|
On June 11 2011 05:27 Lomak wrote:Show nested quote +On June 11 2011 05:23 DarkPlasmaBall wrote:On June 11 2011 05:18 Lomak wrote:+ Show Spoiler +Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever. He listed 5 pairs of games (not 6), and they aren't matchups. ZP =/= ZvP ZT =/= TZ, ZP =/= PZ because of the fact that they're pairs of games (getting Zerg then Terran, as opposed to Terran then Zerg). You clearly didn't even attempt to read the OP. You literally could not be more wrong with the brain teaser's information. Player 1 is Nestea (always playes zerg). So player 1 is constant as Z so the Zv? is the only question there is. IF one player is playing zerg and one is playing random, the chances of it being ZvZ are 1/3 not 1/5. Its not my fault this thread is full of condescending people bickering over a poorly written "brain teaser" If the OP wanted people to understand WTF he was asking he would say. 2 players played 2 starcraft games as Random Vs. Random. The first game was ZVZ, AND THEN ask his question, it would make some sense. But he doesnt. He lead's you to believe that ONE player is ZERG and the OTHER is RANDOM. It's not a brain teaser if he can't even ask the question properly.
Yea this is my conclusion as well.
|
On June 11 2011 05:42 Logo wrote:Show nested quote +On June 11 2011 05:41 Sasashi wrote:On June 11 2011 05:38 Logo wrote:On June 11 2011 05:36 Sasashi wrote:On June 11 2011 05:34 Logo wrote:On June 11 2011 05:32 Sasashi wrote:On June 11 2011 05:31 Logo wrote:On June 11 2011 05:27 Lomak wrote:On June 11 2011 05:23 DarkPlasmaBall wrote:On June 11 2011 05:18 Lomak wrote:+ Show Spoiler +Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever. He listed 5 pairs of games (not 6), and they aren't matchups. ZP =/= ZvP ZT =/= TZ, ZP =/= PZ because of the fact that they're pairs of games (getting Zerg then Terran, as opposed to Terran then Zerg). You clearly didn't even attempt to read the OP. You literally could not be more wrong with the brain teaser's information. Player 1 is Nestea (always playes zerg). So player 1 is constant as Z so the Zv? is the only question there is. IF one player is playing zerg and one is playing random, the chances of it being ZvZ are 1/3 not 1/5. Its not my fault this thread is full of condescending people bickering over a poorly written "brain teaser" That's not the question though. The question is over a 2 game series what's the chance of there being 2 ZvZs played given that at least one game was a ZvZ. Nope. The question is over a 2 game series what is the probability of getting zerg twice when playing as random, if at least one time you get zerg, and the order you play them in matters. Uh that's the exact same thing as what I said. Not really, you got the same result, but the question wasn't about playing a specific match like zerg vs zerg. It was about playing zerg as a race twice while playing random. It's not the same thing, despite the same answer. It breaks down to the same probability question of P(A|B) where A = at least one game as a zerg and B = both games as zerg since one side of the match-up is constant in the OP's example. Yeah I realize that, and stated that. Was just correcting you as the original question wasn't about matchups. Well yeah... sort of. The OP has posted 2 versions of the same question (after his edits) which is what's causing a lot of confusion.
Yeah =\
|
On June 11 2011 05:40 eluv wrote:
The most disturbing thing about this is that this thread is 21 pages long. This is not ambiguous. It is not a paradox. It is possible counterintuitive, but the answer is unquestionably right - get over it and get on with your lives.
It is ambiguous for reasons that are well documented. This is a rephrasing of the ambiguous question originally posed as part of the "Boy or Girl Paradox."
|
On June 11 2011 05:45 Mente wrote:Show nested quote +On June 11 2011 05:27 Lomak wrote:On June 11 2011 05:23 DarkPlasmaBall wrote:On June 11 2011 05:18 Lomak wrote:+ Show Spoiler +Then I am restricted to ZZ, ZP, ZT, PZ, TZ. ZZ is one out of five possible choices
So the answer is 1/5 even though you listed 6 matchups and ZvZ is in there twice as well as ZvP and ZvT. Call me names if you wish but this doesn't make any sense whatsoever. He listed 5 pairs of games (not 6), and they aren't matchups. ZP =/= ZvP ZT =/= TZ, ZP =/= PZ because of the fact that they're pairs of games (getting Zerg then Terran, as opposed to Terran then Zerg). You clearly didn't even attempt to read the OP. You literally could not be more wrong with the brain teaser's information. Player 1 is Nestea (always playes zerg). So player 1 is constant as Z so the Zv? is the only question there is. IF one player is playing zerg and one is playing random, the chances of it being ZvZ are 1/3 not 1/5. Its not my fault this thread is full of condescending people bickering over a poorly written "brain teaser" If the OP wanted people to understand WTF he was asking he would say. 2 players played 2 starcraft games as Random Vs. Random. The first game was ZVZ, AND THEN ask his question, it would make some sense. But he doesnt. He lead's you to believe that ONE player is ZERG and the OTHER is RANDOM. It's not a brain teaser if he can't even ask the question properly. Yea this is my conclusion as well.
