Challenge your brains! Can you solve this problem?

[MBz]

I'd have to agree with Roy Hobbs on this problem and that its wording is fairly confusing/not worded properly.  "What is the probability that the other one is a male?"  That is the question being asked.  That answer would be 50% since as stated before the events are independent and when worded like it would be like flipping 2 coins after each other.  You flip a coin heads, the next time you flip the probability to flip a heads is still 50%. Now if from the start, it said you had an equal amount of female and male dogs, what was the probability that you would get 2 males.  That would be 1/3.

[fairweatherfan]

I think it's in the wording.

If the question is "What are the chances that both dogs are male?", it's 1/3.

If the question is "What are the chances that the *other* dog (i.e., the one that could be male or female) is male?", it's 1/2.

You *know* dog #1 is male.  It can't be female, because you're only asked to look for the "other".  Therefore, "female/male" isn't an option (nor, of course, is "female/female".  You're left with either "male/male", or "male/female".

If the answer that was sought was 1/3, the question should have been posed correctly.  The "other" can only be male or female; it can't be a third gender.

If one dog is definitely male, the probability that the other dog is male is the same as the probability that both are male - since one has been established to be male, "the other dog is male" = "both dogs are male"

[Bahku]

I'd have to agree with Roy Hobbs on this problem and that its wording is fairly confusing/not worded properly.  "What is the probability that the other one is a male?"  That is the question being asked.  That answer would be 50% since as stated before the events are independent and when worded like it would be like flipping 2 coins after each other.  You flip a coin heads, the next time you flip the probability to flip a heads is still 50%. Now if from the start, it said you had an equal amount of female and male dogs, what was the probability that you would get 2 males.  That would be 1/3.

But that's not the variable ... the question still began with a complete unknown, ONE was eliminated, therefor there are still 3 variables remaining. (ouch) Trust me, scientists have been arguing this one for a long time, and it completely depends on what you start with, though everyone here has a valid argument. This, by it's very nature, has the possibility of becoming one of the longest-lasting threads on CB, because the paradox allows for more than one answer. (But it's still fun to discuss ... at least for the first hour or so).

[Fan from VT]

It's still a bit ambiguous.

The answer is 1/3 OR 1/2 depending on the selection method.

For example, if you sent out a survey that said "respond if you are a member of a 2 child home and at least 1 of you is a boy," the odds that the other child would be a boy is 50%.

If you just selected for 2 children households, picked one, found that at least one child was a boy, then asked the question, then the odds that the other child is a boy is 1/3.



Example 1:

Boy1, Boy2, Boy3, and Boy4 are all members of 2 children homes. They are arranged as such:

Boy1Boy2, Boy3Girl, Boy4Girl.

If you pick at random, you are identifying Boy1, Boy2, Boy3, or Boy4.

Boy1 and Boy2 are in BoyBoy households, and Boy3 and Boy4 are in BoyGirl households. Obviously you have a 1/4 chance of picking each boy, so the likelihood that the "other" child will be a boy is 1/2.


Example 2:
GirlGirl, BoyGirl, GirlBoy, and BoyBoy are two children homes. Pick one at random. Then you state that at least one child is a Boy. You've therefore eliminated GirlGirl as a choice, so if one is a Boy, then there's a 1/3 chance the other child is a boy.



The question as posed is basically example 2, so I have to admit I was wrong.

The gender of the other child is completely independent of the gender of the first, and if there was one male puppy and another yet to be born, the next puppy born has a 50/50 shot of male or female. However, knowing it's a two puppy litter and knowing at least 1 is male means there's a 1/3 chance the other is male.

As I thought about it more, this made sense logically, and not just mathematically.

The two puppies could be GG, BG, GB, or BB. We know that overall, a population is about 50/50 Boy/Girl. Well, if we played the above scenario a hundred times and guessed 50/50 each time:

BG
BB
BG
BB
BG
BB
BG
BB
BG
BB

we quickly end up with a population that is 75% male, which can't happen!


Good question.

[bdm860]

See I read this whole dog question as what is the probability of A given B.

What is the probability of a dog being male, given that one of the dogs is already male.

Dog 1 = M or F
Dog 2 = M or F

4 possible outcomes.

