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... and we've not had a plane on a treadmill for a while:
A man sees a sign in a window advertising two Beagle puppies for sale. He goes in and tells the shopkeeper he will only take the puppies if there’s at least one boy.
The shopkeeper phones his wife who is bathing the dogs and asks her if there’s at least one boy. She says yes.
What is the chance there are two boys?
If they are dogs then 100% certain. If there are bitches though...
Must......be........trick.......
50% unless she’s bathing different dogs.
edit - who gives a shit, the buyer is happy as he’s got what he wants already.
Is it 1/3?
Four options, boy-boy, boy-girl*, girl-boy*, girl-girl.
information supplied rules out one option, chance of two boys is this one out of the three remaining options.
1/3.
*these are different.
Twelfty baby robins?
Which is the trans dog?
It could be any of the following options
both dogs are male
one dog is male and one dog is female
So 50% chance.
Not seen this problem, but seems simple....probability is 1/3....I think.
Edit....yep....
Original options are:
X=Male, Y=Male
X=Male, Y=Female
X=Female, Y=Male
X=Female, Y=Female
Option 4 is out, so 1 in 3.
The chance of being male is 50:50 the gender of the other one doesnt affect it.
1/3 unless there is a trick. Puppies can't be boys as boys are ikle humans?
2/3
It could be any of the following options
both dogs are male
one dog is male and one dog is female
So 50% chance.
after reading all the posts so far im going with this one...... (thats after an initial thought of 75%)
Genuine question - How can it be 1/3? Surely the chances of it not been either a boy or girl are so slim as to be negligible? I’m missing some sort of weird mathematical shit aren’t I?
As said if she is bathing the dogs in question, and there are only two of them, then with the information surely the chance both are boys is 50%
She has confirmed at least one dog is a boy. The other dog can only be a boy or a girl. 50/50.
It cant be that simple can it????
I should clarify that for the sake of argument, there's no trick here involving differences in birth rates; terminology of dogs vs bitches; terminology of words like "boy" when dogs aren't boys; gender-fluid beagles or any other random gittery.
Genuine question – How can it be 1/3? Surely the chances of it not been either a boy or girl are so slim as to be negligible? I’m missing some sort of weird mathematical shit aren’t I?
Sockpuppet has a good explanation above.
50/50, th sex of th first dog is irrelevant to the equation now it’s known.
th sex of th first dog is irrelevant to the equation now it’s known.
The wife did not state which of the two dogs she was talking about.
Is this not the same as the deal or no deal thing/monty hall problem? If it is then I still won't understand it.
*these are different.
Not in this case surely. Only 2 options for the other non defined as male dog.
His wife & 2 dogs in the bath : )
50/50, th sex of th first dog is irrelevant to the equation now it’s known.
Yes, but either dog could be either sex. Adds another option.
Someone will be faster. The question is about a pair of pups. We already know something about the pair - one is male.
Gah. Sock puppet explains it perfectly but I'll do it again anyway as that's what I do. Possibilities are MM, FM, MF, FF, equal one in four possibility of each, until we learn that one is M. So we then know it can't be FF, meaning chances of MM become one in three. The end.
Same as flipping a coin, if you flip once and it's heads, logically the chance of the second flip being heads is still 50:50 but by the laws of probability the chance of the second flip being heads is reduced.
Genuine question – How can it be 1/3? Surely the chances of it not been either a boy or girl are so slim as to be negligible? I’m missing some sort of weird mathematical shit aren’t I?
Sockpuppet has a good explanation above.
The question states that one is definitely a boy though. I'd agree with the 50%.
MF and FM are one and the same so do not count twice.
X=Male, Y=Male
X=Male, Y=Female
X=Female, Y=Male
X=Female, Y=Female
??? Eh??Female wont make Y eggs, females dont have Y chromosomes.
So all female eggs are X 50% sperm Y to make a dog, 50% X to make a bitch.
