Blind Punk Riot: ...really now... please draw and cut out four dog shapes.
write male on two of them, write female on the other two.
Hold in your hand one called male. Thats the one you have been told is male.
look at the three you have remaining, you should have two females and one male. So the dog is either female, REALLY REALLY FEMALE, much like it is female already, and male.
thats ignoring the fact you should have removed the one saying female on it anyway. since you have a choice of male or female.
Is that a better way to explain it?
Do it at home yourself. but becareful with the scissors, and probably use a soft pencil like a 2b or something, you don't want to hurt yourself.
Probably best to have an adult help you too. Health and Safety is key!
Correction: Take 8 cardboard cutouts. Call 4 male, 4 female. Remove two females. Now work out probability of two males.
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.
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.
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.
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.
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.
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.
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.
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.
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)
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.
One dog is male. Pick the one to be male. It doesnt matter which
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.
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.
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%.
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?
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.
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.
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%.
orannis62: 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.
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.
Saphatorael: [quote=Saphatorael post=18.73797.810811] 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.
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.
orannis62: 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.
The same thing can be expanded to include coin tosses. If you toss two coins, you will have a distribution of possible results analogous to the pup gender question. The result of each coin toss is independent from the other. 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. As Dirtface said, it's the Monte Hall problem.
The thing though is that there was an extra factor in the Monte Hall problem--if behind three doors there were two booby prizes and one real prize, Monte Hall was never allowed to knock out the real prize. No one ever explains this! It drove me *nuts* for the longest time!
So here's how Montie Hall worked: when you selected, you had a 2/3 shot at picking the booby prize, and only 1/3 shot at picking the real prize. Now when Montie picks, that means that it's 2/3rds probable that his set contains the real prize. However, you know he can't pick the real prize, so that's why you increase your chances: you know Montie eliminated a booby prize from two options and the odds were you didn't pick--and therefore render unavailable to Montie--the real prize.
Is "Excluding the case of two girls, what is the probability that two random children are of different gender?" *REALLY* an equivalent way of saying "A random two-child family with at least one boy is chosen. What is the probability that it has a girl?" What about "If we know one child is a boy, what is the probability that the other child is not?" Is that also eqivalent?
Because if "If we know one child is a boy, what is the probability that the other child is not?" is equivalent, wouldn't the matrix for that be:
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).
Here's the thing, though: why don't we recalculate the odds when we get new knowledge?
We start out with a matrix like this:
Puppy 1/ Puppy 2 M M M F F M F F
Right? But then we get new information, that the puppy the bath-giving guy is referring to is male. So don't we now have to use this matrix:
Referred To Puppy/ Other Puppy M M M F
(EDIT) in other words, isn't saying 33% making the mistake of only eliminating the possibility of F/F, when really, the new knowledge also eliminates the possibility of F/M(end EDIT) so that we don't have a .25 to redistribute to get back to 1, but rather a .50 to redistribute? Remember, the question doesn't ask us to calculate the odds at any point prior to finding out that one of the puppies is male. And the guy giving the bath isn't under the same constraints as Montie Hall--Montie Hall couldn't pick the real prize if it was there, while this guy has to 'pick' puppies to look at until he gets to a male one before making his statement.
Isn't the real probability we need to estimate that of whether the Bath Giving Man is referring to the first puppy he looked at which turned out to be male and therefore put an end to his search, or whether the first puppy he looked at was female and he had to look at the other puppy before making his statement: "Yes!" and is therefore referring to the second puppy he looked at?
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.
You're right that the situation is exactly the same if you use coins. And if you flip a coin, and it's heads, then the probability that the other coin will be heads is 50%. But: If someone flips two coins and keeps the result hidden from you, and you ask him if there's a 'heads', and he says yes, then the probability that they're both heads is one-third.
25% ok heres how sex chromosones are xx & xy therefore through the use of a punnet square you can resolve this X X therefore i could randomly select another male at a 50%*50% chance in other words 25% X|XX XX Y|XY XY
All right, let's try this. If you have two dogs, the chances that at least 1 will be male are 75 percent, which is the probability of both being male added to the probability of only one being a male. The probability that both will be a male is 25 percent, while the probability that only 1 will be a male is 50 percent.
Since the probability of only 1 being a male (the male the dog washer saw) is twice that of both being males, the odds are, respectively, 66 percent and 33 percent.
crepesack : 25% ok heres how sex chromosones are xx & xy therefore through the use of a punnet square you can resolve this X X therefore i could randomly select another male at a 50%*50% chance in other words 25% X|XX XX Y|XY XY
This works, except that it doesn't eliminate the situation which is eliminated in the problem. There can't be two females. This looks at the complete set:MF, FM, MM, FF. But there can't be two females. Therefore, there are only three outcomes, and only in 1 is the other dog a male.
