Poll: A little math problem

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geizr

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Cheeze_Pavilion post=18.73797.836379 said:
geizr post=18.73797.836353 said:
Cheeze_Pavilion post=18.73797.836339 said:
geizr post=18.73797.836332 said:
This alludes to an advanced subtlety that one can introduce called degeneracy. Suppose, for example, we considered the M/F and F/M configurations to be the same thing, say a M/F combination. Some may use this as a justification that the probability must be 50%, because the two configurations are really the same.
Some may, but I am not. I consider those combinations to be the same thing because we know that a specific dog is male--the one that is not "the other one" otherwise the problem makes no grammatical sense.
Okay, you claim that we know which specific dog is male. So, from the information given in the problem, which one is male, dog1 or dog2?
Which do you want to represent the dog that is not "the other one"?

Actually, I just thought of a better way to get my point across that forcing the label is an error.

Suppose you walk into the pet-shop, and there are two pup in two different bins, one on the left and one on the right. You ask the pet-shop own the exact same question as the person in the problem, "is at least one pup a male?" Now, the pet-shop own has already looked at the pups and knows the gender of both pups. He answers you with a "yes", and this is all the information you are given. Without examining either pup(because that would add new information that the original problem does not provide), can you tell which pup is male and which is female? The answer is no. You can not assign the gender of male or female to the pup on the left(which is what you are trying to do), and likewise for the pup on the right. All you know is that at least one of the pups is male. So, now you have three possible outcomes if you go to examine the pups directly

Left Right
M M
M F
F M

M/M only occurs in 1 possibility out of 3, and thus has a probability of 33%.
 

Samirat

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Cheeze, what if you performed this problem in real life?

Two random dogs.

Admittedly, the washer woman would not always answer yes, and if this occured, you would have to restart. When she did, though, if you asked her if the other dog was male, she would not be like "What's the other dog, hahaa, they are both male" No. If the first dog examined is male, then the "other dog" is the other dog. If the first dog examined is female, and the second is male, the "other dog" is the first dog. This is the way you determine order.

If that doesn't work for you, scratch it off the record. Here's a new explanation.

In 2/3rds of the cases, the "other dog" is clearly defined, correct? It is a female. Only in 1 third of situations is there any confusion about "the other dog," and in this case, does it matter which is the examined dog and which is the "other dog?" Obviously, the washer woman is going to pick the dog that she didn't examine, but whichever dog you take as the "male dog," the other one is male. But this only occurs in 1 third of the cases. So even if you defined your space as:

MF
FM
MM
MM,

It doesn't matter, because the male male solutions only happen in one third of cases anyway.

That's a bit garbled, but anything goes, I suppose.
 

Samirat

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Cheeze_Pavilion post=18.73797.836454 said:
geizr post=18.73797.836409 said:
Cheeze_Pavilion post=18.73797.836379 said:
geizr post=18.73797.836353 said:
Okay, you claim that we know which specific dog is male. So, from the information given in the problem, which one is male, dog1 or dog2?
Which do you want to represent the dog that is not "the other one"?
I can't make such a fiat because that would be a logic error. This is a classical problem; so each of the dogs is considered uniquely labeled and distinct, i.e. dog1 and dog2.
No they are not--read the last line of the problem: the dog labeled "the other one" is not the dog by which the Puppy Washing Man gained his knowledge that at least one of the dogs was male, or else the problem makes no grammatical sense. Not only is it a logic triumph, it is a Mr. Spock in a sixty-nine with Commander Data logic triumph.

And that I think concludes the possibility of me making meaningful contributions to this--I'm just explaining the same thing over and over again. I'm sorry, I know it's pretty cool when knowing some extra math or having a powerful mind allows one to see the trap in a word problem, but, this is not one of those word problems.