The question states that one is definitely a boy though.
The question states one, but not which one. It matters.
MF and FM are one and the same so do not count twice.
Exactly this and why sockpuppets explanation didn’t make sense to me regardless of it being well written and reasonable. We already know one is a boy, so that just leaves 50/50. No amount of mathematical jiggery pokery will convince my simple mind otherwise.
The question states one, but not which one. It matters.
How (genuinely)? That dog, whichever one it is, is now out of the equation.
Possibilities are MM, FM, MF, FF, equal one in four possibility of each, until we learn that one is M. So we then know it can’t be FF, meaning chances of MM become one in three. The end.
but thats not the possibilities is it, as MF is the same as FM?
EDIT: typing at the same time, great minds funkmaster 😉
Genuine question – How can it be 1/3? Surely the chances of it not been either a boy or girl are so slim as to be negligible? I’m missing some sort of weird mathematical shit aren’t I?
The question does not give enough info to know which dog is the boy, so there are four out comes.
.....the plane takes off.
Imagine the dogs have stickers on prior to washing. Options are :-
Dog A Dog B
Girl Girl
Boy Girl
Girl Boy
Boy Boy
If the wife confirms Dog A is a boy there are only 2 possible solutions. (50/50)
See below if she confirms dog B is a boy
Dog A Dog B
Girl Girl
Boy Girl
Girl Boy
Boy Boy
Again this time if the wife confirms Dog B is a boy, there are still only 2 possible solutions (50/50)
It does not matter which dog the wife confirms is a boy. We can assume that the only options we have is that there are 2 boys OR 1 boy and 1 girl.
Same if I go to work knowing I have 2 children, but during the day I forget their sexes so I phone home to the wife.
Me 'Can you tell me if one of my children is a son?'
Wife 'Yes'
From that I can only either have a son and a daughter OR two sons. To have a son and a daughter is EXACTLY the same option as to have a daughter and a son.
MF and FM are one and the same so do not count twice.
So if the wife had not been asked a question - what do you think the odds of there being one female and one male puppy were?
Not seen this problem, but seems simple….probability is 1/3….I think.
Your answer is straight out of the blog the puzzle came from.
If the dog and it’s kennel costs £110 and the dog costs £100 more than the kennel, how much is the kennel on it’s own
typing at the same time, great minds funkmaster
High 5 sadexpunk. If we’re wrong (which we probably are) then I can categorically state I’m glad to be shite at maths 😀
Please tell me some of you lot didn't take maths beyond O'level/GCSE 🙂
but thats not the possibilities is it, as MF is the same as FM?
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.
50/50 is the 'logical' conclusion, but it's wrong. Honest 😉
Detailed explanation (or spoiler if you prefer) here:
https://scienceblogs.com/evolutionblog/2006/12/28/a-probability-puzzle-part-two
So if the wife had not been asked a question – what do you think the odds of there being one female and one male puppy were?
But she was asked, so the odds changed.
Anybody have a rocking chair, a corner and a dunce hat I can borrow?
So if the wife had not been asked a question – what do you think the odds of there being one female and one male puppy were?
not clever enough to tell you, but thats a different puzzle. she has been asked and it makes a difference. theres one dog left that we dont know the sex of, so 50%.
EDIT: funkmaster again!! 😀
This is like those Look Around You maths questions.
But if you put them in a sack and say one is male? It’s like Schrodinger’s Dog.
Exactly. If I told you I'd tossed a coin twice, and asked what are the odds l'd tossed heads both times, you'd say "one in four". That's because there are equal chances I'd tossed heads then tails, tails then heads, heads then heads, or tails then tails. So odds of heads then heads is one in four.
But if I asked the same question again and told you that one of the tosses was a head, you'd know that I hadn't tossed tails then tails. So odds of heads then heads are one in three.
Pretty clear? Or a lot of toss? Surely by now someone will have gone through the Bayes' Theorem way if doing this...
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