It was funny in that it was a smart self-parody of American "manliness."
The post-KC version was awful in that it actually celebrated/perpetuated hypermasculinity instead of pointing out how childish it is.
I think you are the only guy I know that actually understood it was a parody.
I sure never caught on to that. I thought it was just the former.
Maybe im just retarded. Adam Corolla seemed pretty reasonable when he was on "Loveline"
JebusUD on
and I wonder about my neighbors even though I don't have them
but they're listening to every word I say
0
Options
GoslingLooking Up Soccer In Mongolia Right Now, ProbablyWatertown, WIRegistered Userregular
edited October 2008
So.... I'm making the call out to any New Yorkers, or anyone in or near the New York metro, that would like to take in a cat.
Some [epithet] left this cat in a carrier on 9th Street in Park Slope on Friday and didn't come back for him. Because I apparently don't have enough drama in my life, I took him to the Park Slope Veterinary Clinic and got him checked out, and he's okay; he's got a thyroid issue, so he needs half a pill twice a day, but no mites or FIV or anything communicable (Dr. Parker said he had a slightly elevated white count, but thinks that's from stress).
What he doesn't have is teeth — and I mean no teeth at all — but he eats like a champ and uses his litter box. He's a good boy.
I'm calling him Nikolai (…don't ask), and he's really very friendly and sweet. He's staying in a rehearsal space at the moment, and he's okay with it, but he's…you know. Bored. He's probably 10 or 12 years old, but except for the thyroid issue, he's in fine shape and would make someone a wonderful fuzzy pet.
Anyone? Orange cat? Eight pounds of toothless love?
If this actually is already your cat, please email me ASAP. I would love to hear all about the TERRIBLE MIX-UP that caused you to leave your FAMILY MEMBER on the street IN A BOX. And if you did it on purpose, you'd better pray to every god there is that I never find out who you are, because I will beat your ass with a flaming beehive, you irresponsible, soulless piece of shit who owes me $300 in bloodwork fees.
Gosling on
I have a new soccer blog The Minnow Tank. Reading it psychically kicks Sepp Blatter in the bean bag.
0
Options
AegeriTiny wee bacteriumsPlateau of LengRegistered Userregular
edited October 2008
Moar writing I feel. Whichever campaign though and I should possibly write that paper on influenza I've been meaning to.
"Of three cards, one is painted red on both sides; one is painted black on both sides; and one is painted red on one side and black on the other. A card is randomly chosen and placed on a table. If the side facing up is read, what is the probability that the other side is also red?"
So at first I think that if you are looking at a red side, then it can be only one of two cards, so 1/2 is the probability.
Is this right? This seems like a deceptive question.
Its 1/2.
You start with a pool of three cards. After the first one is pulled, and its red, you completely discount the all black card. There are two possible cards left, and one of those possible cards has red on the bottom.
1/2
No, no. You've made the same mistake that I did.
You have a 1/3 chance of drawing the red/black card, but a 1/2 chance of flipping it red side up. If you flipped the black side up, then you wouldn't be in that case.
Whereas there's a 1/3 chance of drawing the red/red card, but a 100% chance of flipping it red side up.
So 2/3 in favour of red/red.
I think you're mis-reading the question.
There is a card in front of you. The side you see is red. You are going to flip this over. What is the probability that side is red?
There are only two options here. The card is the black/red card, or the red/red card. Its a 50% chance that one or the other was picked.
Wait, wasn't the question "If the side facing up is red, what is the probability that the other side is also red?" Doesn't that phrasing then assume that you've already got a card red-side-up?
If that's the case, you have three cards. A black/black, a red/black, and a red/red. The card side you're looking at is red. That means the black/black card no longer need be considered. So there are two cards you could possibly be looking at: the red/black or the red/red. In order for the other side of the card to be red, it obviously must be the red/red card that is placed on the table. There is one red/red and one black/red so the chances that the other side is also red are 1/2.
There is a card in front of you. The side you see is red. You are going to flip this over. What is the probability that side is red?
There are only two options here. The card is the black/red card, or the red/red card. Its a 50% chance that one or the other was picked.
I'm not entirely certain of that. Saying 1/2 implies that if the card has a red side, then the red side will always be the side that is face up after randomly drawing it from the bag.
You don't have to make that assumption. The phrasing of the question made it for you.
