Seeing as this is currently breaking Facebook…

but to me, if i am physically presented with these two dogs and am told one is a boy and the other is unknown i only need to physically check one of them to find out what sex both of them are…

How would you do that? You may need to check both to know the sex of both. Either the one you check is female, in which case you do know the sex of both, or the one you check is male, in which case you have no way of knowing the sex of the other without checking it.

You have one dog in each hand

Why?

You have a dog that is male and a dog that is male or female.

In the 1/3rd scenario there is a female dog which doesn’t exist in the question. There is no female dog. There is a male dog, and a dog that is either male or female.

aaaargh! The ‘maths answer’ is the real world logical answer! Never play poker for money.
played last night and won £25 actually!
I say real world and maths answer, but to me, if i am physically presented with these two dogs and am told one is a boy and the other is unknown i only need to physically check one of them to find out what sex both of them are…

So if you check one and it is a boy, what sex is the other one?

copy text, click the speech marks in the reply box which indents the cursor an inch or so.  paste.  then hit return twice to get cursor to start your reply.

ah ok like that thanks spunk

if i am physically presented with these two dogs and am told one is a boy and the other is unknown i only need to physically check one of them to find out what sex both of them are…

This would only help you if you found it to be female…

Edit: missed a couple of posts there…

Why?

You have a dog that is male and a dog that is male or female.

Because doing it with hands, as I suggest, is a good way to make it clear when you are applying knowledge you do not have.

You know at least one dog is male, but you don’t know which one. That is important.

There are two routes to a mixed pair and only one route to get two males. The mixed pair is twice as likely.

Why?
You have a dog that is male and a dog that is male or female.
In the 1/3rd scenario there is a female dog which doesn’t exist in the question. There is no female dog. There is a male dog, and a dog that is either male or female.

For the sake of the explanation, let’s say the dogs are called Rover and Spot.

From the information we have, we know that Rover and Spot are not both female. They could both be male or they could be a male and a female.

Three possibilities therefore exist:

1) Rover male, Spot female

2) Rover female, Spot male

3) Rover male, Spot male.

Assuming a random dog has a 50/50 chance of being male or female, that means the three possibilities are each equally likely to be the true position, ergo the probability of it being option 3 is 1 in 3.

You know at least one dog is male, but you don’t know which one. That is important.

It’s only important if coming from the false premise that one dog is female which you need to get your four outcomes and your 1/3 conclusion.

Neither dog is female.

Edit, dodgy forum double post

I say real world and maths answer, but to me, if i am physically presented with these two dogs and am told one is a boy and the other is unknown i only need to physically check one of them to find out what sex both of them are…

No no no no no no. Only if you are told which one is a boy. Anyway that not the question that the thread’s about.

if i am physically presented with these two dogs and am told one is a boy and the other is unknown i only need to physically check one of them to find out what sex both of them are…

if you are physically presented with them, and told that one is a boy *but not told which one*, and you then look at the back end of one of the dogs and it’s a boy dog, you still don’t know for sure if there’s one or two boys. You might have checked the same dog, you might have checked the other, but you don’t know.

knowing which dog makes all the difference.

if the wife said “the first dog I checked is a boy, I didn’t look at the other one” then it’s 50:50 on the other.

BUT THAT’S NOT WHAT THE OP SAID,

Assuming a random dog has a 50/50 chance of being male or female

You’re assuming that both dogs have a 50/50 chance of being either male or female and then discarding the F/F option to give you 1/3.

Trouble is the question states that the probability of one dog being male = 1.

1×0.5=0.5

0.5×1=0.5

No, probability of *at least one dog* being male = 1. Which is a different thing.

“one dog being male” can only happen one way

“at least one dog out of two (and maybe both) being male” can happen three ways.

You have a dog that is male and a dog that is male or female.

No, you don’t.  You have two dogs and you know that they aren’t both female.  You do not have one of anything at any point in this riddle.

You claim to know that “one dog is male” so if I were to show you a photo of the two dogs would you be able to point to the known-male one?  You’re including extra information which you don’t have.

Neither dog is female.

This isn’t quantum physics. Both dogs have a gender.

Given the 50:50 binary cisgendering there are only four possible equally-likely scenarios when you don’t know any of the genders. At that point the chances that you are in the scenario with two male dogs is 1 in 4.

