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On June 10 2011 15:36 Tektos wrote:
To everyone who says 1/5 is the correct answer:
Are you assuming getting Z P is a different event as getting P Z? Well, yes, because they are different events, aren't they?
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
I play 2 games as Zerg
I play game 1 as Zerg and game 2 as Protoss.
I play game 1 as Protoss and game 2 as Zerg.
I play game 1 as Zerg and game 2 as Terran.
I play game 1 as Terran and game 2 as Zerg.
5 possible outcomes; playing Zerg twice is one of the possible outcomes.
Why aren't you counting ZP and PZ separately?
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Here's a similar problem that my stats teacher gave me some years ago. It's something along these lines.
You are at a party chatting with a guest and they share that they have 2 children. Right after they tell you this a girl walks up to the person you are talking to and the person says "oh, here is my daughter now." The question is what is the probability that the other child is also a girl. The answer I was given was 1/3 for the same reason, 3 possibilities, BG, GB, and GG so 1 out of 3, right?
Here's my problem.. The fact that the girl walked up is a completely random occurrence. It could have just as easily been a boy. If it was a boy, you would also say that the chance of the person having 2 boys is 1/3. Now you must be saying that the probability the unknown offspring is the opposite gender of the first offspring you see is 2/3. We know that the probability of having 2 different gender children vs 2 same-sex children is 50-50. What am I missing?
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This is very simple as many others have said. Given that I played either game 1 as zerg or game 2 as zerg, what are the odds that i played zerg two times in a row.
First off, there are 9 possibile combinations pp pz pt tp tz tt zp zz zt and only in 1 scenario is zz. Get rid of any scenarios in which you don't play zerg and there are 5 left, thus 1/5.
There are many ways to do this problem and they all result in 1/5.
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On June 10 2011 15:51 Tektos wrote:Show nested quote +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. By your reasoning, I can prove to you that you have a fifty percent chance of rolling a 1, using a fair die.
There are two possibilities: you roll 1 or you don't roll 1. 2, 3, 4, 5, 6 are the same result: not 1. That consolidates your possible outcomes down to 2 results, giving 1/2 chance that you roll a 1.
You see the flaw in this, don't you?
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On June 10 2011 15:52 Stropheum wrote:
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.
Since you played 1 game already and got zerg 1 time, your probability for 2 zerg games remains 2/6, yet you factor in that you've already played one game, reducing it to 2/5, and you factor in the fact that you've gotten zerg, making the probability 1/5 that you will get zerg a second time.
Do i win?
No the probability of getting zerg in two consecutive games is 1/3 * 1/3 = 1/9
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On June 10 2011 15:17 Tektos wrote:Show nested quote +On June 10 2011 15:12 EmeraldSparks wrote: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.
So it's based on a huge set of assumptions and extra information.
Let's say that it happened like this:
Your friend sits down to watch TLO play a game. He randoms.
He only manages to watch one of the games.
After the game, he tells you, "holy shit, TLO's zerg is fucking awesome."
Or "TLO's protoss is fucking awesome."
Or "TLO's terran is fucking awesome."
Then, he asks you, because he's a clever sadistic bastard,
"What's the chance that TLO was zerg twice?"
Or "What's the chance that TLO was protoss twice?"
Or "What's the chance that TLO was terran twice?"
The answer to the first question is one third.
The answer to the second question is one third.
The answer to the third question is one third.
So the answer is one third if the backstory happened like this. But it's 1/5th if the backstory happened another way, say that your friend is this disgustingly patriotic zerg fanboy.and would only tell you that TLO played a zerg game and he doesn't give a shit about terran or protoss games.
But since you don't know the backstory, you can't answer the question.
The same way you can't answer my question without knowing if I ate a sandwich today.
Solution: + Show Spoiler +I haven't had a sandwich in weeks.
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United States10774 Posts
On June 10 2011 15:51 Tektos wrote:Show nested quote +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.
ZP and PZ are different outcomes. TZ and ZT are different outcomes.
Getting a Zerg in the first game and Protoss in the second game is different from getting Protoss in the first game and then Zerg in the second. Same for ZT/TZ.