This...is just a badly worded problem, probably a corruption of another problem is worded such that you would all be right and the 50% people would have been snagged by not thinking deeply enough about it. But not this problem.
If you performed this experiment with random dogs, as the question asks for, and received about those random dogs the information that at least one was male, the answer would be 33 percent. It may be mathematically hard to figure out, it may be difficult to understand the equation, but the bottom line is that if this should actually occur, the answer would be, finally, irrevocably, 33 percent. It can't be both 50 percent and 33 percent because of semantics. Only a difference in understanding of the premise could accomplish that (for instance, if you thought that in repetition of this experiment, you would have to get an answer of "yes," 100 percent of the time.) Barring that, the answer is set in stone.
 

geizr

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Cheeze_Pavilion post=18.73797.836454 said:
geizr post=18.73797.836409 said:
Cheeze_Pavilion post=18.73797.836379 said:
geizr post=18.73797.836353 said:
Okay, you claim that we know which specific dog is male. So, from the information given in the problem, which one is male, dog1 or dog2?
Which do you want to represent the dog that is not "the other one"?
I can't make such a fiat because that would be a logic error. This is a classical problem; so each of the dogs is considered uniquely labeled and distinct, i.e. dog1 and dog2.
No they are not--read the last line of the problem: the dog labeled "the other one" is not the dog by which the Puppy Washing Man gained his knowledge that at least one of the dogs was male, or else the problem makes no grammatical sense. Not only is it a logic triumph, it is a Mr. Spock in a sixty-nine with Commander Data logic triumph.

And that I think concludes the possibility of me making meaningful contributions to this--I'm just explaining the same thing over and over again. I'm sorry, I know it's pretty cool when knowing some extra math or having a powerful mind allows one to see the trap in a word problem, but, this is not one of those word problems.

This...is just a badly worded problem, probably a corruption of another problem is worded such that you would all be right and the 50% people would have been snagged by not thinking deeply enough about it. But not this problem.
I think you may be the one that needs to reread the problem carefully.

A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?
The last line of the problem says that we want to know what is the probability that both dogs are male give that we know at least one dog is male. The problem is that which one is male is unknown to us, even though the Puppy Washing Man knows. It is the fact that it is unknown to us that we get 3 configurations, not 2, leading to a probability of 33%. I'm not sure how many more ways I can say this to convince you that the correct probability is 33%. If you want, try this experiment:

Take 2 six-side dice and roll them. Whenever you get a 1,2, or 3, mark that as female; whenever you get a 4,5, or 6, mark that as male. Re-roll all instance when both dice are female. Do this about a hundred times and count the number of instances you get male/male and the number of instances you get male/female. You will find the male/male instances occur roughly 33% of the time. This experiment is no different from the combinations you get for the puppies.

And just to answer your following post, no, I did not change the problem at all. It's exactly the same. The pet-shop owner already knows everything about both pups. However, you only know that at least one pup is male. The Puppy Washing Man has access to examine both pup, exactly the same as the pet-shop owner I refer to. So, the problem is exactly the same.
 

Samirat

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Cheeze_Pavilion post=18.73797.836536 said:
Samirat post=18.73797.836513 said:
It can't be both 50 percent and 33 percent because of semantics. Only a difference in understanding of the premise could accomplish that
Semantics have nothing to do with the understanding of a premise? In a word problem?

Next you'll tell me decimal points have nothing to do with the understanding of a base value in mathematical equation! ;-D
Right, right. But this particular misunderstanding over semantics, about "the other dog," doesn't affect how you set up the problem. And if you can set up the problem correctly, you can solve it. As long as you don't assume the first dog is male, which most of the 50 percenters did, in casual error, you should be fine.
 

geizr

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Samirat post=18.73797.836551 said:
Cheeze_Pavilion post=18.73797.836536 said:
Samirat post=18.73797.836513 said:
It can't be both 50 percent and 33 percent because of semantics. Only a difference in understanding of the premise could accomplish that
Semantics have nothing to do with the understanding of a premise? In a word problem?

