Poll: A little math problem

[NewClassic_v1legacy]

k, i'll stick around until i'm banned or whatever just to piss you dumb motherfuckers off.

now why the hell are you saying they mean everything in this problem. no disagreement there, but why are you bringing it up. nevermind i don't want to know you guys are retarded.

"this guy is right"
"NO UR WRONG HERES A LINK TO A SONG"
"no im right. here's a link explaining why"
"DONT CARE"
"w/e"
"GENDER DOESNT AFFECT THE PROBABILITY OMG UR WRONG"
"dude wtf im trying to go away and you keep talking to me"
"I WANT TO GET THE LAST WORD IN"
"fuck you"
"AGKHBILRHGLIEHW"
"that wasn't a word"
"FUCK YOU"
you guys are just fucking egging this shit on, trying to get a fucking point in or some shit.

I highlighted the important part of the rules for you, they are as follows:

Grammar and spelling. Everyone who moderates the boards is an editor. Bad spelling and improper grammar physically hurts us, precious. Respect yourself and others; take the time to check your spelling and how your post is worded. Also, one exclamation point or question mark is enough at the end of a sentence.

Flaming and trolling. Let's keep things civil, here. Responding to someone's post just to attack him is unacceptable.

Act like an adult, and I don't swing the banstick. We're intelligent, responsible people, all of us. We know how to treat one another, and we know how to disagree respectfully without things turning nasty. This is our lawn, we invite you to play here, but be polite and follow our rules or find somewhere else to hang.

Here's the deal. You disagree, we disagree, and we're done with it. That's how it could work, but instead, you've endeavored to berate us for posting just to get the last word in when you yourself are posting to get the last word in.

Here's the deal, you're making a bigger deal out of this than there really should be. How about we all drop it, and continue with the mathematical problem like this thread is really all about.

A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?

The problem doesn't provide enough information to assess the situation. If both beagles are from the same mother, the chance is 33%, if they are from different mothers, it is 50%, if the question is calling for whether or not they're male, female, or both male, then the percentage is 33%.

Which specifically, isn't identifiable, therefore I voted with the option that had the most possible answers, which is 33%.

 

[Saskwach]

But the probability of the set of both coin tosses follows a rigid structure. Knowing one coin's result changes the probability of the other coin's result.

But it's independent from the other one. It's a stupid trick.

Or this might be way over my head and I should just leave. I take it you learn these evil and useless tricks in later years.

They're not tricks: they're tests of whether you can determine from the information you're given specifically what kind of statistics problem you have on your hands. Real world stats problems won't kindly tell you what probability distribution you should use, for instance - poisson, normal, binomial, exponential, etc. Half of what you're taught in (good) probs and stats classes is what elements you should look for in any given problem to find the solution method before you find the answer. If, to return to probability distributions, you're told that you're looking at a continuous set of variables (as in the numbers are not discrete - there are no jumps in value, so, say, every whole number) you know that poisson and binomial distributions are immediately off the table.
What else are you told, though? Are we mapping population growth? Then it's exponential distribution you want. Now kindly start using the equations we've given you for exponential probability distributions.
My Year 12 Probs and Stats unit was practically made on these "trick" questions. You were given a written explanation of the problem, but not told what type of problem it was; if you couldn't figure that out then you didn't know the subject anyway, and didn't deserve the marks.

Edit: I see this post was already replied to. My bad.

Edit 2: You're all wrong; it's 33.333 recurring. So there.

 

[werepossum]

But the probability of the set of both coin tosses follows a rigid structure. Knowing one coin's result changes the probability of the other coin's result.

But it's independent from the other one. It's a stupid trick.

Or this might be way over my head and I should just leave. I take it you learn these evil and useless tricks in later years.

