Cheeze_Pavilion: SNIP I've read it, and you're missing the point. The question "What is the probability that the other one is a male?" makes no sense unless we can distinguish between puppies.
Based on the information in the question, in what way can you distinguish between the puppies other than by calling one 'the male' and the other 'the one we don't know about'?
That's at the heart of the problem, I think. Clearly, the question should have read "What is the probability that both pups are male." That I think would have been clear to everyone
-given that at least one s male. Let's not open up 25% as an option for debate. And besides, where's the fun in making things obvious? That's just not how maths is done.
What we need is language that somehow screens out the FF pups from the predicted 25/50/25 breakdown, without making it obvious that they have been screened.
Doug: The question states that at least one pup is male (because the person doing to phoning asked and the answer was 'Yes')
If that's what it states, on what basis is the end question referring to "the other one"? What information do we have that allows us to identify one puppy as "the other one" if not information that one specific pup is male, and "the other one" is the one we don't know anything about?
It's interesting to do this while replaying God of War. You all probably are thinking of cute little puppies. I'm thinking of little hell hounds that I want to pound into the ground for the extra health.
I think that the issue has moved from one of mathematics to one of semantics. If I'm understanding you right, Cheeze, then you'd agree with the following:
If the guy looks at both dogs before making his statement, and upon checking the dogs says that there is a male, either on the basis that one of the dogs is a male or both are, then the probability that they're both male is 33%.
Or:
The guy picks up a dog, it's male, he may look at the other, or he may leave it unchecked, either way, the chance of them both being male is 50%.
And that what we're really discussing is whether the language in the question (in particular, the phrase 'the other dog') implies the first or the second; or indeed that the language of the question is ambiguous about what he's done, and therefore how do we calculate the probability with that ambiguity.
Or am I wrong, and you'd still say that in the first instance, i.e. he checks both and then reports that there is a male, the chance is 50%?
Cheeze_Pavilion: SNIP I've read it, and you're missing the point. The question "What is the probability that the other one is a male?" makes no sense unless we can distinguish between puppies.
Based on the information in the question, in what way can you distinguish between the puppies other than by calling one 'the male' and the other 'the one we don't know about'?
That's at the heart of the problem, I think. Clearly, the question should have read "What is the probability that both pups are male." That I think would have been clear to everyone
-given that at least one s male. Let's not open up 25% as an option for debate. And besides, where's the fun in making things obvious? That's just not how maths is done.
What we need is language that somehow screens out the FF pups from the predicted 25/50/25 breakdown, without making it obvious that they have been screened.
Yeah, I don't know how to do that either!
If the question were phrased: "There are two puppies. We are informed that at least one of them is male. What is the probability that they are both male?" Would we all agree the answer is 33%?
Doug: The question states that at least one pup is male (because the person doing to phoning asked and the answer was 'Yes')
If that's what it states, on what basis is the end question referring to "the other one"? What information do we have that allows us to identify one puppy as "the other one" if not information that one specific pup is male, and "the other one" is the one we don't know anything about?
It's interesting to do this while replaying God of War. You all probably are thinking of cute little puppies. I'm thinking of little hell hounds that I want to pound into the ground for the extra health.
I think that the issue has moved from one of mathematics to one of semantics.
I would say that the issue was always one of semantics, because it was always a word problem. Just like calculating the chance of catastrophic failure of a building is always an issue of construction, or calculating the chance of winning at the casino is always a matter of the house's rules. There's no such thing as the 'math of roulette'; there's the math of 36 numbers split into black and red, but there's also the math when there's a green zero added, and another math for when a green double zero is added to that.
If I'm understanding you right, Cheeze, then you'd agree with the following:
If the guy looks at both dogs before making his statement, and upon checking the dogs says that there is a male, either on the basis that one of the dogs is a male or both are, then the probability that they're both male is 33%.
Or:
The guy picks up a dog, it's male, he may look at the other, or he may leave it unchecked, either way, the chance of them both being male is 50%.
...
Or am I wrong, and you'd still say that in the first instance, i.e. he checks both and then reports that there is a male, the chance is 50%?