On June 10 2011 15:51 ZombiesOMG wrote:Show nested quote +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?
What do you mean? The opponent's races don't have to do anything with the problem.
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On June 10 2011 15:57 qrs wrote:Show nested quote +On June 10 2011 15:51 Tektos wrote: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. By your reasoning, I can prove to you that you have a fifty percent chance of rolling a 1, using a fair die. There are two possibilities: you roll 1 or you don't roll 1. 2, 3, 4, 5, 6 are the same result: not 1. That consolidates your possible outcomes down to 2 results, giving 1/2 chance that you roll a 1. You see the flaw in this, don't you?
ZP vs. PZ is comepletely different from P(A) vs. P(not A) you seem to have no clue what you're doing.
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The correct way to phrase the problem is
"Given that I played 2 games and that at least one of them was played as zerg, what is the probability that both were played as zerg?"
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On June 10 2011 15:59 darmousseh wrote:
The correct way to phrase the problem is
"Given that I played 2 games and that at least one of them was played as zerg, what is the probability that both were played as zerg?"
100%, if you're Idra.
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On June 10 2011 15:58 OneOther wrote:Show nested quote +On June 10 2011 15:51 Tektos wrote: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. ZP and PZ are different outcomes. TZ and ZT are different outcomes. Getting a Zerg in the first game and Protoss in the second game is different from getting Protoss in the first game and then Zerg in the second. Same for ZT/TZ.
No they are the same event because the games are independent. Everyone is confusing dependent event conditional probability with independent event conditional probability.
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On June 10 2011 16:00 EmeraldSparks wrote:Show nested quote +On June 10 2011 15:59 darmousseh wrote:
The correct way to phrase the problem is
"Given that I played 2 games and that at least one of them was played as zerg, what is the probability that both were played as zerg?" 100%, if you're Idra.
I've seen idra offrace as protoss......
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On June 10 2011 16:01 darmousseh wrote:Show nested quote +On June 10 2011 16:00 EmeraldSparks wrote:On June 10 2011 15:59 darmousseh wrote:
The correct way to phrase the problem is
"Given that I played 2 games and that at least one of them was played as zerg, what is the probability that both were played as zerg?" 100%, if you're Idra. I've seen idra offrace as protoss......
Looks like somebody made a bad assumption.
That somebody's me.
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On June 10 2011 15:59 Tektos wrote:Show nested quote +On June 10 2011 15:57 qrs wrote:On June 10 2011 15:51 Tektos wrote: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. By your reasoning, I can prove to you that you have a fifty percent chance of rolling a 1, using a fair die. There are two possibilities: you roll 1 or you don't roll 1. 2, 3, 4, 5, 6 are the same result: not 1. That consolidates your possible outcomes down to 2 results, giving 1/2 chance that you roll a 1. You see the flaw in this, don't you? ZP vs. PZ is comepletely different from P(A) vs. P(not A) you seem to have no clue what you're doing. What I'm doing is showing how your "consolidation process" can be abused 6 ways till Sunday. You're the one who doesn't seem to know what you're doing.
Playing a Zerg in game 1 and playing a Zerg in game 2 are different, mutually exclusive events. Saying that every event that includes a Zerg is the same event, is precisely as (in)valid as saying that every event that is non-1 is the same event.
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On June 10 2011 16:03 qrs wrote:Show nested quote +On June 10 2011 15:59 Tektos wrote:On June 10 2011 15:57 qrs wrote:On June 10 2011 15:51 Tektos wrote: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. By your reasoning, I can prove to you that you have a fifty percent chance of rolling a 1, using a fair die. There are two possibilities: you roll 1 or you don't roll 1. 2, 3, 4, 5, 6 are the same result: not 1. That consolidates your possible outcomes down to 2 results, giving 1/2 chance that you roll a 1. You see the flaw in this, don't you? ZP vs. PZ is comepletely different from P(A) vs. P(not A) you seem to have no clue what you're doing. What I'm doing is showing how your "consolidation process" can be abused 6 ways till Sunday. You're the one who doesn't seem to know what you're doing. Playing a Zerg in game 1 and playing a Zerg in game 2 are different, mutually exclusive events. Saying that every event that includes a Zerg is the same event, is precisely as (in)valid as saying that every event that is non-1 is the same event.