Samirat: All right, let's try this. If you have two dogs, the chances that at least 1 will be male are 75 percent, which is the probability of both being male added to the probability of only one being a male. The probability that both will be a male is 25 percent, while the probability that only 1 will be a male is 50 percent.
Since the probability of only 1 being a male (the male the dog washer saw) is twice that of both being males, the odds are, respectively, 66 percent and 33 percent.
The problem is though that we get new information--we get the Bath Giving Man searching among the puppies for a male. The 75 percent you're talking about is not only M/M and M/F, but also F/M. Once the Bath Giving Man goes searching and finds a male, we can eliminate not only F/F but also F/M because the questions changes from "Two Unknown Puppies" to "One Known Puppy and One Unknown Puppy."
There's a difference between 'what's the probability of at least one male among a set of two unknown puppies' and 'what's the probability of at least one male among a set of two puppies, one known to be male'. The former is 75%, but the latter is 100%.
EDIT: I think the crux of this is that you're talking about two *unknown* dogs while the problem is talking about one known dog and one unknown dog. There's where the issue is.
Cheeze_Pavilion: Stuff I'd love to quote but feel would add too much length to this post.
I'd already put some thought into that and had a hefty piece of doubt, but I think I have the answer. So we're talking, really, about the ordering of these dogs. When we say that M/F is a different outcome to F/M we're saying that one dog will be checked first and the other second: this is a permutative question, not a combinative one. So let's say we painted those numbers on the dogs' sides so no one could miss it. This man is asked, is dog 1 and/or dog 2 a male, not, is dog 1 a male? This is crucial. If the question were "Is dog 1 a male?" then you would be right; we'd have to drop F/M as a possibility and the answer would become 50%. But these dogs are being ordered, or to be more correct - and a bit more whimsical - one is an individual dog named Jesse and the other is also an individual dog named Other Sexually Ambiguous Name - so if we called out Jesse, OSAN wouldn't come running. Instead, the man (or woman?) was asked: "Is Jesse and/or OSAN a male?", meaning that, conceivably, either M/F or F/M are still on the table because if Jesse (Dog 1) is male then M/F or M/M would be the result and we'd still have to check OSAN's nether regions, but if OSAN (Dog 2) were male then F/M or M/M would be the result and we'd be looking between Jesse's legs. This man has said, "Maybe Jesse's a boy, maybe OSAN's a boy, but I ain't saying which groin I checked." To use coin tosses as the example, a man has flipped two coins. You asked him whether one or both came up tails and he said "Yes." What he didn't say was that his first toss came up tails, which is the same mistake people made in the dog example.
Edit: I don't think I summed up too well so I'll make clear wherein lies the rub. If the question were: "Is Jesse/Dog 1 (ie, a specific dog) a male?" the probability for the next dog being male would be 50%, because OSAN's machismo is totally independent of Jesse's masculinity. But the question is: "Is Jesse/Dog 1 and/or OSAN/Dog 2 (ie, at least one of the two) a male?" which leads to 33%, because OSAN and Jesse and their genitalia haven't been made independent of each other (ok, I've gone too far this time. I'll stop now.).
The problem is though that we get new information--we get the Bath Giving Man searching among the puppies for a male. The 75 percent you're talking about is not only M/M and M/F, but also F/M. Once the Bath Giving Man goes searching and finds a male, we can eliminate not only F/F but also F/M because the questions changes from "Two Unknown Puppies" to "One Known Puppy and One Unknown Puppy."
There's a difference between 'what's the probability of at least one male among a set of two unknown puppies' and 'what's the probability of at least one male among a set of two puppies, one known to be male'. The former is 75%, but the latter is 100%.
Okay, here's the problem. We're all starting with:
Dog A/Dog B M,M F,M M,F F,F each outcome is equally likely.
but when the information that one of the dogs is male is added, the 33%-ers are moving to: Dog A/Dog B M,M F,M M,F each outcome is equally likely.
and the 50%-ers are moving to: known dog/unknown dog M,F M,M
But by changing the heading, the two options are no longer equally likely.
If you toss two coins a hundred times, you'll a pair of heads about 25 times, a pair of tails about 25 times, and one of each about 50 times. But if you disregard any pairs of tails, you're left with 25 pairs of heads, and 50 doubles.