They're not tricks: they're tests of whether you can determine from the information you're given specifically what kind of statistics problem you have on your hands. Real world stats problems won't kindly tell you what probability distribution you should use - poisson, normal, binomial, exponential, etc. Half of what you're taught in (good) probs and stats classes is what elements you should look for in any given problem to find the solution. If, to return to probability distributions, you're told that you're looking at a continuous set of variables (as in the numbers are not discrete - there are no jumps in value, so, say, every whole number) you know that poisson and binomial distributions are immediately off the table.
What else are you told, though? Are we mapping growth in population? Then it's exponential distribution you want. Now kindly start using the equations we've given you for exponential probability distributions.
My Year 12 Probs and Stats unit was practically made on these "trick" questions. You were given a written explanation of the problem, but not told what type of problem it was; if you couldn't figure that out then you didn't know the subject anyway, and didn't deserve the marks.

Well said. What they don't teach you in school is that the equations are easy - framing the problem is the hard part.

NewClassic, the problem assumes a 50-50 distribution between sexes; without that assumption the problem is indeterminate. Therefore there is no difference between same and different mothers. The only reason the odds change is because the two are a set and you have gained additional information about that set.

EDIT: If it's not obvious, the two beagle pups are considered a set ONLY because they were selected together without knowing their sexes; even if they are from the same mother, the distribution of sexes within the set of two pups is considered random. The distribution of puppy sexes therefore follows a known distribution function, namely a 50% chance for either sex. That allows us to make more educated guesses as we gain additional information, a function which clearly approaches unity. When we know nothing of either pup, the second pup has a 50% chance of being male; we can guess no better. When we are holding both pups, each's chance of being male is still 50%, but because we have gained additional information (i.e. physical examination) we can determine each pup's sex with certainty. Between these two extremes is an area where, as we learn more about the two pups, we can make better guesses as to the sex of the second puppy.

 

[werepossum]

But the probability of the set of both coin tosses follows a rigid structure. Knowing one coin's result changes the probability of the other coin's result.

But it's independent from the other one. It's a stupid trick.

Or this might be way over my head and I should just leave. I take it you learn these evil and useless tricks in later years.

Each coin toss is independent, but the distribution of results within the set is a known function. If you knew an honest, unbiased coin was flipped 10 times and the first 9 times came up heads, you'd guess the 10th time would be tails even though as an honest, unbiased coin the chances of that toss were 50-50 just like all the other tosses. Grouping things in sets allows you to use probability functions to make better guesses about the others. The more members whose values you know, the better you can guess the values of the others.

It's not over your head, you're a smart girl. It only looks difficult and illogical because you haven't yet learned it. In any case, understanding the logic is even more important than understanding the actual set probability. Although probably the middle of the night isn't the best time to learn set probability or logic anyway.

As to when, I learned it in eighth grade from my advanced science teacher, but I think in math class it was covered in high school, probably freshman advanced math. I can't remember the year; I only remember the eighth grade teacher because he was my best primary school teacher ever, because he later left teaching and went into construction and I now work with him sometimes, and because he later married my second cousin.

 

[Saskwach]

I didn't want to jump on the "lol I no maths" bandwagon, but there are some who are still confused, so I'll give an in-depth answer - the one I used to explain this to myself.

Firstly, forget this business about "more than one male". What are the possible configurations before this?
Two males, two females, one of both.
The probability of either dog being a male or a female is 50% either way.
Therefore:
Bearing in mind that the total probability of something happening is 1,
Prob of two males = 0.5 x 0.5 = 0.25
Prob of two females = 0.5 x 0.5 = 0.25
Prob of one each = 1 - (sum of other possibilities) = 1 - 0.25 x 2 = 1 - 0.5 = 0.5
OR = 0.5 (using the reasoning that, no matter whether the first dog is male or female, we now just look at the probability of the other dog being the other sex. There are many ways to skin a cat.)
This part is simple, and I don't expect anyone was confused; it just had to be stated.
Now let's take our next piece of information: there is at least one male. What does this mean?
It means that one of our three different possibilities is no longer a possibility: we cannot have two females any more. So we strike that probability out:
Prob of two males = 0.25
Prob of one each = 0.5
Prob of two females = 0 (no longer possible)
BUT WAIT. The probability of getting our only two remaining possibilities is 0.75. What is this other 0.25 probability? That we have Schrodinger's Dog? This is the step that has thrown everyone. The probability of something happening must be 1 (unless "nothing happens" is given as an option in a particular problem, in which case that would also technically be 'something' - but this isn't happening here).
Clearly we have to go back to our probabilities.
Prob of two males (before two females was discounted) = 0.25
Prob of one each (before two females was discounted) = 0.5
Sum of both possibilities = 0.75
So we need to make this 0.75 become 1, as it is the new probability that "something happens".
Prob of two males = 0.25/0.73 = 0.333333333333333...
Prob of one of each = 0.5/0.75 = 0.66666666666666...
Now what was the question? Assuming that at least one dog was male, what is the probability that the other dog is male?
So we take one male from both of these probabilities and we are left with two cases: for one of each we are left with a female; for the two males we now have a second male. The second answer is the outcome we were asked to calculate and its probability, as we have shown, is 1/3.
*pumps fist in triumph. Now rests easy knowing he hasn't lost all his maths mojo*

