Poll: A little math problem

[Lukeje]

I'm pretty sure it will.

(All together now...) "Oh no it won't!"

 

[geizr]

I'm pretty sure it will.

(All together now...) "Oh no it won't!"

Haha! Maybe if we keep doing this it will.

 

[Lukeje]

I'm pretty sure it will.

(All together now...) "Oh no it won't!"

Haha! Maybe if we keep doing this it will.

Indeed, though there is the worry that the Banhammer may striketh down anyone who strays too far off topic...

 

[chromewarriorXIII]

I used science and math...

-Females carry 2 X chromosomes
-Males carry 1 X and 1 Y chromosome so...

X Y
X XX|XY
X XX|XY
If the dogs had 4 babies then it is most likely that 50% would be males and 50% females
So take those numbers and divide them by 2 and you get 2 babies at 25% chance each.

I think that's right... not sure...

 

[Lukeje]

I used science and math...

-Females carry 2 X chromosomes
-Males carry 1 X and 1 Y chromosome so...

X Y
X XX|XY
X XX|XY
If the dogs had 4 babies then it is most likely that 50% would be males and 50% females
So take those numbers and divide them by 2 and you get 2 babies at 25% chance each.

I think that's right... not sure...

Except that that leads to a 1/4 chance that both puppies are female, which we know is an impossibility.

 

[man-man]

I don't want to spend hours reading the many many posts in here, so it's likely someone has said this long ago, but there's an important difference between "there is a puppy, what are the odds of it being male" and "there are 2 puppies, at least one of which is male, what are the odds of them both being male"

The answer to the first is 1/2 (assuming no funky biology going on) but the answer to the second isn't so simple. As people have doubtless been saying through this whole thread, you need to consider all the possible pairings of 2 puppies. 2 males (25% probability) 2 females (25% probability) and one of each (50% probability) You've been given the fact that at least one is male, so that rules out the 2 females scenario, leaving the odds of both being male as 25/75 = 1/3.

You have information about "at least one" of 2 puppies, you can't then ignore one puppy and just look at the odds of the other puppy. Regardless of all else, the odds of 2 puppies being the same sex are less than those of them being 1 male and 1 female, and that still persists when you're given the fact that at least one is male.

 

[Lukeje]

I don't want to spend hours reading the many many posts in here, so it's likely someone has said this long ago, but there's an important difference between "there is a puppy, what are the odds of it being male" and "there are 2 puppies, at least one of which is male, what are the odds of them both being male"

The answer to the first is 1/2 (assuming no funky biology going on) but the answer to the second isn't so simple. As people have doubtless been saying through this whole thread, you need to consider all the possible pairings of 2 puppies. 2 males (25% probability) 2 females (25% probability) and one of each (50% probability) You've been given the fact that at least one is male, so that rules out the 2 females scenario, leaving the odds of both being male as 25/75 = 1/3.

You have information about "at least one" of 2 puppies, you can't then ignore one puppy and just look at the odds of the other puppy. Regardless of all else, the odds of 2 puppies being the same sex are less than those of them being 1 male and 1 female, and that still persists when you're given the fact that at least one is male.

Its not that that leads to amiguity... it is the fact that the statement does not read 'what are the chances both are male', but instead asks 'what is the probability that the other one is male'.

 

[man-man]

If you're given that one is male, then the other being male means that both are male. There's no difference there... "one is male and the other is male" and "both are male" have the same meaning.

If you ignore all the rest of the problem and just look at the chances of a puppy being male, then yes, 1 in 2. But for it to be "the other puppy" you need to take into account the first one, and hence the rest of the situation, which includes the diminished odds of 2 randomly selected puppies being the same sex.

 

[hemahemahema]

Only 34 posts to go, Let's just start talking about something else.

Does anyone know if Braid has ever been released as PC-CD rom, or do you have to pay and download it?

 

[hemahemahema]

If you're given that one is male, then the other being male means that both are male. There's no difference there... "one is male and the other is male" and "both are male" have the same meaning.

If you ignore all the rest of the problem and just look at the chances of a puppy being male, then yes, 1 in 2. But for it to be "the other puppy" you need to take into account the first one, and hence the rest of the situation, which includes the diminished odds of 2 randomly selected puppies being the same sex.

Hey, someone's saying the right thing! Why can't you all listen to the man-man?

 

[Lukeje]

If you're given that one is male, then the other being male means that both are male. There's no difference there... "one is male and the other is male" and "both are male" have the same meaning.

If you ignore all the rest of the problem and just look at the chances of a puppy being male, then yes, 1 in 2. But for it to be "the other puppy" you need to take into account the first one, and hence the rest of the situation, which includes the diminished odds of 2 randomly selected puppies being the same sex.

Not necessarily; it could be read that if both dogs are male, then there is no 'other' and as such the chance of both being male diminishes to zero...

 

[hemahemahema]

If you're given that one is male, then the other being male means that both are male. There's no difference there... "one is male and the other is male" and "both are male" have the same meaning.

If you ignore all the rest of the problem and just look at the chances of a puppy being male, then yes, 1 in 2. But for it to be "the other puppy" you need to take into account the first one, and hence the rest of the situation, which includes the diminished odds of 2 randomly selected puppies being the same sex.

Not necessarily; it could be read that if both dogs are male, then there is no 'other' and as such the chance of both being male diminishes to zero...

Come on, let's stop. Anyone know the answer to my Braid question?

 

[man-man]

What... in hell? Both being male is just one of the possible scenarios, how could it possibly imply that there isn't a second puppy? For that matter, how can "both dogs are male" lead you to "the chance of both being male diminishes to zero"? If both are male, the odds of both being male are 100%, because you just said they're both male.

