Poll: A little math problem

[Captain Wes]

dirtface is correct.

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

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

we aren't working with a graph so they are the same ( http://www.youtube.com/watch?v=GlKL_EpnSp8 ) my link's better too

 

[Jumplion]

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. heres a site to help you not be so dumb. [http://www.mathgoodies.com/lessons/vol6/intro_probability.html]

But, if one is male than why would it matter if the first one is female?

Male, female would mean that one is male, but the other one isn't
Female, male is the exact same thing only with them switched over, one is still male but the other one isn't.

It's asking what's the probability of the other one being male, we know one is male but what about the other one? It doesn't matter which one's first...does it?

Blahch, this is getting annoying, give us the answer Fud!

 

[geizr]

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 is the correct answer, 33%, because there are only 3 outcomes with at least one beagle being male. Of those three, only one outcome has both being male, which is the outcome being asked for in the problem.

 

[Captain Wes]

dont fuck with me and make me go all mathematical on your ignorant ass.

http://www.mathgoodies.com/lessons/vol6/intro_probability.html

i can link to shit too. coordinates aren't just for graphs, mister never went past calculus. fuck, what you think is a graph isn't even a graph. i can fuck with your head and shit and show you what a graph really fucking is man.

but the first dog is irrelevant, why can't you see that if you're so smart. And by the way if you bothered to read your link you'd see that it supports my point not yours

 

[dirtface]

...

The wiki article states it's 1/3, we're just looking for a reason why.

 

[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.

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

No, 33% is correct because the two beagles are part of a set. If you randomly select two pups, each has a 50% chance of being male. Therefore the chances for the set are 25% male and female, 25% male and male, 25% female and male, and 25% female and female. If you examine one pup and confirm that it is male, then you eliminate the 25% chance for the set that both are female. Now your chances (rounded to integers) are 33% male and female, 33% male and male, and 33% female and male. Technically speaking you're not calculating the chance the second pup is male, you are calculating the chance that two males were randomly selected in this set of two. The odds of that were originally 25%, but after eliminating the possibility that two females were selected the odds change to 33%.

If you selected one pup from an infinite, evenly divided pool of beagle pups (note: this is the definition of rabbit hell) and determined that it was a male, the chance that the next pup selected would also be male is 50% because the first pup's sex cannot affect the second pup's sex. But if you select them together, then the odds are as above. Dirtface explained it correctly, he or she was just a bit uncertain as to why.

 

[Captain Wes]

fuck, what you think is a graph isn't even a graph. i can fuck with your head and shit and show you what a graph really fucking is man.

you are acting like it's a graph mr. probably is still in highschool. don't put shit in you dont understand.

 

[Limos]

Your math problem is illogical. Here is a much better one.

The speed of a projectile over time is represented by
y = t^2 +3t

find the rate of change.

Spoiler: The Answer

2t + 3

 

[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.

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?

No, 33% is correct because the two beagles are part of a set. If you randomly select two pups, each has a 50% chance of being male. Therefore the chances for the set are 25% male and female, 25% male and male, 25% female and male, and 25% female and female. If you examine one pup and confirm that it is male, then you eliminate the 25% chance for the set that both are female. Now your chances (rounded to integers) are 33% male and female, 33% male and male, and 33% female and male. Technically speaking you're not calculating the chance the second pup is male, you are calculating the chance that two males were randomly selected in this set of two. The odds of that were originally 25%, but after eliminating the possibility that two females were selected the odds change to 33%.

If you selected one pup from an infinite, evenly divided pool of beagle pups (note: this is the definition of rabbit hell) and determined that it was a male, the chance that the next pup selected would also be male is 50% because the first pup's sex cannot affect the second pup's sex. But if you select them together, then the odds are as above. Dirtface explained it correctly, he or she was just a bit uncertain as to why.

dirtface (HE) knows why, he's just not able to communicate with the slightly less mathematically inclined as well as you ^^ rofl.

 

[LV Solace]

The gender of Beagle A which we know to be male, has no effect whatso ever on the gender of beagle B. Unless there is some unknown about the birthing patterns of beagles, but we'll leave that in the closet until we can agree that there is a 50/50 chance that beagle B is male.

Now if you flip a coin and it turns up heads. The second flip of that coin, in no way depends on the first flip. The third wont depend on the second, and the fourth wont depend on the third. That is what chance is. Things that are completely independent of one another.

The only way any one could get anything other than 50%, is if you belive that beagle B depends on beagle A. which it does not. Thats just simple truth.

 

[Fire Daemon]

Werepossum is right. Damn, I couldn't believe I didn't see that at first! I feel so stupid now.

I suppose trying to play games and solve maths isn't a winning combination.

 

[Shivari]

Maths

Meh, I've never been all that great at math, my elementary school really sucked, so I went into 7th grade not knowing stuff that should have been review. I've been playing catchup ever since, and I've managed to start getting As consistently since freshman year (now a sophomore). But no doubt that if the Catholic school I went to through 6th grade had taught me better I'd be in Algebra 2 instead of Geometry this year.

