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If you still don't agree, I can't help you. Try wikipedia.
In the wikipedia entry on the boy and girl problem the question is worded "what is the probability that BOTH children are the same gender". You say "the other game" which is a single game (as evidenced by the fact that the noun is singular). If you're talking about a single game, the race you draw is not impacted by what you drew in another game. Had you said "the other game as well", THEN it would have been clear you were asking the chances that both were zerg. The only people who are saying it's fine are the ones who were familiar with the problem prior to looking at this post and so immediately knew what you were going for regardless of the wording.
Just say "both games" instead of implying that we're all stupid for reading what you wrote instead of psychically knowing what you meant without prior knowledge of the problem.
EDIT: Sorry didn't catch where you already changed the wording. The poll still says "the other game" so you're still going to get people saying 1/3.
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The two games are independant events, and their order doesn't matter. Saying that it's twice as likely you'll get P because ZP and PZ are two different possibilities should intuitively make no sense to anyone. PZ and ZP are the same thing because the order of the games played has no effect on the solution. The odds of getting zerg in "the other game," REGARDLESS of what happens in "a game" are 1/3.
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You have made the odds of getting zerg in at least 1 game 1.
Therefore the odds you will be zerg twice in a row are only dependant on the odds of getting zerg in the remaining game 2.
The odds are 1/3 for getting a specific race in any single matchup.
so 1/3
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Independent probability. Conditional probability. I feel so sorry for the OP.
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On June 10 2011 10:11 theDreamStick wrote:
It is correct that I am asking a conditional probability question:
Given that at least one of my games was Zerg, what is the probability that both of my games are zerg.
It's either 100%, or 0%, depending on whether or not both games were zerg.
I'll give you a puzzle: I either ate a sandwich today, or I didn't.
What is the probability that I ate a sandwich?
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I've never learned about conditional probability before, so I ran a few calculations, and interesting to find out, 1/5 was indeed the correct answer.
For the couple posts above me that still don't understand, 1/3 would be the answer if the two games are individual events, but they are not. They are linked with the condition that if one game is P or T, then the other cannot be. If you factory in this condition into your calculations, you end up with 1/5.
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On June 10 2011 15:12 EmeraldSparks wrote:Show nested quote +On June 10 2011 10:11 theDreamStick wrote:
It is correct that I am asking a conditional probability question:
Given that at least one of my games was Zerg, what is the probability that both of my games are zerg. It's either 100%, or 0%, depending on whether or not both games were zerg. I'll give you a puzzle: I either ate a sandwich today, or I didn't. What is the probability that I ate a sandwich?
With the information provided nobody is able to tell you what the probability that you eat a sandwich is, as eating a sandwich vs. not eating a sandwich is not a fixed probability event. Some people eat sandwiches every day, some people eat sandwiches once a week, some people never eat sandwiches. The only information that can be presented is that the probability that it is either of those two events is 100% because the probability of an event added to the compliment of that event is always 100%.
You could, however, give us a various sample of consecutive days stating whether you ate a sandwich on that day or not and an approximate probability to a certain confidence level based on sample size could be predicted provided the event of eating a sandwich is not pattern based.
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You should specify the question that you are playing random. As for the Nestea vs TLO example, isn't that just not applicable? Because you know that Nestea will play zerg. So one side of the matchup will allways be zerg, hence it would be 1/3 in that case. But maybe you just added a non related fact to confuse people?
Also just because people understand what you mean with your question does not mean that the wording is good. In fact it could be better.
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This whole thread makes me feel dumb. Never was good at math.
Anyway,
Here's a real probability mind screw for you: It's a real pic a friend of mine uploaded to facebook during her a recent trip to Vegas. I call it: 34 Red (lol) -
+ Show Spoiler +
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Sigh that wording threw me off majorly. But after reading it a few times it is definetly 1/5th. I think a better way of wording said question is,
If in one of my 2 games i played i played as zerg, what is the chance that i play as zerg?.
Wow reading back on that i can't even word it correctly lol
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On June 10 2011 15:15 Chairman Ray wrote:
I've never learned about conditional probability before, so I ran a few calculations, and interesting to find out, 1/5 was indeed the correct answer.
For the couple posts above me that still don't understand, 1/3 would be the answer if the two games are individual events, but they are not. They are linked with the condition that if one game is P or T, then the other cannot be. If you factory in this condition into your calculations, you end up with 1/5.
They ARE individual events. I said this few pages ago and I'll say it again:
If the Bnet server told you that in the next two games you play as random, it will ensure that you play at zerg at least once, then the two games are related, and this would in fact be true.
But this question says this: I played two games as random. I got zerg once. Whats the chance that I got zerg the other game?
Also look at it this way: Assume he got zerg first game. Whats the probability of getting zerg second game? Vice versa? Add them? Still 1/3.
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To everyone who says 1/5 is the correct answer:
Are you assuming getting Z P is a different event as getting P Z?
