Poll: A little math problem

[man-man]

But when looking at the odds for 2 puppies, you have to consider them as separate to assign equal odds to each possibility.

If you prefer, the possibilities for 2 puppies are 2 males (25% chance) 1 of each (50%) chance and 2 females (25% chance). Remove the option for it to be 2 females and you're left with the possibility of 2 males or 1 of each, and 2 males has the odds 25/75. Which is... 1/3

 

[Fraught]

But when looking at the odds for 2 puppies, you have to consider them as separate to assign equal odds to each possibility.

If you prefer, the possibilities for 2 puppies are 2 males (25% chance) 1 of each (50%) chance and 2 females (25% chance). Remove the option for it to be 2 females and you're left with the possibility of 2 males or 1 of each, and 2 males has the odds 25/75. Which is... 1/3

That's the point. You can't add any "equal odds to each possibility", because we have established that one is a male. The only possibility we are trying to figure out, is whether the other one is male or female. Thus making it 50%.

 

[man-man]

That was for _all_ the possibilities involving 2 puppies. Then when you add the information that one is male you can eliminate all the options that don't meet that condition. Leaving either one of 2 ways to say the same thing:

1M2M, 1M2F and 1F2M at 25% probability each
"both male" at 25% chance, "one of each" at 50% chance.

Either way, the option where both are male has a chance of 25 out of 75, which is 33%

 

[Dboyz-x.etown]

It's 33.3%. Since you know that one of them is a male, and the other has a .5 probability of being a male, there is a 66.6% chance that they'll both be male.

Er, wait, no, since they're exclusive, it's only a 50 percent chance.

Or are they exclusive?

GAHH!! I DON'T KNOW!

Great. This is going to keep me up at night. Note to self: Stop trying to solve math problems when you're totally spent. At least it's not the airplane on a treadmill argument this time...
::EDIT::

Alright, I'm fairly confident that they're exclusive. And I'm also fairly certain that this post adds nothing to the thread. I'm also fairly disappointed that there is no "delete post" option for me to solve this.

::EDIT 2::
Wait, no, after rereading the OP, it's not. Also, after reading many of the posts before me, I find that the common fallacy is that people label the puppies, with 1 puppy being a "known" male, and that the other puppy can go either way. The thing is, we don't know which puppy is the "other". The "other" puppy could either be the one that fulfilled the "at least 1 is male" requirement, or it could be the "undefined" puppy as well. However, most label the "other" puppy as the one which is unknown. Also, I seem to remember a similar probability problem like this. Also, this reminds me of the movie 21 and that game show question...
http://www.youtube.com/watch?v=14KMJyf5N7E

 

[Dboyz-x.etown]

Actually, I think I've got it.

It's a 66.6% chance that the other puppy is male. When you look at all the possible pairings:
M/F
M/M
F/M
(And yes, there is a difference between M/F and F/M)
you just simply count up all the m's and f's

-so it's 4 m's, 2 f's

-so it's 6 total, and out of them, 4 m's

-so it's a 4/6 probability that the "other" is male.


Everyone else who was doing the pairings were on the right track, however often they made the mistake of either not eliminating the F/F choice (which is an impossibility), or they just counted M/F and F/M as the same thing (which is not the case, as seeing that we're not looking at a pairing here, but only one unit of that pairing. If you look back at the possible pairings I listed, you'll see that in each column, there are two M's and one F for the possibilities, where each column could represent the assignment of "other").

And as for everyone who complained about the wording: In my opinion, the wording of the question is what makes it so beautiful. It wouldn't be nearly as complex if it weren't for that, it would just be a boring old exclusivity or compounded exclusivity based probability problem.

 

[Dboyz-x.etown]

Another way of thinking this problem through could be to use 6 parallel universes.

UNIVERSE 1:
You walk into a pet store. You see two dogs, one on the left and one on the right. You walk over to the one on the RIGHT, and find that it is a MALE. The one on the left was also a MALE.

UNIVERSE 2:
You walk into a pet store. You see two dogs, one on the left and one on the right. You walk over to the one on the LEFT, and find that it is a MALE. The one on the left was also a MALE.

UNIVERSE 3:
You walk into a pet store. You see two dogs, one on the left and one on the right. You walk over to the one on the RIGHT, and find that it is a FEMALE. The one on the left was a MALE.

UNIVERSE 4:
You walk into a pet store. You see two dogs, one on the left and one on the right. You walk over to the one on the LEFT, and find that it is a MALE. The one on the left was a FEMALE.

UNIVERSE 5:
You walk into a pet store. You see two dogs, one on the left and one on the right. You walk over to the one on the LEFT, and find that it is a FEMALE. The one on the left was a MALE.

UNIVERSE 6:
You walk into a pet store. You see two dogs, one on the left and one on the right. You walk over to the one on the RIGHT, and find that it is a MALE. The one on the left was a FEMALE.