Well said. What they don't teach you in school is that the equations are easy - framing the problem is the hard part.

Obscenely easy. I loved high school Probs and Stats more than any other maths subject because once you figured out what the problem was, you just had to put the numbers you were given into the equations you were taught. No proofs or complex integrations here, nosiree.

 

[werepossum]

I didn't want to jump on the "lol I no maths" bandwagon, but there are some who are still confused, so I'll give an in-depth answer - the one I used to explain this to myself.

Firstly, forget this business about "more than one male". What are the possible configurations before this?
Two males, two females, one of both.
The probability of either dog being a male or a female is 50% either way.
Therefore:
Bearing in mind that the total probability of something happening is 1,
Prob of two males = 0.5 x 0.5 = 0.25
Prob of two females = 0.5 x 0.5 = 0.25
Prob of one each = 1 - (sum of other possibilities) = 1 - 0.25 x 2 = 1 - 0.5 = 0.5
OR = 0.5 (using the reasoning that, no matter whether the first dog is male or female, we now just look at the probability of the other dog being the other sex. There are many ways to skin a cat.)
This part is simple, and I don't expect anyone was confused; it just had to be stated.
Now let's take our next piece of information: there is at least one male. What does this mean?
It means that one of our three different possibilities is no longer a possibility: we cannot have two females any more. So we strike that probability out:
Prob of two males = 0.25
Prob of one each = 0.5
Prob of two females = 0 (no longer possible)
BUT WAIT. The probability of getting our only two remaining possibilities is 0.75. What is this other 0.25 probability? That we have Schrodinger's Dog? This is the step that has thrown everyone. The probability of something happening must be 1 (unless "nothing happens" is given as an option in a particular problem, in which case that would also technically be 'something' - but this isn't happening here).
Clearly we have to go back to our probabilities.
Prob of two males (before two females was discounted) = 0.25
Prob of one each (before two females was discounted) = 0.5
Sum of both possibilities = 0.75
So we need to make this 0.75 become 1, as it is the new probability that "something happens".
Prob of two males = 0.25/0.73 = 0.333333333333333...
Prob of one of each = 0.5/0.75 = 0.66666666666666...
Now what was the question? Assuming that at least one dog was male, what is the probability that the other dog is male?
So we take one male from both of these probabilities and we are left with two cases: for one of each we are left with a female; for the two males we now have a second male. The second answer is the outcome we were asked to calculate and its probability, as we have shown, is 1/3.
*pumps fist in triumph. Now rests easy knowing he hasn't lost all his maths mojo*

Well said. What they don't teach you in school is that the equations are easy - framing the problem is the hard part.

Obscenely easy. I loved high school Probs and Stats more than any other maths subject because once you figured out what the problem was, you just had to put the numbers you were given into the equations you were taught. No proofs or complex integrations here, nosiree.

An alternative way of looking at the problem, but probably the best explanation so far.

 

[BlueCheese]

The answer is 50% or 33.33%, these are one of these questions where, if it appeared in a maths test, the answer would be hotly contested to scab that odd mark or two. As dirtface said, there would be three possible situations - 2 males, female and male and vice versa. Is there a difference between female & male and male & female? Apparently, some maths teachers think so.