I'd still say 50% in both instances and here's why--you listed the two possiblites as in your 33% scenario as:
1) on the basis that both of the dogs are male
OR
2) the basis that one of the dogs is a male.
So wouldn't (1) look like:
Pair of Dogs M
And (2) would look like
Basis Dog/Non-Basis Dog M/M (if he liked Basis Dog better and decided to talk about that dog and not the other or about the pair) M/F (if he had to talk about basis dog to say "Yes!")
Right? Because putting anything but an M under Basis Dog...doesn't make any sense, because otherwise, it wouldn't be a sufficient basis for saying "Yes!" to the question.
Cheeze_Pavilion: SNIP I've read it, and you're missing the point. The question "What is the probability that the other one is a male?" makes no sense unless we can distinguish between puppies.
Based on the information in the question, in what way can you distinguish between the puppies other than by calling one 'the male' and the other 'the one we don't know about'?
That's at the heart of the problem, I think. Clearly, the question should have read "What is the probability that both pups are male." That I think would have been clear to everyone
-given that at least one s male. Let's not open up 25% as an option for debate. And besides, where's the fun in making things obvious? That's just not how maths is done.
What we need is language that somehow screens out the FF pups from the predicted 25/50/25 breakdown, without making it obvious that they have been screened.
Yeah, I don't know how to do that either!
If the question were phrased: "There are two puppies. We are informed that at least one of them is male. What is the probability that they are both male?" Would we all agree the answer is 33%?
Still don't think so. Maybe more like: "There are two puppies. They were picked randomly from pairs made up of random puppies. We are informed that any time an FF pair is picked, it is sent to a pet store in Park Slope, and this pet store is in DUMBo. What is the probability that they are both male?"
I'd still say 50% in both instances and here's why--you listed the two possiblites as in your 33% scenario as:
1) on the basis that both of the dogs are male
OR
2) the basis that one of the dogs is a male.
So wouldn't (1) look like:
Pair of Dogs M
And (2) would look like
Basis Dog/Non-Basis Dog M/M (if he liked Basis Dog better and decided to talk about that dog and not the other or about the pair) M/F (if he had to talk about basis dog to say "Yes!")
Right? Because putting anything but an M under Basis Dog...doesn't make any sense, because otherwise, it wouldn't be a sufficient basis for saying "Yes!" to the question.
I phrased it poorly. What I meant for 2) is: the basis that only one of the dogs is male. So 1) would look like:
Pair of Dogs MM
and 2) would look like:
Basis Dog/Non Basis Dog MF
So the question is: how likely is option 1, and how likely is option 2? If we go back to having just two dogs, before the dogs have been examined by the dog-sexer, we have:
Dog 1/Dog 2 (all options are equally likely) MM MF FM FF
The first leads to option 1. The second leads to option 2, with Dog 1 being the basis dog. The third leads to option 2, with Dog 2 being the basis dog. These three possibilities are equally likely. The fourth is ruled out by the sexing. So one third of the time, it's option 1, and two-thirds of the time it's option 2. That how moving from Dog 1/Dog 2 to Basis Dog/Other Dog affects the probabilities.
I'd still say 50% in both instances and here's why--you listed the two possiblites as in your 33% scenario as:
1) on the basis that both of the dogs are male
OR
2) the basis that one of the dogs is a male.
So wouldn't (1) look like:
Pair of Dogs M
And (2) would look like
Basis Dog/Non-Basis Dog M/M (if he liked Basis Dog better and decided to talk about that dog and not the other or about the pair) M/F (if he had to talk about basis dog to say "Yes!")
Right? Because putting anything but an M under Basis Dog...doesn't make any sense, because otherwise, it wouldn't be a sufficient basis for saying "Yes!" to the question.
I phrased it poorly. What I meant for 2) is: the basis that only one of the dogs is male. So 1) would look like:
Pair of Dogs MM
and 2) would look like:
Basis Dog/Non Basis Dog MF
So the question is: how likely is option 1, and how likely is option 2? If we go back to having just two dogs, before the dogs have been examined by the dog-sexer, we have:
Dog 1/Dog 2 (all options are equally likely) MM MF FM FF
The first leads to option 1. The second leads to option 2, with Dog 1 being the basis dog. The third leads to option 2, with Dog 2 being the basis dog.
