Poll: A little math problem

[Jamanticus]

Ok this is like the monty hall problem.
If atleast one is male, you have three possible situations
Beagle1 Beagle 2
Male Male
Male Female
Female Male

Of these three possible outcomes, only one of them results in the second beagle being male.
Thats where the 33% comes from...
That's the logic behind the problem, however there's also the idea of independent events... the mind boggles.

This.

Then again, I don't remember very much about probability and the last time I studied that subject was 5 years ago.

 

[Anarchemitis]

Yay Probability!

If you flip a coin and it comes up Heads 7/20 tosses, what are the chances the 21st toss will be Heads again?

 

[Imitation Saccharin]

-deleted (not worth it)

 

[Shivari]

Ok this is like the monty hall problem.
If atleast one is male, you have three possible situations
Beagle1 Beagle 2
Male Male
Male Female
Female Male

Of these three possible outcomes, only one of them results in the second beagle being male.
Thats where the 33% comes from...
That's the logic behind the problem, however there's also the idea of independent events... the mind boggles.

So is this thing right or were the rest of us right?

 

[cleverlymadeup]

based on the question stated in the pole it's 50% the other one is male

simply because you can disregard anything about the other dog, the lady is unsure if it's male or female but one is definitely male. so this means the other dog has a 50% chance of being male or female irregardless of the sex of the other dog

 

[werepossum]

Ok this is like the monty hall problem.
If atleast one is male, you have three possible situations
Beagle1 Beagle 2
Male Male
Male Female
Female Male

Of these three possible outcomes, only one of them results in the second beagle being male.
Thats where the 33% comes from...
That's the logic behind the problem, however there's also the idea of independent events... the mind boggles.

Give this person a banana. The two beagles are already selected as part of a set; thus the 33%. If you selected one beagle, discovered it was male, then selected another beagle, the chance would be 50%. But if you select them two at a time and identify one as male, you remove the chance that both are female before the second chance.

 

[Fud]

Ok this is like the monty hall problem.
If atleast one is male, you have three possible situations
Beagle1 Beagle 2
Male Male
Male Female
Female Male

Of these three possible outcomes, only one of them results in the second beagle being male.
Thats where the 33% comes from...
That's the logic behind the problem, however there's also the idea of independent events... the mind boggles.

Well, this is actually why I posted this topic. I think that this is the mathematically proven solution, atleast according to Wikipedia. However, if you think about the problem as the person randomly grabbed one and it happened to be male, instead of looking at both and then replying, I think the answer is actually 50%. This answer seems right to me because if they just randomly grab one, there's a 50% chance that he would have grabbed a male in one of the two m/f pairs, and a 100% chance in the m/m pair. So, at least to my tired brain, there's a 50% chance that the male is from the m/m pair and a 50% from the m/f pair, therefore a 50% of the other being male. That seemed to be vaguely like the Bertrand's box paradox.

 

[Captain Wes]

Ok this is like the monty hall problem.
If atleast one is male, you have three possible situations
Beagle1 Beagle 2
Male Male
Male Female
Female Male

fixed: that says the same as the one above it so it's irrelevant
we know that the first is male so it would look like this
male/ male
male/female

 

[Shivari]

Ok this is like the monty hall problem.
If atleast one is male, you have three possible situations
Beagle1 Beagle 2
Male Male
Male Female
Female Male

Of these three possible outcomes, only one of them results in the second beagle being male.
Thats where the 33% comes from...
That's the logic behind the problem, however there's also the idea of independent events... the mind boggles.

Give this person a banana. The two beagles are already selected as part of a set; thus the 33%. If you selected one beagle, discovered it was male, then selected another beagle, the chance would be 50%. But if you select them two at a time and identify one as male, you remove the chance that both are female before the second chance.

But the gender of the second one isn't dependent on the first one. It's still 50% right?

 

[Jamanticus]

Ok this is like the monty hall problem.
If atleast one is male, you have three possible situations
Beagle1 Beagle 2
Male Male
Male Female
Female Male

fixed: that says the same as the one above it so it's irrelevant

Ah, but there's a reason it was like that- it was the order.

 

[dirtface]

Ok this is like the monty hall problem.
If atleast one is male, you have three possible situations
Beagle1 Beagle 2
Male Male
Male Female
Female Male

Of these three possible outcomes, only one of them results in the second beagle being male.
Thats where the 33% comes from...
That's the logic behind the problem, however there's also the idea of independent events... the mind boggles.

So is this thing right or were the rest of us right?

... The wikipedia article states its 1/3, i was just explaining it.
For the other guy who said that male female was the same as female male - you're sortof correct, but not really. they are different cases that can be grouped...

 

[Captain Wes]

Ok this is like the monty hall problem.
If atleast one is male, you have three possible situations
Beagle1 Beagle 2
Male Male
Male Female
Female Male

fixed: that says the same as the one above it so it's irrelevant

Ah, but there's a reason it was like that- it was the order.

I came back for three mins. after leaving to have fun so im not getting into this cirle logic again

 

[Fud]

I don't really think that there is a single right answer with the wording. It leaves the fact about if the person on the phone knows the gender of both beagles, or if they just grabbed one at random unknown.

 

[CaesarsSalad]

Male/Female and Female/Male is the same in this situation but it is twice as probable as Male/Male. So 1/3 is correct.

 

[Shivari]

I believe the correct answer is "Fuck Math".

 

[Imitation Saccharin]

Male/Female and Female/Male is the same in this situation but it is twice as probable as Male/Male. So 1/3 is correct.

1 dog is male.
1 dog is male or female.

There are 2 possible combinations as order is not specif iced as important.

Unless the position of each slot is named (i.e.- the 1st beagle is King Beagle and the 2nd beagle is Vice President Beagle) the positions of each gender are irrelevant.

 

[Jamanticus]

Ok this is like the monty hall problem.
If atleast one is male, you have three possible situations
Beagle1 Beagle 2
Male Male
Male Female
Female Male

fixed: that says the same as the one above it so it's irrelevant

Ah, but there's a reason it was like that- it was the order.

I came back for three mins. after leaving to have fun so im not getting into this cirle logic again

Oh- sorry.... I just looked at your post again and realized that I forgot to read the whole thing before posting.

And I still say that the answer to the math question is either 33% or 50%.....But I'm leaning toward 50% right now......

 

[cleverlymadeup]

Male/Female and Female/Male is the same in this situation but it is twice as probable as Male/Male. So 1/3 is correct.

it's a bit more simple than that, you are reading to much into it

1 is male
1 is unknown

the other one has a 50% chance of being male of female, the sex of the first one has no bearing on the sex of the other one. so all the male/male, male/female, female/male outcomes don't apply, even so both male/female and female/male are the same, meaning one does get removed and you are left with only 2 options male/male and male/female

most people are reading too much into a word logic problem

 

[Imitation Saccharin]

Slot 1/Slot 2
Male/Female -negative

Slot 1/Slot 2
Female/Male -positive

Slot1/Slot 2
Male/Male- positive

So assuming this, shouldn't the answer be 2/3 and not 1/3 anyways?

 

[dirtface]

dirtface is correct.

(male,female) and (female,male) are two different outcomes.

proof: (0,1) does not equal (1,0).


edit -- taxi driver is wrong in the above post. its no trick. you guys just don't understand how this translates into math. your dumb.

For those that know a bit of maths, I could be wrong but it reminds me of linear algebra, Group theory, ie. it's not commutative.