Once you find out that at least one of the dogs is male then you can narrow down which scenario you could be in by ruling out the all-female scenario. The chances that you are in the scenario with two males dogs increases to 1 in 3.

If that doesn’t work for you then I’d urge you to look at the simulation on my spreadsheet where it empirically shows this to be true by simulating 1000 cases.

Cougar….tut tut. I have now read the solution linked to at the bottom of page 1 and you changed the genders of the shop keeper and the washer of dogs when you posted in the OP. Gender stereotyping right there with your male shop owner and female skivvy dog washer.

I really haven’t, that’s how it was presented when I came across it. (I’d never have done that anyway, I actually considered removing all references to the gender of the dog-washer).

You’re assuming that both dogs have a 50/50 chance of being either male or female and then discarding the F/F option to give you 1/3.

Trouble is the question states that the probability of one dog being male = 1.

1×0.5=0.5

0.5×1=0.5

This is now a personal journey of discovery for you to establish why that is not the correct answer. I am afraid we are long past the point where you being right is a possibility.

Go flip some coins. Flip two together and discard any where you get two tails. Count how many valid flips (discounting the two tails) it takes you to get 10 double heads. I’ll eat my hat and you’ll win the internet if it does not take you somewhere between 27 and 33 attempts. At that point come back and ask us again to explain how it works.

I don’t know how often I can explain the same thing in a slightly different way, but.

Consider these three situations (or read the previous pages):

1: I’ve just flipped two coins.  What is the probability that both are heads?

2: I’ve just flipped two coins.  Looking at them I can see that the one on the left is heads.  What is the likelihood that both are heads?

2: I’ve just flipped two coins.  Looking at them I can see that at least one of them is heads.  What is the likelihood that both are heads?

In case 1 I don’t think anyone is suggesting that the answer is anything other than 1 in 4 (and if they are they should probably step away from the thread).

In case 2 we prescribe a value to a specific coin, at which point it becomes irrelevant to the value of the other one.  The other coin is either H or T, so the probability of two heads is 50:50.

In case 3 we know one is a head but not which one.  With two coins, there are three permutations which could give us a tableau containing at least one head.  So the chances of getting two heads is one in three.

Pretty much everyone who is saying the dogs answer is 50:50 is interpreting it as though it’s the second scenario here.  But there’s nothing in the riddle to say that one specific dog is male, just that there’s at least one somewhere in the mix.  It’s outside information which they’re erroneously including.  One of the dogs is black and the other is brown, can you tell me which is the male one?  If you can’t then it’s demonstrably scenario 3 not 2.

You claim to know that “one dog is male” so if I were to show you a photo of the two dogs would you be able to point to the known-male one? You’re including extra information which you don’t have.

I do know that one dog is male.

I do not know which dog is male. You’re ascribing extra information to my claim that I haven’t made.

I’m more of as cat person really.

 I’ll eat my hat and you’ll win the internet if it does not take you somewhere between 27 and 33 attempts.

Possibly a little precise to wager for pilleusphagia, but yes, it should be roughly around those figures.

That is very much what my speadsheet does: It flips two coins/genders/whatevers 1000 times and examines the results.

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?

I’ve just re read it and the answer is 1, is at no point on the question stated “1 dog is male” nor is it asked “what are the chances both dogs are male” it’s just inference.

Clearly there is at least one boy, clearly there are two boys, i can see them from my desk. The question has nothing to do with dogs or maths it’s an obscured logic problem with a very simple answer but very good distraction.

The dogs are in fact not beagles but red herring and we’re all wrong (unless this has already come up in pages 2-6, in which case i apologise).

That is very much what my speadsheet does: It flips two coins/genders/whatevers 1000 times and examines the results.

[whisper] Yes, I know. I love your spreadsheet. I’m going to be thinking about as I go to sleep tonight in a naughty way. But – monkey see, monkey do, learning through doing etc I think we have more chance of him coming round if he does it himself [/whisper]

Gosh, aren’t these threads irritating?  A group of people that have the right answer (and are sure of it) and another group of people who are completely wrong but are equally sure they’re right.

Have we mentioned Dunning-Kruger yet?

It’s no wonder we’ve got a bloody Brexit on our hands.

Go flip some coins

Why is everyone insisting on using a model that is proven to give an impossible outcome?

All your coin tosses/spreadsheets do is prove that they are not applicable to the problem.

this has to be deliberate?

2: I’ve just flipped two coins.  Looking at them I can see that the one on the left is heads.  What is the likelihood that both are heads?