In each of these situations, the variables are UNIQUE, and accounts for every possibility of outcome. In each universe, either the sex of the other puppy changes, or the identity of the "other" does. When you lay it out, you see that there is a 4/6 probability of the other being male.

@ Cheese:

The Question is not broken, and no, that wouldn't be nonsensical. It took me a while to figure out what you were saying, but yes, you do have to include the probability of the puppy in question BEING the known male in the first place.

 

[Narthlotep]

So, we're just leaving out the possibility of hermaphroditic dogs? Because if we aren't then wouldn't the odds then become something like this?
puppy 1/2
M/M
M/F
M/H
F/M
H/M
and then the odds would be more in the range of 20% for both puppies to be male?

And to the above poster, if your willing to use 6 parallel universes, then you should use all of the theoretically possible (read: infinite number of) parallel universes that m- and string theorists currently think make up our multiverse. Then you can have the fun of doing quantum jumps or phasing through solid objects, or have the same puppy be in both locations at once. And if we go that far, then the chance would either be 0% or an infinite% chance of the other puppy being male, quite possibly at the same time, in the same universe no less.

 

[Lukeje]

So, we're just leaving out the possibility of hermaphroditic dogs? Because if we aren't then wouldn't the odds then become something like this?
puppy 1/2
M/M
M/F
M/H
F/M
H/M
and then the odds would be more in the range of 20% for both puppies to be male?

And to the above poster, if your willing to use 6 parallel universes, then you should use all of the theoretically possible (read: infinite number of) parallel universes that m- and string theorists currently think make up our multiverse. Then you can have the fun of doing quantum jumps or phasing through solid objects, or have the same puppy be in both locations at once. And if we go that far, then the chance would either be 0% or an infinite% chance of the other puppy being male, quite possibly at the same time, in the same universe no less.

But you are not taking into account the incredibly low probability for hermaphroditic puppies, which would effectively make the probabilities for anything with an 'H' zero.

And with regard to the 'infinite universes', I've already suggested that we should technically take a weighted average of all the possibilities.

 

[Narthlotep]

So, we're just leaving out the possibility of hermaphroditic dogs? Because if we aren't then wouldn't the odds then become something like this?
puppy 1/2
M/M
M/F
M/H
F/M
H/M
and then the odds would be more in the range of 20% for both puppies to be male?

And to the above poster, if your willing to use 6 parallel universes, then you should use all of the theoretically possible (read: infinite number of) parallel universes that m- and string theorists currently think make up our multiverse. Then you can have the fun of doing quantum jumps or phasing through solid objects, or have the same puppy be in both locations at once. And if we go that far, then the chance would either be 0% or an infinite% chance of the other puppy being male, quite possibly at the same time, in the same universe no less.

But you are not taking into account the incredibly low probability for hermaphroditic puppies, which would effectively make the probabilities for anything with an 'H' zero.

And with regard to the 'infinite universes', I've already suggested that we should technically take a weighted average of all the possibilities.

I'm just trying to push the thread up to 1,000 posts while being more or less on topic, and to annoy everyone who took stats in school, as I've posted somewhere in here that I went the calculus/trig route for mathematics.

 

[Lukeje]

I'm just trying to push the thread up to 1,000 posts while being more or less on topic, and to annoy everyone who took stats in school, as I've posted somewhere in here that I went the calculus/trig route for mathematics.

Less than 10 to go now...

 

[Dommyboy]

I'm just trying to push the thread up to 1,000 posts while being more or less on topic, and to annoy everyone who took stats in school, as I've posted somewhere in here that I went the calculus/trig route for mathematics.

Less than 10 to go now...

Achievement whores are slowly destroying the Escapist.

For me though I would say it would be 50% because it seems slightly logical.

 

[Lukeje]

I'm just trying to push the thread up to 1,000 posts while being more or less on topic, and to annoy everyone who took stats in school, as I've posted somewhere in here that I went the calculus/trig route for mathematics.

Less than 10 to go now...

Achievement whores are slowly destroying the Escapist.

For me though I would say it would be 50% because it seems slightly logical.

Don't you mean Badge whores? And just stating 50% without explanation is blasphemy around these here parts.

 

[leafsnail]

50%. It is independant of the other dogs gender.

 

[Narthlotep]

Nah, I like my quantum mechanics answer of it being more or less pointless to try to determine, as it could be the same dog occupying multiple locations at once. Though that then degrades into there might be only one dog, and the others are merely quantum madness, or there might be an infinite number of dogs occupying the same point in space-time.

 

[Samirat]

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 now. If you asked the washer woman this question in real life, she's not going to be like "Oh, but they're both male, there is no other dog, haha, I'm so much cleverer than you." No, asking if the other dog is male as well is the same as asking if both dogs are male, in slightly less specific wording. Your argument is almost Cheezesque.