Poll: A little math problem

[Blind Punk Riot]

ok ok ok.
Looking at it another way


Four dogs.
1/4 chance for each.
One of them Is male.
1/3 chance that the other dog is male.


One hundred dogs.
1/100 chance for each.
99 of them are male.
1/99 chance that the other dog is male.

X dogs.
1/x chance for each.
X-1 of them are male.
1/x-1 chance that the other dog is male.


Mathematicians about you will realise that that is wrong.

 

[phenaxkian]

Binomial distribution. we have 2 "trials" we'll say 50% chance of being male, and that the sex of each one is independeant (e.g. the sex of one doesn't affect the sex of the other).

X~B(2, .5) X=Male.
We want P(X>=1).
That's the same as 1 - P(X<=0).
If you look up P(X<=0)in a table of binomial distribution statstics, you get the P(X<=0)=0.25.
so 1-0.7 5= .25 = 25%.

 

[Blind Punk Riot]

I understand your point, you're just wrong. there is no other way to explain it.

One has to be male...
One is male...

I just don't get why you're saying that since one is male, the other is female. Why are you repeating yourself with that statement. I know you'll say "because it could be either dog"
unless the dog is changing sex during your counting, you are wrong. Theres no other way to put it.

Check the wikipedia link I put above (here it is again http://en.wikipedia.org/wiki/Boy_or_Girl_paradox ), it gives the correct explanation better than I can.

That doesn't say that one of them is a boy. This one does.

 

[Lukeje]

I understand your point, you're just wrong. there is no other way to explain it.

One has to be male...
One is male...

I just don't get why you're saying that since one is male, the other is female. Why are you repeating yourself with that statement. I know you'll say "because it could be either dog"
unless the dog is changing sex during your counting, you are wrong. Theres no other way to put it.

Check the wikipedia link I put above (here it is again http://en.wikipedia.org/wiki/Boy_or_Girl_paradox ), it gives the correct explanation better than I can.

That doesn't say that one of them is a boy. This one does.

Question 2 does... A random two-child family with at least one boy is chosen. What is the probability that it has a girl? (2/3)

 

[irrelevantnugget]

I understand your point, you're just wrong. there is no other way to explain it.

One has to be male...
One is male...

I just don't get why you're saying that since one is male, the other is female. Why are you repeating yourself with that statement. I know you'll say "because it could be either dog"
unless the dog is changing sex during your counting, you are wrong. Theres no other way to put it.

Because you're leaving an option out.

The options to you are:
Male - Male
Male - Female

The option you are forgetting is Female - Male. It never disappeared, and should not be left out.
Also, no sex changes. It never needed to be changed. The third option has Dog B tagged as the male, changing the question to: Is Dog A male or female? Instead of the automated "Is Dog B male or female"

Again, you are presuming Dog A is the male one. From that point of reasoning, Dog B has a 50% chance of being male, yes. But it is never stated that Dog A IS the male one. Many people just presume that one dog is male and label it Dog A automatically, solely calculating the odds that Dog B is male or not.

But the shopkeeper says: Yes, there is a male dog. The male dog can be either A or B, and starting from this point, there is a possibility of 1/3.

 

[Blind Punk Riot]

I understand your point, you're just wrong. there is no other way to explain it.

One has to be male...
One is male...

I just don't get why you're saying that since one is male, the other is female. Why are you repeating yourself with that statement. I know you'll say "because it could be either dog"
unless the dog is changing sex during your counting, you are wrong. Theres no other way to put it.

Check the wikipedia link I put above (here it is again http://en.wikipedia.org/wiki/Boy_or_Girl_paradox ), it gives the correct explanation better than I can.

That doesn't say that one of them is a boy. This one does.

Question 2 does... A random two-child family with at least one boy is chosen. What is the probability that it has a girl? (2/3)

Sorry i meant, they are randomly choosing one of them, rather than suggesting the odds of the other one. Accepting that one is male.

 

[Blind Punk Riot]

I understand your point, you're just wrong. there is no other way to explain it.

One has to be male...
One is male...

I just don't get why you're saying that since one is male, the other is female. Why are you repeating yourself with that statement. I know you'll say "because it could be either dog"
unless the dog is changing sex during your counting, you are wrong. Theres no other way to put it.

Because you're leaving an option out.

The options to you are:
Male - Male
Male - Female

The option you are forgetting is Female - Male. It never disappeared, and should not be left out.
Also, no sex changes. It never needed to be changed. The third option has Dog B tagged as the male, changing the question to: Is Dog A male or female? Instead of the automated "Is Dog B male or female"

Again, you are presuming Dog A is the male one. From that point of reasoning, Dog B has a 50% chance of being male, yes. But it is never stated that Dog A IS the male one. Many people just presume that one dog is male and label it Dog A automatically, solely calculating the odds that Dog B is male or not.