No, the options rule out anything that doesn't look like them, including their converses. Otherwise, the options do not cover all possible outcomes.
Remember, this is under the heading you called "If the guy looks at both dogs before making his statement"; if he's looked at both dogs, F/M doesn't add a row, it just switches Basis Dog/Non Basis Dog to Non Basis Dog/Basis Dog
Dog 1/Dog 2 (all options are equally likely) MM MF FM FF
The first leads to option 1. The second leads to option 2, with Dog 1 being the basis dog. The third leads to option 2, with Dog 2 being the basis dog.
No, the options rule out anything that doesn't look like them, including their converses. Otherwise, the options do not cover all possible outcomes.
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?
Cheeze_Pavilion: Remember, this is under the heading you called "If the guy looks at both dogs before making his statement"; if he's looked at both dogs, F/M doesn't add a row, it just switches Basis Dog/Non Basis Dog to Non Basis Dog/Basis Dog
But if you fold F/M into M/F that doubles the effective probability of M/F.
Remember, this is under the heading you called "If the guy looks at both dogs before making his statement"; if he's looked at both dogs, F/M doesn't add a row, it just switches Basis Dog/Non Basis Dog to Non Basis Dog/Basis Dog
Okay, so you've got:
Pair of Males: M/M
Basis Dog/Non Basis Dog M/F
Non Basis Dog/Basis Dog F/M
The last two options give the same result, but came from different equally likely options, and are still equally as likely.
Now, here is where things change. In the problem, we are told that at least one of the dogs is male. But, we don't know which one.
Actually, we do. Look at the question the word problem asks: "What is the probability that the other one is a male?"
If the only info we have for discriminating between one dog and the other is that one is male, and we're asked about the "other" dog, that means our info must pertain to not just the set but one of the dogs, the non-other one, so things don't actually change, other than our labels for the matix, maybe.
No, you really don't. Here's what the question says:
"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?
First off, the problem explicitly asks if at least one is male(which is answered in the affirmative), but there is no indication if we are talking about dog1 or dog2. Consequently, we don't really know which one is the "other" one-is it dog1 or dog2? It is only if the problem explicitly labels dog1 or dog2 as being that one that we are told is male do we get a probability of 50%. But, because we don't know which one is being referenced when we are told at least one is male, we obtain a probability of only 33%. It is invalid logic to presuppose that it MUST be dog1(or dog2) that the problem is referencing as the known male dog.
geizr: SNIP First off, the problem explicitly asks if at least one is male(which is answered in the affirmative), but there is no indication if we are talking about dog1 or dog2. Consequently, we don't really know which one is the "other" one-is it dog1 or dog2? It is only if the problem explicitly labels dog1 or dog2 as being that one that we are told is male do we get a probability of 50%. But, because we don't know which one is being referenced when we are told at least one is male, we obtain a probability of only 33%. It is invalid logic to presuppose that it MUST be dog1(or dog2) that the problem is referencing as the known male dog.
Quite true. "The other one" refers to the pup which was not identified as male; both the pup which has been identified as male and "the other one" have equal probability of being either pup in the set, in any position in the set, by any ordering of the set, because we have no information about any particular pup, only about the set. In other words, if we are looking down at the two pups in the pet store, we do not know if they are male and female, female and male, or male and male; only female and female can be ruled out by the information we have available. This statement is true regardless of how we order the pups within the set; we have no conclusive knowledge about either pup, only about the set of pups.
This is demonstrably true because either pup being male OR both pups being male will produce the exact same answer, "Yes", from the pet store lady to the question "Is at least one a male?" Given that we do not hear the groomer's side of the conversation, we have no idea if he said: "I just looked at the pup in my hand and it is a male." (50% chance of two males.) "I remember that at least one was a male, maybe both." (33% chance of two males.) "I remember that one was a male, but not which one." (Indeterminate - could mean only one male, or that he only specifically remembers one pup.) "There is one female and one male." (0% chance of two males.) "They are both males." (100% chance of two males.) "The brown one is a male, but the black one is a female." (0% chance of two males.) "The little one is a female, but the bigger one is a male." (0% chance of two males.)
