011 12:42 synapse wrote:
Show nested quote +
On June 10 2011 12:30 Tektos wrote:
It is clearly 1/4 cause he can be Zerg, Protoss, Terran or Random trololol.

1/4 + (1/4)(1/3) = 1/3
Nice try


Yeah but you forgot that Blizzard's algorithm for random isn't 1/3 chance for each race, it gives you zerg 50% of the time.
ixi.genocide
Profile Blog Joined June 2010
United States981 Posts
June 10 2011 03:47 GMT
#122
I answered 1/3. The reason why I answered 1/3 is because the wording made it seem that you played Z first and the # that you were looking for was the probability that you would get Z again. If I hadn't read your op incorrectly (which is prolly my fault) I would have answered 1/5 (only 1 ZZ option out of all Zx pairings)
Beez
Profile Joined October 2010
Canada18 Posts
June 10 2011 03:51 GMT
#123
the answer doesnt rely on knowing the race of your opponent so you can be T, P, or Z. theres no reason to even think about matchups.
EchelonTee
Profile Joined February 2011
United States5261 Posts
Last Edited: 2011-06-10 03:55:29
June 10 2011 03:53 GMT
#124
the wording of the question is fine, and it is clearly 1/5. focusing too much on the order of reading question leads to problems in understanding....basic probablity = HS education


pleaseee restore the original wording of the Q and put ur rebuttal in spoilers or at the bottom of the post...I enjoyed it as a Q
aka "neophyte". learn lots. dont judge. laugh for no reason. be nice. seek happiness. -D[9]
Deleted User 123474
Profile Joined November 2010
292 Posts
June 10 2011 03:53 GMT
#125
This is a good question.

I spent about 20-30 minutes mulling it over in my mind, thinking, it has to be 1/3. But then why would he post it? I couldn't get behind the idea of it being anything else though, so I decided to brute force it in a slightly different way than OP.

OP got Zerg at least once, so possible permutations are: ZP,ZT,ZZ; PZ,TZ,ZZ. The duplicate "ZZ" is crossed out. Now we have one "ZZ" left and the others are ZP,ZT,PZ,TZ -- meaning the chance of getting Zerg in the other game as well once already getting Zerg once has to be 1/5.

Probability will stick you in the egg boiler and make you delicious.
lyAsakura
Profile Blog Joined March 2010
United States1414 Posts
June 10 2011 03:58 GMT
#126
On June 10 2011 12:51 Beez wrote:
the answer doesnt rely on knowing the race of your opponent so you can be T, P, or Z. theres no reason to even think about matchups.


indeed you are correct, but everybody knows this and nobody is arguing about matchups
it is simply "i play two games as random, i spawn as zerg atleast once, what's the probability of me spawning as zerg in both games"
WeMade FOX would be a deadly SC2 team.
Zarathusta
Profile Joined September 2010
United States114 Posts
June 10 2011 04:00 GMT
#127
The edited question is incredibly confusing, next time don't give into peer pressure.
Jinsho
Profile Joined March 2011
United Kingdom3101 Posts
June 10 2011 04:01 GMT
#128
You are taking an incredibly simple problem, obscuring it with imprecise language and presenting it to a casual populace, then defending yourself with "but three people understood it".

That is not how you should perform a serious survey.
MangoTango
Profile Blog Joined June 2010
United States3670 Posts
June 10 2011 04:04 GMT
#129
The level of confidence that people have in their wrong opnions is hilarious. Much like the Monty Hall problem.
"One fish, two fish, red fish, BLUE TANK!" - Artosis
LunaSaint
Profile Blog Joined April 2011
United Kingdom620 Posts
Last Edited: 2011-06-10 04:10:42
June 10 2011 04:09 GMT
#130
On June 10 2011 12:39 DragonDefonce wrote:
You are treating PZ and ZP separate. That's not correct. If one of your game was played as zerg, then the probability of getting any one of the races on the other is 1/3.

1/5 would be the answer if the Bnet server tells you that you are about to play two games, and it will make it so that you play zerg at least once. In that case, PZ and ZP would be different, but in this case, they are the same.

Pretty sure this is correct.

The order of the games has absolutely no effect. At least one game is the same as there is a game, is it not?
W2
Profile Blog Joined January 2011
United States1177 Posts
June 10 2011 04:13 GMT
#131
On June 10 2011 13:01 Jinsho wrote:
You are taking an incredibly simple problem, obscuring it with imprecise language and presenting it to a casual populace, then defending yourself with "but three people understood it".

That is not how you should perform a serious survey.


Yea, if you are going to post a brainteaser, I'd rather they be fun riddles where you have to think outside the box (like those prisoner/island/village population ones). Re-visiting math isn't fun for me. Anyone got some good riddles to share?
Hi
Tektos
Profile Joined November 2010
Australia1321 Posts
Last Edited: 2011-06-10 04:19:51
June 10 2011 04:19 GMT
#132
This isn't even a brain teaser it is a simple probability question with added convoluted and ambiguous use of the English language to throw people off.