But the shopkeeper says: Yes, there is a male dog. The male dog can be either A or B, and starting from this point, there is a possibility of 1/3.

It should be removed.

One dog is male. Pick the one to be male. It doesnt matter which, but you have to stay continuous throughout.


Here is my explaination of why you are wrong;

Four dogs.
1/4 chance for each.
One of them Is male.
1/3 chance that the other dog is male.

One hundred dogs.
1/100 chance for each.
99 of them are male.
1/99 chance that the other dog is male.

X dogs.
1/x chance for each.
X-1 of them are male.
1/x-1 chance that the other dog is male.


That is what you are saying. That is why you are wrong.

 

[Lukeje]

I understand your point, you're just wrong. there is no other way to explain it.

One has to be male...
One is male...

I just don't get why you're saying that since one is male, the other is female. Why are you repeating yourself with that statement. I know you'll say "because it could be either dog"
unless the dog is changing sex during your counting, you are wrong. Theres no other way to put it.

Check the wikipedia link I put above (here it is again http://en.wikipedia.org/wiki/Boy_or_Girl_paradox ), it gives the correct explanation better than I can.

That doesn't say that one of them is a boy. This one does.

Question 2 does... A random two-child family with at least one boy is chosen. What is the probability that it has a girl? (2/3)

Sorry i meant, they are randomly choosing one of them, rather than suggesting the odds of the other one. Accepting that one is male.

But since we can't distinguish them (they're in a 'quantum state of superposition') it's exactly the same. As soon as we started giving them identifying marks, we lose this superposition, and the wavestate comes crashing down into two possibilities.
Edit: we can label them as long as we show each individual microstate, as above posters have done, and give them weighted probabilities, i.e. only one dog male W=0.5 (two separate
yet equivalent microstates)
both dogs male W=0.25(only one microstate)

 

[irrelevantnugget]

I understand your point, you're just wrong. there is no other way to explain it.

One has to be male...
One is male...

I just don't get why you're saying that since one is male, the other is female. Why are you repeating yourself with that statement. I know you'll say "because it could be either dog"
unless the dog is changing sex during your counting, you are wrong. Theres no other way to put it.

Because you're leaving an option out.

The options to you are:
Male - Male
Male - Female

The option you are forgetting is Female - Male. It never disappeared, and should not be left out.
Also, no sex changes. It never needed to be changed. The third option has Dog B tagged as the male, changing the question to: Is Dog A male or female? Instead of the automated "Is Dog B male or female"

Again, you are presuming Dog A is the male one. From that point of reasoning, Dog B has a 50% chance of being male, yes. But it is never stated that Dog A IS the male one. Many people just presume that one dog is male and label it Dog A automatically, solely calculating the odds that Dog B is male or not.

But the shopkeeper says: Yes, there is a male dog. The male dog can be either A or B, and starting from this point, there is a possibility of 1/3.

One dog is male. Pick the one to be male. It doesnt matter which

It does.

 

[Ancalagon]

X dogs.
1/x chance for each.
X-1 of them are male.
1/x-1 chance that the other dog is male.


That is what you are saying. That is why you are wrong.

If I understand what you think we're saying, then in the situation we're discussing, with two dogs, we'd by saying that the chance the other dog is male is 1/2-1 or 1.

 

[chomesuke]

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.

He's right. the chance is 25 per-cent. It's called a treediagram.

Each choice has got 2 outcomes, the second time the choice occurs there are still 2 possibilities(but now 2 choices to be made), so you get 4 outcomes. And since the chances are the same you get this

MM:25%
MF:25%
FM:25%
FF:25%

So if the first is male, the cance that the second is male will be 25%.

...right?

 

[Blind Punk Riot]

X dogs.
1/x chance for each.
X-1 of them are male.
1/x-1 chance that the other dog is male.


That is what you are saying. That is why you are wrong.

If I understand what you think we're saying, then in the situation we're discussing, with two dogs, we'd by saying that the chance the other dog is male is 1/2-1 or 1.

I meant x+1

and then the other example is

99 dogs
98 are male
what are the odds that they are all male?
1/100?

 

[Jack Spencer Jr]

So, we haven't figured out if the second dog is male or not, but we have found out why mathematicians have such a hard time getting laid. I read the Wikipedia article on this and it's just shows how divorced from reality this shit is. How is it at all possible for a woman to have a 1 out of 3 chance to have two sons while a man have a 1 out of two chance when both a man and a woman are necessary to conceive in the first place? It's numbers and logic while having nothing at all to do with common sense. Death to the eggheads, I say.

 

[Lukeje]

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.

He's right. the chance is 25 per-cent. It's called a treediagram.

Each choice has got 2 outcomes, the second time the choice occurs there are still 2 possibilities(but now 2 choices to be made), so you get 4 outcomes. And since the chances are the same you get this

MM:25%
MF:25%
FM:25%
FF:25%

So if the first is male, the cance that the second is male will be 25%.