To use coin tosses as the example, a man has flipped two coins. You asked him whether one or both came up tails and he said "Yes." What he didn't say was that his first toss came up tails, which is the same mistake people made in the dog example.
No, we didn't ask him if either of two things are true, those two things being 'at least one is tails' OR 'both are tails'. We only asked "Is at least one a male?" not "Is at least one a male OR are both male."
Your correct that he's not being asked "is dog 1 a male?" but it's also true he's not exactly being asked "is dog 1 and/or dog 2 a male"; he's being asked "start searching among the puppies and say 'yes!' as soon as you encounter a male."
Think of it this way--the probability of at least one dog being male was 75% before we called up Puppy Bathing Man, right? M/M, M/F, F/M, but not F/F.
However, that 75% was based on the probability of *either* dogs being male being 50%. Once we know that one dog is male, one of the events becomes 100% probable.
That's what I think the issue is--saying 33% is not taking into account the fact that once an event actually occurs, the probability of it occurring goes up to 100%. It's like the lottery--the chances of winning a future lottery may be .8353498574357%, but if you *actually do win the lottery* your chances of winning go up to 100%.
And in this case, Puppy Bathing Man won the Male Puppy Finding 'Lottery'--once an event occurs, however improbable it may have been estimated to be before it happened, the probability of it happening is 100%
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 question wasn't, "Tell me when you find a male," but rather, "Is at least one a male?" I think what you're saying is that any normal person would say "Yes," as soon as she found a pecker (yes, apparently it's a Peeping Tina), but this isn't a normal person - this is a literal person (we're assuming). Edit: Actually, even then it'd be 33%. So Tina checks both these dogs and if Jesse is a male she says yes. If OSAN is a male she says yes. If Jesse and OSAN are males she says yes. There are three possibilities here, all equally likely, so the possibility of both being male is 1/3.
Cheeze_Pavilion: No, we didn't ask him if either of two things are true, those two things being 'at least one is tails' OR 'both are tails'. We only asked "Is at least one a male?" not "Is at least one a male OR are both male."
Those last two questions, when added together, amount to the same question as just the first.
Cheeze_Pavilion: That's what I think the issue is--saying 33% is not taking into account the fact that once an event actually occurs, the probability of it occurring goes up to 100%. It's like the lottery--the chances of winning a future lottery may be .8353498574357%, but if you *actually do win the lottery* your chances of winning go up to 100%.
I think that's exactly the issue the 33% percent crowd is taking into account - but I can't understand your logic here, so feel free to explain it again.
but when the information that one of the dogs is male is added, the 33%-ers are moving to: Dog A/Dog B M,M F,M M,F each outcome is equally likely.
and the 50%-ers are moving to: known dog/unknown dog M,F M,M
But by changing the heading, the two options are no longer equally likely.
Isn't that the weird thing here? That by phrasing the question differently, we move to different matrices. So the real question is who has the better phrasing.
If you toss two coins a hundred times, you'll a pair of heads about 25 times, a pair of tails about 25 times, and one of each about 50 times. But if you disregard any pairs of tails, you're left with 25 pairs of heads, and 50 doubles.
True, but in this case, we're no longer flipping two coins. We're only flipping one--the Unknown Dog. The Known Dog is like a coin left on the table facing heads up while we flip the other one. We're not reducing the possibility of two pairs of tails to Zero by 'disregarding any pairs of tails'; we're reducing the possibility by putting one coin heads up and only flipping the other. Which means we're doing more than just 'disregarding any pairs of tails' we're actually making it physically impossible for there to be a pair of tails. The thing is, the act that makes it physically impossible for there to be a pair of tales ALSO makes it physically impossible for another of the possibilities, the one that the 33%ers are talking about.
In other words, when you eliminate Tails/Tails by not flipping one of the coins, you have to either eliminate Heads/Tails OR (and this is an exclusive or) Tails/Heads, otherwise your model does not accurately capture the reality it is describing.
Ahh... now I understand. There are two questions. 1.If the person sexes one of the dogs, finds it to be male, and then puts it down, what is the probability, that if the this person then picks up a dog at random, the dog is male? The answer being 2/3. 2.If the person sexes one of the dogs, finds it to be male, and then keeps hold of it, what is the probability that the the other dog is male? The answer in this case being 1/2. Edited, as 1. =2/3, not 1/3.
Correction: Take 8 cardboard cutouts. Call 4 male, 4 female. Remove two females. Now work out probability of two males.