It's not random vs random. It's zerg vs random for 2 games. The question is how often does random roll zerg both games when your friend tells you that he saw an awesome ZvZ?
|
On June 11 2011 05:50 scorch- wrote:Show nested quote +On June 11 2011 05:40 eluv wrote:
The most disturbing thing about this is that this thread is 21 pages long. This is not ambiguous. It is not a paradox. It is possible counterintuitive, but the answer is unquestionably right - get over it and get on with your lives. It is ambiguous for reasons that are well documented. This is a rephrasing of the ambiguous question originally posed as part of the "Boy or Girl Paradox."
It's mostly ambiguous for external factors to the intention of the problem, as in stuff like you are not equally likely to have a girl as you are a boy and selecting random isn't truly random.
Either way people should accept that the answer (in the OP's example) can range from 1/5 to 1/3 inclusive depending on all of the underlying factors, assumptions, and wording. Claiming it's 1/3rd absolutely and not acknowledging that the general idea of the problem can produce a 1/5 probability is not understanding the problem fully.
|
United States10774 Posts
EDIT: Haha never mind, it's not even really worth it.
|
Czech Republic11293 Posts
Nice brainteaser, almost got me! The easiest way I solved it with is just writing all the possible outcomes that the random player could get, if at least one of his game was as zerg: PZ, TZ, ZZ, ZP and ZT ^^
|
I was already familiar with the boy or girl problem, and I still fell for it.
|
Details on this (original problem):
1 player play 2 games of starcraft as random. In at least one of the games he got zerg but it is unknown which one. What is the probability that the player played zerg in both games?
All of the possibilities would be TT, TP, TZ, PP, PT, PZ, ZZ, ZT, ZP. We know this to be a fact because without any conditions every race has 1/3 chance of being chosen, so every outcome here has 1/3*1/3=1/9 chance of being positive. There are 9 possible outcomes and 1/9*9=1 which ensures that we have every outcome listed.
Now we know that zerg was played in at least one of the games which would be the outcomes
TZ, PZ, ZZ, ZT, ZP
the chance of that happening is 1/9+1/9+1/9+1/9+1/9=5/9 but that is not what we want to solve.
Our condition is that zerg was played at least once with a probability of 1, which means that
Pr(TZ) + Pr(PZ) + Pr(ZZ) + Pr(ZT) + Pr(ZP) = 1, and all the probabilities are equal.
If you solve the above equation you get that Pr(ZZ) = 1/5
Hope this is helpful for some, who had problems understanding the op.
|
Ouch, I got tricked. Thank you for the very clear explaining though 
|
The part where you introduced Nestea and TLO to give some hints just made it more confusing for some people.  DON'T think of his opponent at all, only focus on him, the player who's playing random.
2 games, 9 sets of possible outcomes (for his own race only, still ignore the opponent).
4 of these don't include zerg at all.
Left are T+Z, P+Z, Z+T, Z+P and Z+Z.
Only Z+Z fulfils the requirements of him getting zerg both games. ^^ 1/5.
|
On June 11 2011 06:05 Logo wrote:Show nested quote +On June 11 2011 05:50 scorch- wrote:On June 11 2011 05:40 eluv wrote:
The most disturbing thing about this is that this thread is 21 pages long. This is not ambiguous. It is not a paradox. It is possible counterintuitive, but the answer is unquestionably right - get over it and get on with your lives. It is ambiguous for reasons that are well documented. This is a rephrasing of the ambiguous question originally posed as part of the "Boy or Girl Paradox." It's mostly ambiguous for external factors to the intention of the problem, as in stuff like you are not equally likely to have a girl as you are a boy and selecting random isn't truly random.