...right?

No, because the probability of FF is zero.

 

[Samirat]

Guys, it would be 33.3(repeating)% if the question was asking for a set, but it's not. It's asking for the other puppy, not how they are together. As such, the first puppy might as well not even be there, as it has no bearing on the gender of the second. Therefore, it's 50%.

No, because it doesn't say the "first" puppy is male. It could be either one. So this means that instead of just having two choices: Male and Female, and Male and Male, you have three: Male and Female, Male and Male, and Female and Male. In only one of these is the other dog male. Hence, 1 out of 3.

In hindsight, I don't know why I'm bothering to write this, since those that say 50 percent aren't even reading the explanations.

 

[Ancalagon]

and then the other example is

99 dogs
98 are male
what are the odds that they are all male?
1/100?

If you've taken each dog out one by one, and they've just happened to be male, then the probability that the final one is male is 50%.

If you've gone looking for males, and found 98, the probability that the last one is:

1/2^100/((1/2^100)+(100/2^100)) = 1/101

since the probability of having 100 males is 1/2^100, and having the probability of having 99 males and one female is 100/2^100.

 

[Samirat]

X dogs.
1/x chance for each.
X-1 of them are male.
1/x-1 chance that the other dog is male.


That is what you are saying. That is why you are wrong.

If I understand what you think we're saying, then in the situation we're discussing, with two dogs, we'd by saying that the chance the other dog is male is 1/2-1 or 1.

I meant x+1

and then the other example is

99 dogs
98 are male
what are the odds that they are all male?
1/100?

Yes, something like that. There is only one case where they are all male. But a female dog could be in the first position, the second position, the third position, and so on. Not just in the last position. Think how much more likely it is to have 98 male dogs and 1 female. I'm not going to list all the cases, but suffice to say there is only 1 situation where all the dogs are male, while there are some 99 situations where you have 1 female in amongst the group.

You look kind of foolish trying to put down everyone for being wrong when it's you yourself who haven't mastered the concepts of probability.

Consider it in terms of coins. If you flip 2 coins, the chances that you will have 1 tails and 1 heads are 50 percent, right? If you don't believe me, flip the damn coins. Now, if we're assured that at least one of them landed on heads, that removes only one option, tails tails. There are still double the chances of having one tail and one heads instead of having two heads. Only 1 3rd of the time, will you have two heads.

I really don't know how to explain it any clearer, and I suspect others have done a better job than myself, so I suspect you're not going to see the solution. But oh well.

 

[Samirat]

That is where you went wrong in your reasoning.

Ok without wanting to quote your entire thing.

if you have decided Dog A is male, why do you still have a Dog A is female in your remaining odds?



I suggest, you read that carefully, rethink your choice.
I have read what you are saying, and it is unfortunately wrong. I don't get a kick out of correcting people, I just get upset when people don't understand things. And I have really tried to help you out here and explain to you how it is.

But if you don't want to rethink your own statement then thats fine.
Be wrong. Just please go through the humiliation of finding someone you know who is incredibly good at maths and probability for them to tell you you are wrong. Since you can't really trust me, some random person of the internet, that you are wrong.


I feel a throbbing pain in the front of my brain.

Does it say Dog A is male? No, it doesn't. It says at least one dog is male. So the case where dog A is female and dog B is male is still valid. The question doesn't specify that the "first dog" is male, just that one of the two is. I'm not sure why you'd think it did. Perhaps you misread.

 

[thedoclc]

Saph, you've either run into a troll or someone too smug to listen to any ideas outside his/her own ungrounded and undefended certainty that their intuitions are right. Might as well just let it go.

 

[Avatar Roku]

Guys, it would be 33.3(repeating)% if the question was asking for a set, but it's not. It's asking for the other puppy, not how they are together. As such, the first puppy might as well not even be there, as it has no bearing on the gender of the second. Therefore, it's 50%.

No, because it doesn't say the "first" puppy is male. It could be either one. So this means that instead of just having two choices: Male and Female, and Male and Male, you have three: Male and Female, Male and Male, and Female and Male. In only one of these is the other dog male. Hence, 1 out of 3.

In hindsight, I don't know why I'm bothering to write this, since those that say 50 percent aren't even reading the explanations.

No, we are reading your explanations. Mathematically, you're right. Logically, you're not. To use the tired analogy of a coin toss, if you flip a coin twice, the the first flip has no bearing on the second. That is this exact situation, seriously, you could have replaced "Male puppy" with "Heads coin" in the original problem. The main discrepancy is cause by the fact that the problem is worded vaguely enough that the answer could be either 50% 0r 33%.

EDIT: To address the other thing you brought up, it doesn't matter whether the known puppy is first or second, we just assign it the title "first" to make it quicker to refer to one or the other.