[Closed] Seeing as this is currently breaking Facebook...

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?