Next you'll tell me decimal points have nothing to do with the understanding of a base value in mathematical equation! ;-D
Right, right. But this particular misunderstanding over semantics, about "the other dog," doesn't affect how you set up the problem. And if you can set up the problem correctly, you can solve it. As long as you don't assume the first dog is male, which most of the 50 percenters did, in casual error, you should be fine.
This is what I was trying to get at with Cheeze about the degeneracy of the M/F combination if we use his interpretation of the problem. Even if you force-label dog1 as the male dog, the occurrence of the "other" dog being female has a 2-fold degeneracy. So, that case has to be counted twice if you are going to setup the problem the way Cheeze is proposing. But, even then, you still get 33% for the probability of obtaining both dogs as male. The problem with this method is that it is treacherous because the degeneracy is not immediately obvious. Whereas, just explicitly writing down the different configurations with dog1 and dog2 allows you to find the correct answer without having to make strange assertions(like force-labeling) or clobbering your brain with hidden degeneracies.
 

Ancalagon

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Cheeze_Pavilion post=18.73797.836327 said:
kailsar post=18.73797.836172 said:
So say Dog 1 is an English Beagle, and Dog 2 is an American Beagle, then the American Beagle couldn't be male while the English one is female, since that configuration (FM) doesn't look like any option available in the new set of configurations?
I don't know, because that is a different problem. We can make up any kind of adjectives we like to describe the pups and differentiate them, but, unless there's a basis for doing so in this word problem, you're answering a different question.
The reason that I put the adjectives in there is to demonstrate that you have to know which specific puppy is male to get the answer 50%. You say that by putting in the adjectives I'm changing the nature of the question, so:

Question 1:

"There are two puppies. We are informed that at least one of them is male. What is the probability that they are both male?"

Question 2:

"There are two puppies, an American Beagle and an English Beagle. We are informed that at least one of them is male. What is the probability that they are both male?"

Do these two questions have a different set of probabilities of outcomes?
 

Ancalagon

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Cheeze_Pavilion post=18.73797.837747 said:
kailsar post=18.73797.836953 said:
Question 1:

"There are two puppies. We are informed that at least one of them is male. What is the probability that they are both male?"

Question 2:

"There are two puppies, an American Beagle and an English Beagle. We are informed that at least one of them is male. What is the probability that they are both male?"

Do these two questions have a different set of probabilities of outcomes?
Maybe, but, I don't see the relevance of those questions to this one:

A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?

which is the one we were asked :-D
The reason I asked those two questions, rather than the original one, is that I'm trying to understand where we're diverging. Question 1 is how I see the original question with all of the ambiguity and filler taken out. If you think that the answer to Question 1 is 50%, it's a maths thing. If you think it's 33% then we're understanding what's going on in the original question differently. If it is about the maths, Question 2 is to see if the difference is resolved if it is made more obvious that the dogs cannot be interchanged for the convenience of any calculations. What would you say the answers to Question 1 and Question 2 are?
 

FrcknFrckn

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Yikes, I check back to see if anyone has come to their senses, and find 3 more pages... now we're arguing about semantics? Never mind that the question doesn't actually impart any new information to the situation, it really doesn't change the outcome at all:

-

Ok, allow me to play the part of the washer.

I have 2 dogs here at my washing shop, A and B. You ask me whether one is male. I answer yes. You then try to determine the probability that the OTHER ONE is male.

There are 4 possible dog combinations here:

1. A is female, B is female. (00)
We can automatically ignore this - as I have already told you that one dog is male, this could not possibly be the case.

2. A is male, B is female. (10)
So yeah, this is a valid case. So the dog I was talking about is dog A, and the 'other dog' must be dog B. Since B is not male, the 'other dog' in this case is not male.

3. A is female, B is male. (01)
Again, this is a valid case, just like #2. So the dog I was talking about is dog B, and the 'other dog' must be dog A. Since A is not male, the 'other dog' in this case is not male.

4. A is male, B is male. (11)
Again a valid case. But which dog was I talking about, A or B? IT DOESN'T MATTER. Heck, I might even be done with the washing, and not remember which was which. If I was talking about A, then the 'other dog' was B. If I was talking about B, then the 'other dog' was A. Either way, the 'other dog' was male.