So you're saying that the chance of picking the red/black card and putting it black side up is 0?
And that the chance of picking the red/black card and having it facing red side up is the exactly the same as picking the red/red card and having it face red side up?
It was funny in that it was a smart self-parody of American "manliness."
The post-KC version was awful in that it actually celebrated/perpetuated hypermasculinity instead of pointing out how childish it is.
I think you are the only guy I know that actually understood it was a parody.
I sure never caught on to that. I thought it was just the former.
Maybe im just retarded. Adam Corolla seemed pretty reasonable when he was on "Loveline"
His whole thing was that he's a loser and he knows it. The Man Show was originally about celebrating that loserdom, how deep down many men are easily entertained and can identify with being well-meaning-but-stupidity-prone doofuses. It was about dumb guys laughing at themselves and not feeling so bad about being dumb.
It was about celebrating the guy who got beat up by the fratties. It became about celebrating the fratties themselves. It traded Hooters girls for porn stars and beer jokes became cocaine jokes and shooting a guy in the face became funnier than kicking a guy in the junk. It went from funny and dumb to disturbing and suck.
Zimmydoom, Zimmydoom
Flew away in a balloon
Had sex with polar bears
While sitting in a reclining chair
Now there are Zim-Bear hybrids
Running around and clawing eyelids
Watch out, a Zim-Bear is about to have sex with yooooooou!
You don't have to make that assumption. The phrasing of the question made it for you.
So you're saying that the chance of picking the red/black card and putting it black side up is 0?
And that the chance of picking the red/black card and having it facing red side up is the exactly the same as picking the red/red card and having it face red side up?
No, I'm saying that the question says you picked a card and placed it on the table and the side that is facing up is red.
"If the side facing up is read, what is the probability that the other side is also red?"
That is directly copy and pasted from the original post. Due to the use of the word "also," I'm assuming that "read" is supposed to be "red." If it's not then you're right.
You don't have to make that assumption. The phrasing of the question made it for you.
So you're saying that the chance of picking the red/black card and putting it black side up is 0?
And that the chance of picking the red/black card and having it facing red side up is the exactly the same as picking the red/red card and having it face red side up?
No, I'm saying that the question says you picked a card and placed it on the table and the side that is facing up is red.
"If the side facing up is read, what is the probability that the other side is also red?"
That is directly copy and pasted from the original post. Due to the use of the word "also," I'm assuming that "read" is supposed to be "red." If it's not then you're right.
Hmmm... well, here's my reasoning which follows the question, tell me which part is wrong:
"A card is randomly chosen and placed upon the table."
Chances of drawing the black/black card: 1/3
Chances of drawing the red/black card and red side up: 1/6
Chances of drawing the red/black card and black side up: 1/6
Chances of drawing the red/red card: 1/3
"If the side facing up is red..."
[strike]Chances of drawing the black/black card: 1/3[/strike]
Chances of drawing the red/black card and red side up: 1/6
[strike]Chances of drawing the red/black card and black side up: 1/6[/strike]
Chances of drawing the red/red card: 1/3
"what is the probability that the other side is also red?"
Red/red is 1/3, vs red/black and red side up's 1/6.
So that's why I say 2/3 and that you need to consider the chances of drawing the red/black card with black side up.
The part that's wrong is that you are considering the possibility of drawing a card with the black side up when the problem states you've already drawn a card and that the red side of whatever card it is is the side that's facing up.
It doesn't say you will always draw a card and red-side-up. It says if. It is asking you to assume that this is what has happened and to work from there. It doesn't say "If you randomly choose a card, what are the chances that the side facing down is red," in which case the answer would, I believe, be 2/3. It says "If the side facing up is red, what is the probability that the other side is also red?" The answer, then, is 1/2.
Res on
[SIGPIC][/SIGPIC]
0
Options
SarksusATTACK AND DETHRONE GODRegistered Userregular
edited October 2008
Is this probability question identical to the Monty Hall problem? If so, look it up on Wikipedia, accept what it says, and never talk about this again.
Good fucking god, is there some simple way to assay the chloride content of ethylene glycol? Every option is terrible, or will make a mess. You know what, fuck it, I'm making a mess.
electricitylikesme on
0
Options
SarksusATTACK AND DETHRONE GODRegistered Userregular
Good fucking god, is there some simple way to assay the chloride content of ethylene glycol? Every option is terrible, or will make a mess. You know what, fuck it, I'm making a mess.