 

[Captain Wes]

NOT A {HH,HT,TT}
NOT A [HH,TH,TT}
BUT A {HH,HT,TH,TT} NIGGA
and HH and et cetera is a same as (H,H) where the first part is the first coin and the second is the second coin.

that's asuming that the first dogs sex has any bearing on the seconds but it does not
one dog is irrelevant to the other aside from that they are in the proximity to each other so the options look like this
dog 1 (male)
dog 2 (male /female)

you say the first dog is important but he isn't, in terms of probablity he doesn't exist

 

[Jumplion]

By the way...

Spoiler: Wrong

The answer is what you say it is, but it?s still a trick, as I explained before.

I'm never going to get tired of watching that.

 

[Jumplion]

By the way...

Spoiler: Wrong

The answer is what you say it is, but it?s still a trick, as I explained before.

I'm never going to get tired of watching that.

I have also found that it's a hard hit to the ego and...

Wait a minute...where's my Taxi Driver stick avatar I put in for?!?!?

Shit, I've been compromised!

*jumps out the window*

 

[werepossum]

I believe the correct answer is "Fuck Math".

In Russia, math fucks you!

Actually this sort of math is extremely important in business, even more so in science. Failing to understand such basic statistical functions can lead to poor investments or failed designs. Suppose an available investment is a company with two copper mines, both of which must strike copper to make a profit. From the scientific analysis of similar mines, each mine has a 50% chance of striking a workable copper vein. One of the mines already has struck copper. What's the chance the other will also strike copper? If the investment is $1,000,000 but the possible profit is $2,750,000, is it a mathematically sound investment or not? If you know the chance of a fuel rod failing from a defect is 50% and one of your two fuel rods has failed, what are the odds your new death star is going to suffer complete power failure?

Math is much more likely to fuck you than the reverse.

The trick to the question is that it populates the set with things - namely, a pup and its sex - whose probability spread you intrinsically understand. If the set were sixty randomly selected molecules each with a 0.003% chance of being contaminated, you would probably see the problem as a set problem rather than as simply guessing the sex of a second pup.

Imitation Saccharin, the question was the probability of the second pup also being male, not the probability of either pup being male.

 

[Jamanticus]

you guys are dumb. im gonna go back to my probability homework now. afk

hooray! my first scarring of the newbs. How'd I do?

I say you did very well..... But now I can't scare anyone off- darn you, (ZHU)!

So the solution is 33%?

EDIT: Alright, maybe I can scare someone off....

 

[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.

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

It's not that simple. You don't know that dog one is male and the other is unknown. You know that at least 1 or two is male. That really makes a difference. Suppose the gender of each dog is determined by a coin toss. We have 4 outcomes.

Dog 1
male
female

Dog 2
male
female

or

dog 1 dog 2
male male
male female
female male
female female

Every possibility is equal

Female/female is out. That leaves us with male/male and the possibility that only one of them is male. The second option is twice as probable because the male dog can be either one. If you don't believe it, just try it with some coins and keep the record.

lrn2math (just kidding, many people have problems with stochastic)

English is not my first language so there might be some grammarmistakes or something...

 

[Jamanticus]

edit -- but idgaf i just found this login on a website so i could educate you fags.

and this kid is a 3rd year math major, so fuck you

Are you actively seeking to get yourself banned, SeymourB?

Reported, again.

EDIT: my evilest post yet, at 666

Now, is the solution 33%?

 

[Jumplion]

I believe the correct answer is "Fuck Math".

In Russia, math fucks you!

Actually this sort of math is extremely important in business, even more so in science. Failing to understand such basic statistical functions can lead to poor investments or failed designs. Suppose an available investment is a company with two copper mines, both of which must strike copper to make a profit. From the scientific analysis of similar mines, each mine has a 50% chance of striking a workable copper vein. One of the mines already has struck copper. What's the chance the other will also strike copper? If the investment is $1,000,000 but the possible profit is $2,750,000, is it a mathematically sound investment or not? If you know the chance of a fuel rod failing from a defect is 50% and one of your two fuel rods has failed, what are the odds your new death star is going to suffer complete power failure?

Math is much more likely to fuck you than the reverse.

The trick to the question is that it populates the set with things - namely, a pup and its sex - whose probability spread you intrinsically understand. If the set were sixty randomly selected molecules each with a 0.003% chance of being contaminated, you would probably see the problem as a set problem rather than as simply guessing the sex of a second pup.

Imitation Saccharin, the question was the probability of the second pup also being male, not the probability of either pup being male.

OH, OH, OH! Okay, now I understand, I remember seeing a problem or two like that (most notably in the movie "21"), now I get why it's 33%.

But the one thing that's nagging me is that, lets say, if the other copper mine struck, erm, copper than the other one still has 50% chance without putting the one that struck copper into effect....right? But unless that fuel rod was in direct explosive radius of the other fuel rod, the one that was defective still wouldn't have any real effect on the other one.....

Okay, maybe I don't get it yet.

And Seymor, it's just a math problem calm down, jeez. so get back to your probability homework young man!