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Imagine two dogs sitting in front of you. you do not know their genders what are the possible combinations of gender?
Left dog being a girl and right dog being a boy is not the same as right dog being a girl and left being a boy.
thats not what we're faced with here, accept the puzzle for what it is, we dont have to imagine anything, we're told one dog is a male, we're left with one dog! it cant be anything other than 50%! surely? 😀
I only work on logic
This is a subtle but different question. The order in which the dogs are sat is now relevant.
just (half) read the answer in that link, but its going on the 4 possibilities again, FM and MF being different etc. so its a flawed answer. i hope.
Jean is shorter than Brutus but taller than Imhotep. Imhotep is taller than Jean, but shorter than Lord Scotland. Lord Scotland is twice the height of Jean and Brutus combined but only one-tenth of the height of Millsy. Millsy is at a constant height of x − y. If Jean stands exactly one nautical mile away from Lord Scotland, how tall is Imhotep?
Are chance and probability the same?
The OP asks what is the chance, so is the chance 1/2 but the probability is 1/3?
accept the puzzle for what it is, we dont have to imagine anything, we’re told one dog is a male, we’re left with one dog! it cant be anything other than 50%! surely
But we don't know which one she was looking at when she said it was male, so the original options still stand.
Dog on left could still be male or female, dog on right the same.
You are assuming more information than you have. So take one option from the original 4, you must be left with 3 options. Counterintuitive, but true.
But she was asked, so the odds changed.
Nope. This is a just a different version of the Monty Hall problem.
Are chance and probability the same?
cant see how they can be different. either way its worded, the chance or probability that the second dog is male is 50%.
But we don’t know which one she was looking at when she said it was male, so the original options still stand.
Dog on left could still be male or female, dog on right the same.
You are assuming more information than you have. So take one option from the original 4, you must be left with 3 options. Counterintuitive, but true.
noooooooo! if she looks at dog on left, the chance of dog on right being male is 50%. if she looks at dog on right, the chance of dog on left being male is 50%. it matters not one jot which dog shes looked at. is one (any one) male? yes. that leaves one other dog. 50% chance its male.
So please do the math for me on Monty shall with only two doors in total.
Do pet shops sell even distributions of beagles? Perhaps dogs are more popular than bitches? Or perhaps beagles tend towards an uneven distribution of genders.
I definitely wouldn't buy a dog from a pet shop where the owner had to phone his wife up to tell what sex the animals are, rather than just having a look
thats not what we’re faced with here, accept the puzzle for what it is, we dont have to imagine anything, we’re told one dog is a male, we’re left with one dog! it cant be anything other than 50%! surely? 😀
The point I am making is that you do not know specifically which dog is the boy so there are 4 scenarios. This is why the answer to the question is 1/3. I used the sitting dog example to show that knowing one dog is a boy is not the same as knowing the dog on the left is a boy.
Keep fighting the logical fight sadexpunk 😀 I’m with you all the way. One dog is now accounted for and my tiny mind can’t fathom how that could leave more than two options. I can see the reasoning behind the one in three argument, but it seems to be over complicating a very simple matter. ****ing scientists! Probably testing makeup on beagles whilst coming up with this 😂
The question is “at least one boy pup” do you have to count all options with a boy in as we don’t know which pup she has examined.
1/3
What if one of the dogs is on a treadmill? Or identifies as an attack helicopter?
Cougar, I fear you posted this question knowing fine well the answer and the confusion it would cause 😀.
About 3 posts in sockpuppet gave the answer and a clear explanation in the next post. Now we’re well into the second page and I fear someone thinks 1/2 and 1/3 are the same thing, which most times they’re not.
Maybe you need to look at it another way. She is basically saying that the combo of dogs is NOT girl/girl. Knocking off one of the four original options. Leaves you one option from 3 now so 1/3.
2 dogs. 1 is a boy so can be excluded.
We now have 1 dog. Boy or girl, we don't know. 50/50.
So the answer is 50%
What if one of the dogs is in a treadmill? Or identifies as an attack helicopter?
Ooh! I know this one.
The point I am making is that you do not know specifically which dog is the boy so there are 4 scenarios.
and the point im making is that you dont need to know. so there are 2.
EDIT:
Maybe you need to look at it another way. She is basically saying that the combo of dogs is NOT girl/girl. Knocking off one of the four original options. Leaves you one option from 3 now so 1/3.
except 2 of those options are MF and FM which are the same thing. one male and one female.
There’s a zero chance that they’re both boys.
If you ask anyone if there’s at least one boy then there are only three possible answers.
Yes, there’s at least one boy.
No - in which case they both be bitches
They’re both boys.
Only the most pedantic of STW pedants would fudge the issue by simply answering Yes.
Since it was a man asking the question of a woman there is naturally some interpretation of the answer required which using lady-logic means that there must be one boy and one girl.
I don’t give a shit about the actual maths.