So, we have 3 equally likely situations left after we rule out #1 - 2, 3, and 4. And of those three, only one has 2 male dogs.

1/3 = 33%.
 

FrcknFrckn

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Cheeze_Pavilion post=18.73797.838257 said:
FrcknFrckn post=18.73797.838237 said:
Yikes, I check back to see if anyone has come to their senses, and find 3 more pages... now we're arguing about semantics?
You're right--arguing about semantics when discussing a word problem makes as much sense as arguing about whether there's a green zero on a roulette wheel when discussing casino odds.

Oh wait...

I mean, is it really that difficult to see that when you answer a word problem, you need to answer the actual problem posed by the words?
Is it really difficult to see that you're reading more into the words 'other one' than are there? If you'd read the rest of my post, you'd see that those words make absolutely no difference whatsoever.
 

Alex_P

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And, semantically, I think the parsimonious reading is the one that assumes that "the other" refers to the sex of the second dog independent of which dog it happens to be (analogous to saying "Is there a second male?") rather than the ones that modifies "at least one" to mean "this specific one."

-- Alex
 

FrcknFrckn

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Cheeze_Pavilion post=18.73797.838290 said:
The other what? The question was about a pair--saying 'the other one' makes no sense. That's like if someone asks you if you'd like french fries or baked potato, and you go 'the other one, please'.
No, it's more like if I have a penny and a quarter, one in each of my hands. I tell you one of my hands has a penny in it, and you ask for the coin in the OTHER HAND. Which hand is it, left or right? That's correct, you don't know - because talking about the OTHER HAND doesn't actually tell you which is which. I can't emphasize this enough: it doesn't give you any knowledge whatsoever about which hand I was referring to.
 

Alex_P

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Cheeze_Pavilion post=18.73797.838313 said:
And the parsimonious reading is also that when Puppy Washing Man goes "Yes!" he means that either both dogs are male or a specific dog in a male/female pair is male, not that he has knowledge that they are from Guaranteed Lesbo Free Puppy Pairs Breeder or some other kind of knowledge that there was screening of the FF pairs.
Of course.

... And that's how you get 33%.

-- Alex
 

iain62a

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"she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile"

Could this include the guy that was cleaning them too? maybe it's a trick and none of the dogs are male
 

FrcknFrckn

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Cheeze_Pavilion post=18.73797.838370 said:
FrcknFrckn post=18.73797.838330 said:
Cheeze_Pavilion post=18.73797.838290 said:
The other what? The question was about a pair--saying 'the other one' makes no sense. That's like if someone asks you if you'd like french fries or baked potato, and you go 'the other one, please'.
No, it's more like if I have a penny and a quarter, one in each of my hands. I tell you one of my hands has a penny in it, and you ask for the coin in the OTHER HAND. Which hand is it, left or right? That's correct, you don't know - because talking about the OTHER HAND doesn't actually tell you which is which. I can't emphasize this enough: it doesn't give you any knowledge whatsoever about which hand I was referring to.
Well, if you tell me you've got one and only one coin in each hand, and you're holding pennies and quarters, and you tell me you've got a penny in one hand, then you've got a quarter in the other. If I ask for the coin in the other hand than the one that has penny--regardless of which hand it is--if it's not a quarter, then unless you were holding quantum coin, you were lying when you said you were holding pennies AND quarters.

As long as I know you're holding pennies and quarters, and one and only one coin in each hand, and I ask what's in the other hand from the penny hand, if it's not a quarter, how were you able to tell me you were holding "a penny and a quarter, one in each of my hands" in the first place?

Think about it for a second!
And now you're fixated on quarters. I'll try phrasing it differently so you don't get hung up again:

I have two coins, one in each of my hands. I tell you at least one of my hands has a penny in it, and you ask for the coin in the OTHER HAND. Which hand is it, left or right?

And before you go off talking about quantum this-and-that, try to get what I'm demonstrating here: talking about the 'other hand' doesn't change the coins in my hands, it doesn't give you any more information about the coins in my hands, it doesn't tell you which coin is a penny, or even if both coins are pennies. IT TELLS YOU NOTHING.
 