No it can't be abused 6 ways till Sunday unless you have no clue about a thing to do with probability. Order of occurrence does not play a factor in independent event conditional probability.
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ZZ - Your friend reports, "TLO played at least one game as zerg."
ZP - Your friend reports, "TLO played at least one game as zerg."
ZT - Your friend reports, "TLO played at least one game as zerg."
PZ - Your friend reports, "TLO played at least one game as zerg."
TZ - Your friend reports, "TLO played at least one game as zerg."
TT - Your friend goes to the pub.
TP - Your friend goes to the pub.
PT - Your friend goes to the pub.
PP - Your friend goes to the pub.
Your friend asks you, "what is the chance that TLO played both games with the same race."
The answer to the question in this case is 1/5.
ZZ - Your friend reports, "TLO played at least one game as zerg."
ZP - Your friend reports with 50% chance "TLO played at least one game as zerg" and 50% chance "TLO played at least one game as protoss."
ZT - Your friend reports with 50% chance "TLO played at least one game as zerg" and 50% chance "TLO played at least one game as terran."
PZ - Your friend reports with 50% chance "TLO played at least one game as zerg" and 50% chance "TLO played at least one game as protoss."
TZ - Your friend reports with 50% chance "TLO played at least one game as zerg" and 50% chance "TLO played at least one game as terran."
TT - Your friend reports, "TLO played at least one game as terran."
TP - Your friend reports with 50% chance "TLO played at least one game as protoss" and 50% chance "TLO played at least one game as terran."
PT - Your friend reports with 50% chance "TLO played at least one game as protoss" and 50% chance "TLO played at least one game as terran."
PP - Your friend reports, "TLO played at least one game as protoss."
Your friend asks you, "what is the chance that TLO played both games with the same race."
The answer to the question in this case is 1/3.
You have no idea of knowing your friend's method of coming up with the question.
You cannot answer the question without further information.
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On June 10 2011 16:01 Tektos wrote:
No they are the same event because the games are independent. Everyone is confusing dependent event conditional probability with independent event conditional probability. OK, let's do it this way. Imagine that a bunch of random games are played.
1/9 of them will be ZZ.
2/9 of them will be ZP/PZ
2/9 of them will be ZT/TZ
2/9 of them will be TP/PT
1/9 of them will be PP
1/9 of them will be TT.
Do you agree with the above?
If so, then consider the set of sets of games where at least one Z is played: 1/5 of these are ZZ.
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Thank you OneOther. Starcraft themed math makes the mundane fun.
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On June 10 2011 16:06 Tektos wrote:Show nested quote +On June 10 2011 16:03 qrs wrote:On June 10 2011 15:59 Tektos wrote:On June 10 2011 15:57 qrs wrote:On June 10 2011 15:51 Tektos wrote: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. By your reasoning, I can prove to you that you have a fifty percent chance of rolling a 1, using a fair die. There are two possibilities: you roll 1 or you don't roll 1. 2, 3, 4, 5, 6 are the same result: not 1. That consolidates your possible outcomes down to 2 results, giving 1/2 chance that you roll a 1. You see the flaw in this, don't you? ZP vs. PZ is comepletely different from P(A) vs. P(not A) you seem to have no clue what you're doing. What I'm doing is showing how your "consolidation process" can be abused 6 ways till Sunday. You're the one who doesn't seem to know what you're doing. Playing a Zerg in game 1 and playing a Zerg in game 2 are different, mutually exclusive events. Saying that every event that includes a Zerg is the same event, is precisely as (in)valid as saying that every event that is non-1 is the same event. No it can't be abused 6 ways till Sunday unless you have no clue about a thing to do with probability. Order of occurrence does not play a factor in independent event conditional probability.
By your logic, the chance of flipping two coins and get two heads in a row is a 1/3, because the possible outcomes are HH, TT, and HT (which is the same as TH).
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erp.
1/5 - i don't see how hard this can be lol
All you have to do, if you're confused, is to list out all the possibilities.
As I'm posting right now, 61% have answered 1/3 lol
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EPT