Bad cougar.

Knowing you’ve flipped two coins before revealing the result means the probability of the second is determined. So it’s a 1 in 4 chance the rh coin is a head.

If you flip a coin get a head, then flip a second coin the chances of a head on the second coin are 1 in 2 as it’s wholly independent of the first.

The whole reason the problem in the op is a problem is down to when the probability is locked in.

Why is everyone insisting on using a model that is proven to give an impossible outcome?

OK, go to the dog pound, ask them to pick two random dogs out for you and do the same experiment (check they have roughly 50:50 girls to boys first). Repeat it 100 times and see the results. Its easier with the coins but the result will be the same.

ask them to pick two random dogs

They’re not random in the question.

OK, go to the dog pound, ask them to pick two random dogs out for you…Repeat it 100 times

And if you think that includes an “impossible outcome” then get them to only tell you the results that include at least one male dog.

Spoiler: you’ll get roughly 25 both-male pairs and roughly 50 mixed pairs.

And if you think that includes an “impossible outcome” then get them to only tell you the results that include at least one male dog.

So you want to use a model that gives incorrect answers, but then to just disregard those to fit your own agenda?

Do you work for the government?

Spoiler: you’ll get roughly 25 both-male pairs and roughly 50 mixed pairs.

Only by incorrectly using a 50/50 scenario that doesn’t exist and discarding the answers you don’t like.

Knowing you’ve flipped two coins before revealing the result means the probability of the second is determined. So it’s a 1 in 4 chance the rh coin is a head.

I’m not sure as I follow what you mean?

go to the dog pound, ask them to pick two random dogs out for you and do the same experiment

About 33% of the time I got both heads and the RSPCA are on the phone.

I don’t think this is a modified Monty Hall problem, because you aren’t selecting a dog. This leaves the known M against an unknown dog.

I’m on the 1/3 side and the best way I think about it is that MF and FM are important and different because you don’t know which dog the wife picked up first and if she needed to pick a second one up.

Thus of the 4 combinations there are, only 1 is discounted by the wife leaving 3 equally likely outcomes.

That is, M first, or F then M, or F then M.<span style=”font-size: 0.8rem;”>That means MM is 1/3.</span>

Or alternatively, it’s not the same as saying there is Dog A and Dog B. Dog A is M, what is the likelihood that both are M? As this eliminates FM and FF from the 4 possibilities.

Gosh, aren’t these threads irritating?

Nope, they’re great. One of the threads with the most participation I’ve seen for ages. A bit of debate, a few laughs and irritating people who just have to involve brexit in every bloody thread 😉

Hate to say but I think sbob is lost to us.

Hmmm.. can’t decide if @sbob is just trolling at this point or he really still doesn’t get it.

Perhaps @mrb123’s simplified wording may help?

If you have two dogs and you know that they are not both female, what is the probability of them both being male?

Hate to say but I think sbob is lost to us.

I understand the maths you are using, I’m questioning whether or not you should be using it.

Your coin toss models give outcomes that do not fit with the question, yet you insist on using them. Do you disagree with that fact?

I fear that the time may have come to give up on this one.

Do you disagree with that fact?

Yes.

Before you have gained information you can be in any scenario. At that point all four outcomes are equally likely. As represented by the coin tosses, or spreadsheet results, or whatever.

Once you gain the information that they are not both girls then you can remove one of the possible outcomes as a possibility. But the remaining three outcomes are still equally likely and only one of them is “both boys”.

Do you disagree with that fact?

Vociferously.

If you understand the maths used by the 1/3 gang then hopefully you can see that all the information in the phone call does is discount the two female option. That is exactly like tossing two coins and discarding the two tail option when it occurs. The scenarios are identical.

Keeping it to dogs, but ickle little models of dogs…I’ve got two models of dogs – they could be ickle boy dog models or ickle girl dog models. I have one in each hand out in front of me. I know what gender each is but you don’t. You ask me one question “is one of your model dogs a boy?” – I chearly reply yes! Because it is true. You then ask me to show you.

Reality one….I open my left hand and show you a boy dog and open my right hand and show you a girl dog

Reality two….I open my right right hand and show you a boy dog and my left hand and show you a girl dog

Reality three…I open both my hands and both have got a boy dog.

It took 3 versions of events to show you the options of what could have been hidden in my hands. Ergo 1/3.