Maths test: 1/3
Logically: 1/2

If you still don't get it, draw a probability tree!

 

[Saskwach]

An alternative way of looking at the problem, but probably the best explanation so far.

How probable, would you say?

 

[Cpt. Red]

Here is the probability tree of this event(aright means a male and a left means a female):
./\
/\/\
After finding out that one of them are male we need to remove the one only female
./\
/\/
Now there is three possibilities each of them have a 1/3 chance. Only one of them have two males therefore the possibility is 1/3 of the other being male...
EDIT: Dang, its hard to do some good asci "art"...

 

[werepossum]

An alternative way of looking at the problem, but probably the best explanation so far.

How probable, would you say?

Hah! Well, I'd say almost a certainty; you took the "50%" argument and showed exactly how it becomes 33%. I won't go more precise than that for fear of starting another round of arguing.

 

[The Wooster]

What do you mean? An African or European beagle?

 

[wilsonscrazybed]

I've pruned some of the meaner posts out of this thread and given SeymourB a week to cool off from math related rage. To the other users who were engaging him, please refrain from the "I'm going to tell the teacher." sort of comments, they just egg people on and do nothing to calm. Report things you think are harmful to the discussion and be done with it. If you feel like you need to talk to someone about an incident PM a moderator.

 

[Graustein]

Looks like a variation on the Monty Hall problem, so I'm gonna give a 1/3 chance.

EDIT: Awesome, I was right. Incidentally, I dropped maths in year 11

 

[Fire Daemon]

What do you mean? An African or European beagle?

I don't know that! Waaaaaaaaa!!!

Does anyone else have an interesting maths problem. This one has been solved many times.

 

[werepossum]

I've pruned some of the meaner posts out of this thread and given SeymourB a week to cool off from math related rage. To the other users who were engaging him, please refrain from the "I'm going to tell the teacher." sort of comments, they just egg people on and do nothing to calm. Report things you think are harmful to the discussion and be done with it. If you feel like you need to talk to someone about an incident PM a moderator.

Thank you, and thank you.

 

[Knight Templar]

The chance that both dogs are male is 25%. Flip two coins, whats the chance of both being heads or both being tales? 25% with a 50% that one will be heads one will be tails.

The same thing applies here. I think. I might have read the question wrong.

Yet when one is known only one X remains, therefore it is 50%

 

[Fire Daemon]

The chance that both dogs are male is 25%. Flip two coins, whats the chance of both being heads or both being tales? 25% with a 50% that one will be heads one will be tails.

The same thing applies here. I think. I might have read the question wrong.

Yet when one is known only one X remains, therefore it is 50%

Actually it's 33% (repeating of course). I'm not sure who said it first but someone sloved it and where both wrong. However I was the closest so I am more correct then you. Huzzah!

I still can't believe I read that question so wrong.

 

[mipegg]

Wrong, there is not a 50% chance of the other being female.

Look at it this way, on the first dog you have a 50/50 chance, from that you then have another 50/50 chance. This the chance of getting both males is the 2 probabilities times together, thus making it 25%.

Also, Temple, just because you know 1 it doesn't mean that the probability is reduced, there was still a 50% chance that dog could be female which must be taken into consideration.

EDIT 33%? Hmm, well, there are 3 a few possibilitiies of how to get that, if you count up all the possible pairings

Male Male
Female Male
Male Female
Female Female
Male Male
Female Female

Or, 33%.

 

[N.K]

I'm guessing 50% based on nothing but the fact that there were 2 dogs and one of them is male

I've never been one for math.

 

[Samirat]

It's 33 percent.

You have 4 possible situations:

Female Female
Female Male
Male Female
Male Male

The first, FF, can be ruled out, since one of them was a male. Any of the remaining three will have "at least one male." And since only one out of three of the remaining cases gives you two Males, the probability that the other is a male as well is 1/3.

The two cases FM and MF are not the same thing. They are two different situations, and occupy distinct positions in the sample space. The chances that, out of two dogs, you will get one female and one male are 50 percent, because there are two different cases that yield that result. If you don't believe me, toss two coins a bunch of times, and see how many times you get one head and one tails.