This problem seems to exploit the same flaw in common reasoning as the Monty Hall problem - the simple answer ignores some of the pertinent information. Whether it's simplifying the 3 door problem to a choice between 2 doors (when actually the information about the 3rd door is relevant) or simplifying this 2 dog problem into a question about 1 dog. Same problem.

You could phrase the problem differently - start with "there are 2 dogs, what are the odds that they're both male?" to which the answer is 1/4. Then say "one of them's male, now what are the odds?" To which the answer is that, since the "both female" option has been eliminated, the odds have risen to 1/3. It's an equivalent question, and the answers are the same.

 

[Lukeje]

What... in hell? Both being male is just one of the possible scenarios, how could it possibly imply that there isn't a second puppy? For that matter, how can "both dogs are male" lead you to "the chance of both being male diminishes to zero"? If both are male, the odds of both being male are 100%, because you just said they're both male.

Proof by contradiction; if we assume that both are male, and then find a contradiction, then that means that they can't both be male.
Another way:
Three possibilities
MM
MF
FM
The MM pair has no dog that is not male (given that this is the only criterion we are given to distinguish them), therefore it can't be a valid grouping as it has no 'other'. Both MF and FM pairings have the 'other' dog as female, therefore the probability that the 'other' is male is zero.

 

[hemahemahema]

What... in hell? Both being male is just one of the possible scenarios, how could it possibly imply that there isn't a second puppy? For that matter, how can "both dogs are male" lead you to "the chance of both being male diminishes to zero"? If both are male, the odds of both being male are 100%, because you just said they're both male.

Proof by contradiction; if we assume that both are male, and then find a contradiction, then that means that they can't both be male.
Another way:
Three possibilities
MM
MF
FM
The MM pair has no dog that is not male (given that this is the only criterion we are given to distinguish them), therefore it can't be a valid grouping as it has no 'other'. Both MF and FM pairings have the 'other' dog as female, therefore the probability that the 'other' is male is zero.

Why do we want a dog that's not male? Anyway who cares? Can you get Braid on PC-CD rom?

 

[man-man]

"Other" doesn't mean "a puppy that is not male, it means "a puppy that isn't the first one revealed to be male". The original problem had it revealed that _at least_ one is male. The other one also being male doesn't make it not an "other" puppy.

Both puppies being male is perfectly consistent with one being male, and the other one also being male... no contradiction there.

And to answer the other guy, there will be a PC version of Braid at some point in 2008, but there's no release date as yet, as it says on their website http://braid-game.com/

 

[Lukeje]

"Other" doesn't mean "a puppy that is not male, it means "a puppy that isn't the first one revealed to be male". The original problem had it revealed that _at least_ one is male. The other one also being male doesn't make it not an "other" puppy.

Both puppies being male is perfectly consistent with one being male, and the other one also being male... no contradiction there.

That's your interpretation of the problem. I have to argue that my solution is more beautiful in its solution (I like the poetic justice of everyone fighting over 1/3 / 1/2 while I've found an interpretation that negates both of them), but you are of course entitled to your own opinion.

 

[Fraught]

I don't want to spend hours reading the many many posts in here, so it's likely someone has said this long ago, but there's an important difference between "there is a puppy, what are the odds of it being male" and "there are 2 puppies, at least one of which is male, what are the odds of them both being male"

The answer to the first is 1/2 (assuming no funky biology going on) but the answer to the second isn't so simple. As people have doubtless been saying through this whole thread, you need to consider all the possible pairings of 2 puppies. 2 males (25% probability) 2 females (25% probability) and one of each (50% probability) You've been given the fact that at least one is male, so that rules out the 2 females scenario, leaving the odds of both being male as 25/75 = 1/3.

You have information about "at least one" of 2 puppies, you can't then ignore one puppy and just look at the odds of the other puppy. Regardless of all else, the odds of 2 puppies being the same sex are less than those of them being 1 male and 1 female, and that still persists when you're given the fact that at least one is male.

Yeah, I'm pretty sure the thread asked what is the possibility that the other one is male? Leave the first one out of it, what is the possibility that the other one is male? I mean, we know that the first one is male, so nothing can be changed here. But now we want to know what is the possibility of the other one being a male. Since he can either be a male or a female, I believe that is 50% possibility of being a male.

 

[man-man]

"Leave the first one out of it"

That's exactly it though, you CAN'T leave the first one out, because you've been told things about one of a pair, not 2 separated puppies. Call them Puppy 1 and Puppy 2, at least one is male, but you don't know which.

The possibilities for 2 puppies are (where 1M means "puppy number 1 is male") 1M2M, 1M2F, 1F2M, 1F2F, you're essentially treating it as if we've been told that puppy number 1 is male, so the options remaining are 1M2M and 1M2F, so it's 50/50. Actually we've been told that one of the pair is male, leaving 1M2M, 1M2F and 1F2M as possibilities, so there's a 1/3 chance that the other puppy is male.

I can't think of any other way to say it, you can't just ignore the fact that we've only been told things about the PAIR (one of the pair is male) not about an individual puppy.

 

[NeedAUserName]

The possibilities for 2 puppies are (where 1M means "puppy number 1 is male") 1M2M, 1M2F, 1F2M, 1F2F, you're essentially treating it as if we've been told that puppy number 1 is male, so the options remaining are 1M2M and 1M2F, so it's 50/50. Actually we've been told that one of the pair is male, leaving 1M2M, 1M2F and 1F2M as possibilities, so there's a 1/3 chance that the other puppy is male.

You are an idiot. 1M2F and 1F2M, thats still 1s male 1s female both ways...