Example of an actual brain teaser:

During a visit to a mental asylum, a visitor asked the Director what the criteria is that defines if a patient should be institutionalized.

"Well," said the Director, "we fill up a bathtub. Then we offer a teaspoon, a teacup, and a bucket to the patient and ask the patient to empty the bathtub."

Okay, here's your test:
1. Would you use the spoon?
2. Would you use the teacup?
3. Would you use the bucket?

"Oh, I understand," said the visitor. "A normal person would choose the bucket, as it is larger than the spoon."
What was the director's response?

Answer:
+ Show Spoiler +
"No," answered the Director. "A normal person would pull the plug."
freeloader625
Profile Joined May 2010
United States180 Posts
June 10 2011 04:47 GMT
#133
On June 10 2011 11:51 Warpath wrote:
Guys, the answer is 1/2
It's either he gets Zerg, or he doesn't.


OMG I forgot where this is from.. that stupid teacher right? Someone pls link. BTW you win the thread man.
d.o.c
Profile Joined August 2010
United States49 Posts
Last Edited: 2011-06-10 04:55:51
June 10 2011 04:55 GMT
#134
I love how this is at least the third time this problem has come up on a Teamliquid thread. I think a very small minority of people got this question wrong because they misunderstood probability theory. This isn't a brainteaser. You've just managed to convince yourself that it is. This is a very straightforward question pushed into muddled oblivion with semantics. It's very simple. If at least one of the games is zerg then it is 1/5. If you're asking the probability the second game is zerg it's 1/3.
EDIT: In b4 "it's already posted" several people got here first.
Hamster1800
Profile Blog Joined August 2008
United States175 Posts
June 10 2011 05:02 GMT
#135
On June 10 2011 13:55 d.o.c wrote: If you're asking the probability the second game is zerg it's 1/3.


That is not true. If I said ``I played two games as random today. In at least one of them, I was zerg. What is the probability that I was zerg in the second game?'' then the answer is 3/5, not 1/3. You changed the setup of the problem (to ``I was zerg in the first game'' instead of ``I was zerg at least once''), not the question that it is asking.
D is for Diamond, E is for Everything Else
Mailing
Profile Joined March 2011
United States3087 Posts
June 10 2011 05:02 GMT
#136
"In the question's case, we are not asking the above. We are asking that given that at least 1 game was Zerg, what is the probability that both games are Zerg."


"What is the probability that my other game was as Zerg as well?"


These are two different things, that's not how you ask this question properly in English.
Are you hurting ESPORTS? Find out today - http://www.teamliquid.net/blogs/viewblog.php?topic_id=232866
terr13
Profile Joined April 2007
United States298 Posts
June 10 2011 05:08 GMT
#137
Reading the original post, it seems perfectly clear. It says he played as Zerg at least once. What is the probability that the other game he played is as Zerg as well, it has no mention of what is first or what is second, so the answer is 1/5.
lachy89
Profile Joined November 2010
Australia264 Posts
June 10 2011 05:11 GMT
#138
On June 10 2011 14:08 terr13 wrote:
Reading the original post, it seems perfectly clear. It says he played as Zerg at least once. What is the probability that the other game he played is as Zerg as well, it has no mention of what is first or what is second, so the answer is 1/5.


He re-worded the question...

The initial wording implied that one game was locked as zerg, then what is the probability that the other was zerg. Giving only three posibilities zerg/terran or protoss - therefore 1/3.
jiabung
Profile Blog Joined December 2007
United States720 Posts
June 10 2011 05:12 GMT
#139
http://en.wikipedia.org/wiki/Boy_or_Girl_paradox

The OP's question is just a variation of the "brainteaser" presented above (3 choices instead of 2). The wiki article helps explain why some people might view the question as ambiguous and presents explanations for both sides.
Rammblin Man
Profile Joined April 2011
Canada19 Posts
June 10 2011 05:14 GMT
#140
Well, going by your newly worded question:

What is the probability that you have to watch 2 mirror match games?

This question is still ambiguous. Did your friend, who commented on the ZvZ, specifically set out searching for a ZvZ game? Did he simply pick a two-game series at random which may or may not have had a zerg in it? If its the former then yes, there are 5 possible permutations and the answer is 1/5. If its the latter, then there is no longer any relation between the first and second game and the answer is 1/3.

From your wording, it seems as if your friend simply wanted to see Nestea vs TLO, and so this series very well may not have had TLO as zerg at all. Therefore the answer is 1/3.

The wikipedia page for the boy/girl paradox words this reasoning rather nicely:

From all families with two children, at least one of whom is a boy, a family is chosen at random. This would yield the answer of 1/3.
From all families with two children, one child is selected at random, and the sex of that child is specified. This would yield an answer of 1/2.

There is some further explaining of the problem on the wikipedia page, I suggest you read it.
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