The part that's wrong is that you are considering the possibility of drawing a card with the black side up when the problem states you've already drawn a card and that the red side of whatever card it is is the side that's facing up.
It says "If the side facing up is red, what is the probability that the other side is also red?" The answer, then, is 1/2.
But that's why I cross out the lines which have black side facing up and discard them, leaving me with only the situations where the red card is facing up. I've worked out all the probabilities of getting to the case where the red side is facing up, and then I started working from there.
It doesn't say you will always draw a card and red-side-up. It says if. It is asking you to assume that this is what has happened and to work from there. It doesn't say "If you randomly choose a card, what are the chances that the side facing down is red," in which case the answer would, I believe, be 2/3.
Actually, that one's 1/2. You have 1/3 in drawing red/red, and 1/3 of drawing black/black. There's a 1/2 of the red/black being drawn red face down, so it gets split equally down the center and it ends up 1/2 whether or not any random card will have red face down.
I've worked out all the probabilities of getting to the case where the red side is facing up, and then I started working from there.
You don't. The probability of getting a card red side up is a non-issue because, per the question's phrasing, you are assuming that you have one already. It doesn't matter. For this question's intents and purposes the probability of the card being red-side-up is 1.
There are six possible sides to randomly pick. You know you have a red so throw out the parts of the tree that begin with a black. Now you have three branches that begin with red. Out of these three, two lead to to the Red/Red card thus the probability is 2/3.
Hey guys, if a jet tries to take off from a conveyor belt set to move at the same speed as the jet but in the opposite direction, will the jet be able to take off?
so very sorry
evilbob on
0
Options
ElldrenIs a woman dammitceterum censeoRegistered Userregular
edited October 2008
says the person who originally asked the question.
There are six possible sides to randomly pick. You know you have a red so throw out the parts of the tree that begin with a black. Now you have three branches that begin with red. Out of these three, two lead to to the Red/Red card thus the probability is 2/3.
This has the R/R card existing as two theoretical entities. That is statistics run amok.
Res on
[SIGPIC][/SIGPIC]
0
Options
ElldrenIs a woman dammitceterum censeoRegistered Userregular
Hey guys, if a jet tries to take off from a conveyor belt set to move at the same speed as the jet but in the opposite direction, will the jet be able to take off?
so very sorry
Yes.
Elldren on
fuck gendered marketing
0
Options
ElldrenIs a woman dammitceterum censeoRegistered Userregular
There are six possible sides to randomly pick. You know you have a red so throw out the parts of the tree that begin with a black. Now you have three branches that begin with red. Out of these three, two lead to to the Red/Red card thus the probability is 2/3.
This has the R/R card existing as two theoretical entities. That is statistics run amok.
It has every card that way, because every card has two sides.
Elldren on
fuck gendered marketing
0
Options
ZimmydoomAccept no substitutesRegistered Userregular
There are six possible sides to randomly pick. You know you have a red so throw out the parts of the tree that begin with a black. Now you have three branches that begin with red. Out of these three, two lead to to the Red/Red card thus the probability is 2/3.
This has the R/R card existing as two theoretical entities. That is statistics run amok.
Zimmydoom, Zimmydoom
Flew away in a balloon
Had sex with polar bears
While sitting in a reclining chair
Now there are Zim-Bear hybrids
Running around and clawing eyelids
Watch out, a Zim-Bear is about to have sex with yooooooou!
0
Options
SarksusATTACK AND DETHRONE GODRegistered Userregular
edited October 2008
If I ever take a stats class I am going to print out the Wikipedia page for the Monty Hall problem, staple it to my forehead, and then pass out in my seat each class and I will get an A.
There are six possible sides to randomly pick. You know you have a red so throw out the parts of the tree that begin with a black. Now you have three branches that begin with red. Out of these three, two lead to to the Red/Red card thus the probability is 2/3.
This has the R/R card existing as two theoretical entities. That is statistics run amok.
No, you can get the R/R card two ways. One red side first or the other side first. These are two distinct possibilities.
Posts
Maybe im just retarded. Adam Corolla seemed pretty reasonable when he was on "Loveline"
but they're listening to every word I say
Oh dear.
I think you're mis-reading the question.