All these statements and a near-infinite number of others from the groomer would have caused the pet store lady to say "Yes". From this it should be clear that the ONLY possibility we can eliminate is that of two females. From our oft-quoted charts, we know that this possibility is 25% for a set of two pups regardless if we treat this as a permutation or combination problem, and that the possibility of two males is 25% for a set of two pups regardless if we treat this as a permutation or combination problem. No matter how you work the problem, it is inescapable that the chance of having two males is 1 in 3 or 33%.
Cheeze_Pavilion: Remember, this is under the heading you called "If the guy looks at both dogs before making his statement"; if he's looked at both dogs, F/M doesn't add a row, it just switches Basis Dog/Non Basis Dog to Non Basis Dog/Basis Dog
But if you fold F/M into M/F that doubles the effective probability of M/F.
It would.
Of course, there's no basis on which to 'fold' anything into anything once you've written the problem like that.
First off, the problem explicitly asks if at least one is male(which is answered in the affirmative), but there is no indication if we are talking about dog1 or dog2.
Yes there is--when the problem ends with "What is the probability that the other one is a male?" If the other one is the only dog whose gender is in doubt, then we know a dog that is not the 'other one' is male. As there is only one dog that is not the other one, we know specifically which dog is male.
werepossum: Quite true. "The other one" refers to the pup which was not identified as male; both the pup which has been identified as male and "the other one" have equal probability of being either pup in the set, in any position in the set, by any ordering of the set, because we have no information about any particular pup, only about the set. In other words, if we are looking down at the two pups in the pet store, we do not know if they are male and female, female and male, or male and male; only female and female can be ruled out by the information we have available. This statement is true regardless of how we order the pups within the set; we have no conclusive knowledge about either pup, only about the set of pups.
How do you reconcile:
"The other one" refers to the pup which was not identified as male;
with
we have no information about any particular pup, only about the set.
If one pup was not identified as male, then doesn't it follow that one pup *was* identified as male, and therefore, because there is only one other pup besides the other pup, that we have information about the pup-that-is-not-other?
Dog 1/Dog 2 (all options are equally likely) MM MF FM FF
The first leads to option 1. The second leads to option 2, with Dog 1 being the basis dog. The third leads to option 2, with Dog 2 being the basis dog.
No, the options rule out anything that doesn't look like them, including their converses. Otherwise, the options do not cover all possible outcomes.
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.
werepossum: [quote=geizr post=18.73797.836214] ...From our oft-quoted charts, we know that this possibility is 25% for a set of two pups regardless if we treat this as a permutation or combination problem, and that the possibility of two males is 25% for a set of two pups regardless if we treat this as a permutation or combination problem. No matter how you work the problem, it is inescapable that the chance of having two males is 1 in 3 or 33%.
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. However, we have to be careful because, as it turns out, the M/F combination has a degeneracy of 2. This means it counts twice, while the M/M and F/F combinations count only once each. So, we would still obtain a probability of 33% because of the degeneracy.
M/M + M/F = 1 + 2 = 3 = T
M/F ÷ T * 100% = 2 ÷ 3 * 100% = 66% (probability of one male and one female) M/M ÷ T * 100% = 1 ÷ 3 * 100% = 33% (probability of both males)
Compare this to the original logic in which the configurations are all distinct and nondegenerate:
M/M + M/F + F/M = 1 + 1 + 1 = 3 = T
(M/F + F/M) ÷ T * 100% = (1 + 1) ÷ 3 * 100% = 66% (probability of one male and one female) M/M ÷ T * 100% = 1 ÷ 3 * 100% = 33% (probability of both males)
The only change is how we have to calculate the probabilities, not the final result.
In all honesty, I can see how people can easily get the wrong answer of 50% because we are all taught that there is always a 50% chance of being born male or female. This is true for a single individual; however, it is not necessarily true when we start talking about specific combinations of males and females, as occurs in this problem.
