Poll: A little math problem

[geizr]

?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.

 

[werepossum]

SNIP
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?

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.

 

[geizr]

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.

 

[geizr]

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?

 

[geizr]

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.

 

[geizr]

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]

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]

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]

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.