Those factors are not at all the reason that it is ambiguous. Those factors can be determined. However, the wording of the problem makes it impossible to determine whether having played at least one game of zerg is sufficient for your friend to report that there was an awesome ZvZ. This alters the Probability that your friend will tell you that nestea is amazing ZvZ. If it IS sufficient, the Probability is 5/9 (1/5 of which is ZZ), if it IS NOT, the probability is 1/3 (1/3 of which is ZZ).
|
Extension problems (answers in spoilers):
If player 1 and player 2 played two random vs random games, and you know at least one of the games was a ZvZ, what is the probability that both games were ZvZs?
+ Show Spoiler +
If player 1 and player 2 played two random vs random games, and you know player 1 played zerg in at least one of the games, what is the probability that both games were ZvZs?
+ Show Spoiler +
If player 1 and player 2 played two random vs random games, and you know that at least one of the players played zerg in at least one of the games, what is the probability that both games were ZvZs?
+ Show Spoiler +
|
On June 11 2011 06:34 LastPrime wrote:Extension problems (answers in spoilers): If player 1 and player 2 played two random vs random games, and you know at least one of the games was a ZvZ, what is the probability that both games were ZvZs? + Show Spoiler +If player 1 and player 2 played two random vs random games, and you know player 1 played zerg in at least one of the games, what is the probability that both games were ZvZs? + Show Spoiler +If player 1 and player 2 played two random vs random games, and you know that at least one of the players played zerg in at least one of the games, what is the probability that both games were ZvZs? + Show Spoiler +
How are the bottom 2 not the exact same scenario?
and you know player 1 played zerg in at least one of the games doesn't exclude player 2 from also getting zerg, making it equivalent to "at least one of the players played zerg".
|
The 1/5th is correct. In the post you say the combinations are
ZZ, ZP, ZT, PZ, PP, PT, TZ, TP, TT
Perhaps it will be easier for people to understand if you say there are 3*3=9 combinations listed here
(ZvZ,ZvZ), (ZvZ,ZvP), (ZvZ,ZvT), (ZvP,ZvZ), (ZvP,ZvP), (ZvP,ZvT), (ZvT,ZvZ), (ZvT,ZvP), (ZvT,ZvT)
This is the same thing of course, but some might not understand what ZZ means or why some of those combinations are eliminated when you say a ZvZ was played.
I should also note that there is no confusion with the way the problem is stated originally or now. The confusion is because of people's inability to understand a complicated problem and so they ignorantly and naturally believe it is worded wrong.
|
On June 11 2011 00:29 Ivs wrote:
People are still arguing because OP wanted to present the Boy/Girl paradox, but messed up the wording.
Now there are 3 camps of people
1. People interpreted the OP as the boy/girl paradox, even though OP failed. They say 1/5
2. People who don't really understand whats going on and go with the simplest reasoning. They say 1/3
3. People who are calling out OP's original wording and poor usage of "other", and also get the answer of 1/3.
Chill out guys, no need for name/credential calling. There is no argument here.
Agreed.
The word "other" is the problem. The way the OP restated the problem is much clearer.
|
The answer is clearly 1/2. Either you played both games as Zerg, or you did not play both games as Zerg.
You guys are overthinking this too much XD
|
And the chance of winning the lottery is also 1/2, either you win or you don't, right? --;
|
On June 11 2011 06:50 Shichibukai wrote:Show nested quote +On June 11 2011 06:34 LastPrime wrote:Extension problems (answers in spoilers): If player 1 and player 2 played two random vs random games, and you know at least one of the games was a ZvZ, what is the probability that both games were ZvZs? + Show Spoiler +If player 1 and player 2 played two random vs random games, and you know player 1 played zerg in at least one of the games, what is the probability that both games were ZvZs? + Show Spoiler +If player 1 and player 2 played two random vs random games, and you know that at least one of the players played zerg in at least one of the games, what is the probability that both games were ZvZs? + Show Spoiler + How are the bottom 2 not the exact same scenario? and you know player 1 played zerg in at least one of the games doesn't exclude player 2 from also getting zerg, making it equivalent to " at least one of the players played zerg".
The latter case includes the former. You must consider the case in which player 2 player plays zerg but not player 1.
|
You played two games random of which at least one you were Z and ask what the probobility is that the other be Z aswell. there are five outcomes as many have claimed, but they are not as likly.
ZZ we can be sure
ZN and NZ appear to be four outcomes giving ZZ 1/5 odds
what is forgotten is the possibility of op asking for the N in those cases, e.i ''of which atleast one was P'' that would happen half the time, when this is taken to account we get the real odds 1/3. Good brain teser!
what you want to ask is: I played two games with Z, atlest one was against Z whats the probability the other was aswell.
|
|
|
|
|
|
EPT