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.
"The other one" refers to the pup which was not identified as male;
with
we have no information about any particular pup, only about the set.
If one pup was not identified as male, then doesn't it follow that one pup *was* identified as male, and therefore, because there is only one other pup besides the other pup, that we have information about the pup-that-is-not-other?
Read the rest of the post. The two statements are perfectly reconciled because we cannot assign either role to either pup. No matter how much we learn specifically about either pup, until we learn the sex of both pups we can never know the sex of either pup. Thus we will never be able to say this is the pup that was identified as male and this is the other one.
First off, the problem explicitly asks if at least one is male(which is answered in the affirmative), but there is no indication if we are talking about dog1 or dog2. Consequently, we don't really know which one is the "other" one-is it dog1 or dog2? It is only if the problem explicitly labels dog1 or dog2 as being that one that we are told is male do we get a probability of 50%. But, because we don't know which one is being referenced when we are told at least one is male, we obtain a probability of only 33%. It is invalid logic to presuppose that it MUST be dog1(or dog2) that the problem is referencing as the known male dog.
How do you reconcile:
"The other one" refers to the pup which was not identified as male;
with
we have no information about any particular pup, only about the set.
If one pup was not identified as male, then doesn't it follow that one pup *was* identified as male, and therefore, because there is only one other pup besides the other pup, that we have information about the pup-that-is-not-other?
The problem is that you don't know which is the male pup. Is it dog1 or dog2? If it is dog1, then obviously dog2 is the other one. If it is dog2, then obviously the other one is dog1. So, which one is the male pup? You don't know. So, you also don't know which one is being referred to as the other one. As such, you have 2 statements with one dog male:
dog1 male, dog2 unknown dog1 unknown, dog2 male
or
dog1 dog2 M ??? ??? M
Now, let's suppose that the "other" dog is female. Then, we have 2 configurations one male and one female:
dog1 dog2 M F F M
Then, we have a third statement that both dogs are male. Regardless of the direction we go, the resulting configuration is the same:
dog1 dog2 M M
So, in total, there are 3 possible configurations, but in only one configuration do we get 2 males.
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?
"The other one" refers to the pup which was not identified as male;
with
we have no information about any particular pup, only about the set.
If one pup was not identified as male, then doesn't it follow that one pup *was* identified as male, and therefore, because there is only one other pup besides the other pup, that we have information about the pup-that-is-not-other?
Read the rest of the post. The two statements are perfectly reconciled because we cannot assign either role to either pup. No matter how much we learn specifically about either pup, until we learn the sex of both pups we can never know the sex of either pup. Thus we will never be able to say this is the pup that was identified as male and this is the other one.
Wait, if I specifically learn a specific dog is male--"No matter how much we learn specifically about either pup"--I don't really know that a specific dog is male until I know whether the other dog is male or female?
Like I said, you're talking about quantum puppies now.
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"?
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"?
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. However, it is unknown which of these labels applies to the male dog.
But, now, let's suppose we do it your way for a bit. Suppose I force label the male dog as dog1 and say that the other dog is dog2. Then I get 2 different configurations:
dog1 dog2 M F M M
At first glance, it seems the probability of M/M is then 50%, but such is not the case. The reason is that the M/F combination has a degeneracy of 2. The male dog, dog1, could be the dog picked up and examined, or it could be the dog that remained in the bin, while the dog examined is found to be female. Two cases of a M/F combination. As a result, there are 2 counts of a M/F combination and only 1 count of M/M combination. So, again, we obtain the correct probability is 33%, but only after accounting for the degeneracy. But, this fiat is treacherous logic because we could have miscalculated by missing the degeneracy.
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%.
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.
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.
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.
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: 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.
Samirat: 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.
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?
these are called "independent variables". the outcome of one does not affect the other. so the chance of the 2nd dog being male is equal to the chance first dog being male. cool, i just used my year 12 mathematical methods course in "real life"
What we need is language that somehow screens out the FF pups from the predicted 25/50/25 breakdown, without making it obvious that they have been screened.
Yeah, I don't know how to do that either!