A high 5 for Perchy P
except 2 of those options are MF and FM which are the same thing. one male and one female.
They're not as we don't know the order the wife checked the dogs in.
They’re not as we don’t know the order the wife checked the dogs in.
you dont need to. why on earth does that matter? at least ones a male, whether she looked at left or right dog first. that leaves one other.
starting to wonder if this is a troll now as its soooooo obvious, so ah'm oot for now 😀
except 2 of those options are MF and FM which are the same thing. one male and one female.
They really aren't. Ok, what if one is a dog and one is a cat? Does that help. The options are
M cat - M dog
F cat - F dog
M cat - F dog
F cat - M dog
One is a male (or it's not F cat - F dog). Do you still think F cat - M dog and M cat - F dog are the same thing?
It does not matter the order in which the dogs are checked.
you dont need to. why on earth does that matter? at least ones a male, whether she looked at left or right dog first. that leaves one other.
Because of how the question was asked, it's a probability puzzle.
It's 1/3. everyone move on.....by which I mean post more Kate humbe videos on the country file thread.
grrrrrrrr cant resist. must resist......
Ok, what if one is a dog and one is a cat?
theyre 2 effin beagles! no need for complicated maths. are we sure this really isnt a trick to make poor old sadexpunk smash his laptop???
right! im defo out now, and im only posting again when its been proved that im right!! 😀
Come on you Monty Hall spouting people, do the logic with just two doors.
we don’t know the sex of one dog. There are two choices. Do the numbers.
I’m starting to feel really sorry for the second dog.
She didn’t have to check.
She already knew because she’s the wife of a pet shop owner and she’s bathing two beagle puppies.
Ten seconds after she entered the room she picked them up and said “Who’s a lovely..........boy and is this your.......sister?”
Don’t let maths get in the way of actual logic
theyre 2 effin beagles! no need for complicated maths. are we sure this really isnt a trick to make poor old sadexpunk smash his laptop???
right! im defo out now, and im only posting again when its been proved that im right!!
I AM A FISH
I AM 1/3 OF A FISH
FTFY 😉
Technically the answer is 1/4 because the wife is bathing the dogs not the puppies. Real world answer is 1/2 because m/f is the same as f/m.
The question was written by somebody who is one dimensionally very clever. Sorry but in the real world m/f, f/m is the same thing.
Hi, bye the way. Long time lurker who couldn't resist.
I must be missing something. If the shopkeepers wife turns up with 2 freshly washed dogs for the customer, and we know at least one is a boy, are we really saying that he can possibly walk away with 1 boy and one girl dog, OR walk away with 1 girl dog and one boy dog, and these are considered as 2 separate options?
Similarly I have a 50/50 chance of having a son and a daughter or a daughter and a son??
The fact that these are separate options seems to be the premise of the question. I would have thought that when it is confirmed to the customer that one is a boy, and he chooses to buy them, he has a 50% chance that he will walk out the door with 2 boy dogs.
My head hurts. I will be dreaming beagles now in my sleep.
right! im defo out now, and im only posting again when its been proved that im right!! 😀
Bye then!
The first time I seen a similar puzzle on here it did my head in as I thought 50/50 then after a bit of reading up on it I got why it's 1/3.
are we really saying that he can possibly walk away with 1 boy and one girl dog, OR walk away with 1 girl dog and one dog, and these are considered as 2 separate options?
No, we are staying at least 1 is a boy we just don't know which one is. The first dog she looked at the second.
Nobody asked us to identifywhich dog was going to be which sex. So 50% or you’re mad.
Nobody adkedus to identifywhich dog was going to be which sex. So 50% or you’re mad.
Good night down the pub?
It is only 1/2 if you fix the order...
We did this a couple of years ago, but at that point it involved children... not buying them obviously, but in that question the answer depended on how the question was asked/answered...
If for example in this question the wife had said something which fixed the sex of one of the dogs, by saying the biggest was a boy, or something like that, then the probability of that dog being a boy is 1. At this point the people saying the answer is 1/2 would be correct.
However, the question is specifically worded to avoid that answer, it is a probability trick often used to demonstrate how sneaky probability can be. So in this case the wife's answer does not tell you anything specific about either dog, just about the pair of dogs in general. All you know is that one dog is a boy. So is it the biggest dog? Or is it the smallest?
Without the wife's information the chances that both dogs are boys would be 1/4. Does everyone agree with that?
So, starting from there and then taking the wife's information into account, it tells us that out of the 4 options, we can ignore the one which has no boys. So, how many options are now left?
Out of those that are left, how many of those have two boys?
Come on you Monty Hall spouting people, do the logic with just two doors.
I could explain it to you, but I couldn't understand it for you 🤗🤗
Don’t let maths get in the way of actual logic
Worng way round.
Anyone want to buy a couple of beagles? One of them is pretty traumatised as it thinks it’s three dogs of indeterminate gender. Mathematicians need not apply.