Alex_P

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Cheeze_Pavilion post=18.73797.838382 said:
Alex_P post=18.73797.838363 said:
Cheeze_Pavilion post=18.73797.838313 said:
And the parsimonious reading is also that when Puppy Washing Man goes "Yes!" he means that either both dogs are male or a specific dog in a male/female pair is male, not that he has knowledge that they are from Guaranteed Lesbo Free Puppy Pairs Breeder or some other kind of knowledge that there was screening of the FF pairs.
Of course.

... And that's how you get 33%.
EDIT my bad, thought you were replying to something else.

So for 20+ pages people have been arguing that the reason it's 33% is because we only know that the pair doesn't come from a pool with FFs, and that if we knew he was referring to a specific dog it would be 50%, and now you're saying that even if we know specifically what dog he's referring to, it's *still* 33%?
It all depends on who identifies which dog is the "referent."

There are two dogs, Jesus and Satan.

If you ask "Is Jesus male?" and the answer is "Yes," there is a 50% chance that Satan is male.

If you ask "Is at least one of them male?" and the answer is "Well, I noticed that Jesus is male," there is still only a 33% chance that Satan is male -- even though you specifically know Jesus is the male dog.

-- Alex
 

FrcknFrckn

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Alex_P post=18.73797.838416 said:
There are two dogs, Jesus and Satan.

If you ask "Is Jesus male?" and the answer is "Yes," there is a 50% chance that Satan is male.

If you ask "Is at least one of them male?" and the answer is "Well, Jesus is male," there is still only a 33% chance that Satan is male. Even though you specifically know Jesus is the male dog.

-- Alex
Actually, both those cases are 50%. In both, we end up with the same information: Jesus is male. We know nothing about Satan, so there's a 50% chance he's male or female.

The probability difference arises if you ask "Is at least one of them male?" and the answer is "Yes." In that situation, there is only a 33% chance that both Jesus and Satan is male.
 

Alex_P

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FrcknFrckn post=18.73797.838427 said:
Alex_P post=18.73797.838416 said:
There are two dogs, Jesus and Satan.

If you ask "Is Jesus male?" and the answer is "Yes," there is a 50% chance that Satan is male.

If you ask "Is at least one of them male?" and the answer is "Well, Jesus is male," there is still only a 33% chance that Satan is male. Even though you specifically know Jesus is the male dog.

-- Alex
Actually, both those cases are 50%. In both, we end up with the same information: Jesus is male. We know nothing about Satan, so there's a 50% chance he's male or female.

The probability difference arises if you ask "Is at least one of them male?" and the answer is "Yes." In that situation, there is only a 33% chance that both Jesus and Satan is male.
No, read the second situation carefully.

The trick is that you're inviting the puppy-washer to decide which dog you're referring to, and he can always pick the more favorable answer. If Jesus were female he could've said "Yes, Satan is male" instead.

-- Alex
 

FrcknFrckn

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Alex_P post=18.73797.838442 said:
FrcknFrckn post=18.73797.838427 said:
Alex_P post=18.73797.838416 said:
There are two dogs, Jesus and Satan.

If you ask "Is Jesus male?" and the answer is "Yes," there is a 50% chance that Satan is male.

If you ask "Is at least one of them male?" and the answer is "Well, Jesus is male," there is still only a 33% chance that Satan is male. Even though you specifically know Jesus is the male dog.

-- Alex
Actually, both those cases are 50%. In both, we end up with the same information: Jesus is male. We know nothing about Satan, so there's a 50% chance he's male or female.

The probability difference arises if you ask "Is at least one of them male?" and the answer is "Yes." In that situation, there is only a 33% chance that both Jesus and Satan is male.
No, read the second situation carefully.

The trick is that you're inviting the puppy-washer to decide which dog you're referring to, and he can always pick the more favorable answer. If Jesus were female he could've said "Yes, Satan is male" instead.

-- Alex
Ah, yeah, missed that... good point, objection retracted! :)