I play 2 games as random, at least one is zerg. The possible outcomes given this constraint are:
I play 2 games as zerg
I play 1 game as zerg and 1 as protoss
I play 1 game as zerg and 1 as terran
3 possible outcomes, playing zerg twice is one of the possible outcomes:
1 / 3
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On June 10 2011 15:36 Tektos wrote:
As someone who did quite an extensive amount of probability in my education it utterly offends me when I hear:
"As someone who doesn't know anything about probability... I reread the question multiple times and after thinking about it the answer is definitely 1/5"
/facepalm What offends you about that?
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On June 10 2011 15:38 qrs wrote:
What offends you about that?
Re-read my post I explained it rather than just saying it offends me.
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On June 10 2011 15:38 qrs wrote:Show nested quote +On June 10 2011 15:36 Tektos wrote:
As someone who did quite an extensive amount of probability in my education it utterly offends me when I hear:
"As someone who doesn't know anything about probability... I reread the question multiple times and after thinking about it the answer is definitely 1/5"
/facepalm What offends you about that?
It is offensive because it is wrong.
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Why are both ZP and PZ, and ZT and TZ considered options? As the question states, NesTea will always be Zerg. The only variable is TLO.
Edit: I'm getting the feeling that this is an attempted restatement of the Monty Hall problem. If so, the set-up is entirely wrong.
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United States10774 Posts
Haha wow there seems to be a lot of debate over a simple conditional probability problem. Yes, the wording was a little bit unclear, but his edits clarified what he was talking about. Some people seem to just not understand the concept of a conditional probability, or are still misreading the problem.
"I played two games as random today. In at least one of the games I was Zerg. (NOT the first game, but just in one of the two - this is an important distinction that the OP was a little bit unclear on at first but wasn't necessarily wrong either). Given this, what is the probability that I played both games as Zerg? (Not what the probability of getting Zerg in the second game was)"
Conditional probability P (A l B) = P(A and B) / P(A)
(Probability of having gotten Zerg in two games given Zerg in at least one game) / (Probability of getting Zerg in one game)
The possible combinations are ZZ, ZP, ZT, TZ, PZ, PP, TT, TP, PT.
Probability of getting Zerg in one game = 5/9 (since PP, TT, TP and PT are eliminated)
Given this condition, the probability of getting another Zerg = 1/9 (only ZZ remains out of the 9)
Therefore, using the equation, we arrive at 1/9 divided by 5/9 = 9/45, or 1/5.
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On June 10 2011 15:47 OneOther wrote:
Haha wow there seems to be a lot of debate over a simple conditional probability problem. Yes, the wording was a little bit unclear, but his edits clarified what he was talking about. Some people seem to just not understand the concept of a conditional probability, or are still misreading the problem.
"I played two games as random today. In at least one of the games I was Zerg. (NOT the first game, but just in one of the two - this is an important distinction that the OP was a little bit unclear on at first but wasn't necessarily wrong either). Given this, what is the probability that I played both games as Zerg? (Not what the probability of getting Zerg in the second game was)"
Conditional probability P (A l B) = P(A and B) / P(A)
(Probability of getting Zerg in two games given Zerg in at least one game) / (Probability of getting Zerg in one game)
The possible combinations are ZZ, ZP, ZT, TZ, PZ, PP, TT, TP, PT.
Probability of getting Zerg in one game = 5/9 (since PP, TT, TP and PT are eliminated)
Given this probability, the probability of getting another Zerg = 1/9 (only ZZ remains out of the 9)
Therefore, using the equation, we arrive at 1/9 divided by 5/9 = 9/40, or 1/5.
ZP and PZ are the same outcome (playing once as zerg and once as protoss)
TZ and ZT are the same outcome (playing once as zerg and once as terran)
That consolidates your possible outcomes down to 3 results, giving 1/3 chance that the other game will be zerg.
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On June 10 2011 15:47 OneOther wrote:
The possible combinations are ZZ, ZP, ZT, TZ, PZ, PP, TT, TP, PT.
Probability of getting Zerg in one game = 5/9 (since PP, TT, TP and PT are eliminated)
Given this condition, the probability of getting another Zerg = 1/9 (only ZZ remains out of the 9)
Therefore, using the equation, we arrive at 1/9 divided by 5/9 = 9/45, or 1/5.
I just don't understand how ZP and PZ, or TZ and ZT are different. The OP can only be one of 3 races, why does the opponent's race have any bearing at all? Doesn't that invalidate this calculation?
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Since the probability of getting zerg in 1 game is 1/3, getting zerg in 2 games = 2/6, which simplifies to 1/3 obviously, but is needed for the next step.
The unsimplified probability for a single result out of 6 games looks something like this:
1.Zerg------I
2.Zerg------I
3.Terran----I==========(2 opportunities to get zerg out of 6 games or 1/3 probability)
4.Terran----I
5.Protoss---I
6.Protoss---I
Once you already get zerg as the first game though, the fraction looks more like this:
1.XZergXXI==(this option has already been chosen and is not considered)
2.Zerg------I
3.Terran----I==========(1 opportunity to get zerg out of 5 remaining games or 1/5 probability)
4.Terran----I
5.Protoss---I
6.Protoss---I
So the probability for getting zerg in 2 consecutive games is 1/5
Do i win?
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EPT![[image loading]](http://i56.tinypic.com/u7tg.jpg)