There is a card in front of you. The side you see is red. You are going to flip this over. What is the probability that side is red?
There are only two options here. The card is the black/red card, or the red/red card. Its a 50% chance that one or the other was picked.
3ds friend code: 2981-6032-4118
I don't like quote tree's >.>
3ds friend code: 2981-6032-4118
If that's the case, you have three cards. A black/black, a red/black, and a red/red. The card side you're looking at is red. That means the black/black card no longer need be considered. So there are two cards you could possibly be looking at: the red/black or the red/red. In order for the other side of the card to be red, it obviously must be the red/red card that is placed on the table. There is one red/red and one black/red so the chances that the other side is also red are 1/2.
I'm not entirely certain of that. Saying 1/2 implies that if the card has a red side, then the red side will always be the side that is face up after randomly drawing it from the bag.
I don't think you can make that assumption.
So you're saying that the chance of picking the red/black card and putting it black side up is 0?
And that the chance of picking the red/black card and having it facing red side up is the exactly the same as picking the red/red card and having it face red side up?
His whole thing was that he's a loser and he knows it. The Man Show was originally about celebrating that loserdom, how deep down many men are easily entertained and can identify with being well-meaning-but-stupidity-prone doofuses. It was about dumb guys laughing at themselves and not feeling so bad about being dumb.
It was about celebrating the guy who got beat up by the fratties. It became about celebrating the fratties themselves. It traded Hooters girls for porn stars and beer jokes became cocaine jokes and shooting a guy in the face became funnier than kicking a guy in the junk. It went from funny and dumb to disturbing and suck.
No, I'm saying that the question says you picked a card and placed it on the table and the side that is facing up is red.
"If the side facing up is read, what is the probability that the other side is also red?"
That is directly copy and pasted from the original post. Due to the use of the word "also," I'm assuming that "read" is supposed to be "red." If it's not then you're right.
Hmmm... well, here's my reasoning which follows the question, tell me which part is wrong:
"A card is randomly chosen and placed upon the table."
Chances of drawing the black/black card: 1/3
Chances of drawing the red/black card and red side up: 1/6
Chances of drawing the red/black card and black side up: 1/6
Chances of drawing the red/red card: 1/3
"If the side facing up is red..."
[strike]Chances of drawing the black/black card: 1/3[/strike]
Chances of drawing the red/black card and red side up: 1/6
[strike]Chances of drawing the red/black card and black side up: 1/6[/strike]
Chances of drawing the red/red card: 1/3
"what is the probability that the other side is also red?"
Red/red is 1/3, vs red/black and red side up's 1/6.
So that's why I say 2/3 and that you need to consider the chances of drawing the red/black card with black side up.
It doesn't say you will always draw a card and red-side-up. It says if. It is asking you to assume that this is what has happened and to work from there. It doesn't say "If you randomly choose a card, what are the chances that the side facing down is red," in which case the answer would, I believe, be 2/3. It says "If the side facing up is red, what is the probability that the other side is also red?" The answer, then, is 1/2.
A maxim to live by.
There's always the taste test
I'm partial to the Mila Kunis volume.
What was the name of that segment? Some Like it Flat?
Edit: I'm sorry, that was uncalled for. She is cute. I just need sleep.
But that's why I cross out the lines which have black side facing up and discard them, leaving me with only the situations where the red card is facing up. I've worked out all the probabilities of getting to the case where the red side is facing up, and then I started working from there.
Actually, that one's 1/2. You have 1/3 in drawing red/red, and 1/3 of drawing black/black. There's a 1/2 of the red/black being drawn red face down, so it gets split equally down the center and it ends up 1/2 whether or not any random card will have red face down.
what the hell new guy
do not apologize for well played zingers
Yeah. Let's call this. I need to leave the office and go home now.
Dinner demands cooking!
You don't. The probability of getting a card red side up is a non-issue because, per the question's phrasing, you are assuming that you have one already. It doesn't matter. For this question's intents and purposes the probability of the card being red-side-up is 1.
WUT
We can drop this now, I think.
This has the R/R card existing as two theoretical entities. That is statistics run amok.
Yes.
It has every card that way, because every card has two sides.
Amok!
Amok amok amok amok amok!!! :whistle:
I just knew I was having a hard time fitting the question into the conditional probability equation.
No, you can get the R/R card two ways. One red side first or the other side first. These are two distinct possibilities.