They told us Dog 1 is M, so what we're left with is this:

Dog 1 = M
Dog 2 = M or F
(Or given that Dog 1 is M, what is the probability that Dog 2 is also M?)
 
50%.

[Bahku]

Seriously ... this can go on forever, because empirically, everyone here is correct, depending on how you want to view how the original info is supplied, (the main paradox of statistics). I interpret it to be starting with four variables, because before the original question is aksed, the unknown is total. If you start from after the original info is supplied about the sex of the first puppy, then the variable changes, and it would have to be 1 in 2. But the question begins without any info known, so the remaining variable has to be 1 in 3. (phew!) And the discussion goes on ... and on ... and on ...

[Roy Hobbs]

TP for everyone involved in this dog discussion.  I'm not going to spend any more brainpower on it.

[Bahku]

TP for everyone involved in this dog discussion.  I'm not going to spend any more brainpower on it.

LOL! Agreed ... this thread has gone to the dogs, (granted just little ones) ... and my head hurts!

[Master Po]

[jambr380]

I do have to say this is a great thread- just finished reading everything and I feel a whole lot smarter now...

As for the dog thing, I just can't figure out why you have to eliminate any of the possibilities, since the first dog is independent of the second (the coin theory), but that is what Roy and others have been saying all along...It is neat to see how many statistics fans are on this site, though...

[Master Po]

I made close to a perfect grade in a Masters level statistics class......but yet.....math on Celtics blog site....I couldn't even read any of it.

[Bahku]

I do have to say this is a great thread- just finished reading everything and I feel a whole lot smarter now...

As for the dog thing, I just can't figure out why you have to eliminate any of the possibilities, since the first dog is independent of the second (the coin theory), but that is what Roy and others have been saying all along...It is neat to see how many statistics fans are on this site, though...

I'd have to say that I'm not a statistics "fan" ... it's more of a "love/hate" thing for me ... if you can use it to your benefit, it's great, but otherwise, it's about on the same level for me as accounting, (I'd rather chew broken glass). That's just a joke for all the CPA's out there ... I really adore accounting, (especially at tax time!)

[nickagneta]

This is a very easy answer because of the use of the word "other". In the flawed thinking of the 1/3 answer what you people are leaving out is that you already now that one specific dog(Dog1) is guaranteed to be male.


Dog1/Dog2
Female/Female
Female/Male
Male/Female
Male/Male

So Female/Female is eliminated because because both can't be female. That leaves three options.

Dog1/Dog2
Female/Male
Male/Female
Male/Male

But because Dog1 is guaranteed to be a male you also have to eliminate the Female/Male option. That leaves:

Male/Female
Male/Male

It is all keyed by the use of the word other in the word problem which automatically designates Dog1 as male. That's where most mathematicians and statistically oriented people get fooled. This is not a math or statistics problem. The use of the word "other" makes this a logic problem and logically speaking there is only one of two sexes the "other" dog can be.

Answer is easily 50%

[mgent]

This is a very easy answer because of the use of the word "other". In the flawed thinking of the 1/3 answer what you people are leaving out is that you already now that one specific dog(Dog1) is guaranteed to be male.


Dog1/Dog2
Female/Female
Female/Male
Male/Female
Male/Male

So Female/Female is eliminated because because both can't be female. That leaves three options.

Dog1/Dog2
Female/Male
Male/Female
Male/Male

But because Dog1 is guaranteed to be a male you also have to eliminate the Female/Male option. That leaves:

Male/Female
Male/Male

It is all keyed by the use of the word other in the word problem which automatically designates Dog1 as male. That's where most mathematicians and statistically oriented people get fooled. This is not a math or statistics problem. The use of the word "other" makes this a logic problem and logically speaking there is only one of two sexes the "other" dog can be.

Answer is easily 50%

But you don't know WHICH dog is the male.

[KJR]

Dog Problem

As the question is written, the answer is always 50%.

If you want to make the answer 1 in 3, you have to make the question site or order specific, viz:
at least one of two dogs is male, what is the chance that the dog in box #2 is male?

if the question is not site or order specific, then male/female and female/male are the same, and  the answer is always 50%