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

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.

There are only still three outcomes if you start off with the wrong four.

We’ve been going about this all wrong people. How have we got this far and not used bike parts to explain it?

There are only still three outcomes if you start off with the wrong four.

So explain what you think the possible outcomes are when using convert's ickle little models of dogs?

Your coin toss models give outcomes that do not fit with the question

No.  The coin toss model doesn't fit with your answer.  Which is kinda the point.

There are only still three outcomes if you start off with the wrong four.

Huh?

they could be ickle boy dog models or ickle girl dog models.

That's the same error I've already pointed out.

It's not really the Monty Hall problem but it might be related in that it sounds intuitive.

For Monty Hall it'd be more like you and the dog washer know there are 3 dogs, and both know exactly 1 is male. You say "I'd like the one nearest the plughole", the dog washer knows which of the 3 is the male one, and says "the one furthest from the plughole is a bitch... do you want to stick with the first choice or perhaps change your mind and have the other one?"

Since you really want a male one, you're not going to choose the one you know is female, but you can double the chance of getting the male by correctly choosing from the other 2.

There must be a video on Numberphile youtube channel explaining this?  They probably did it for the general case too with numbers other than 3.

I just went out riding. It was ace.

no dogs, no coins, no (awesome) spreadsheets. It was *ace*

(this post is adding about as much to the substance of the thread as many other offerings, so i though it fair game to report).

That’s the same error I’ve already pointed out.

Explain how you think you pointed it out again? Humour us.

No.

If I toss two coins I could get HH or TT. This is not applicable to the scenario in the question.

The coin toss model doesn’t fit with your answer.

Because it is wrong.

Which is kinda the point.

Precis, as they say in Sweden.

Okay, how much do we actually agree on @sbob?

Do you agree that without any extra information about gender, the chances of randomly being in a male-male pair scenario is 25%?

For Monty Hall it’d be more like you and the dog washer know there are 3 dogs, and both know exactly 1 is male. You say “I’d like the one nearest the plughole”, the dog washer knows which of the 3 is the male one, and says “the one furthest from the plughole is a bitch… do you want to stick with the first choice or perhaps change your mind and have the other one?”

Where’s my giant goat gone? Did the nasty American lady shoot him? 😕

Do you agree that without any extra information about gender, the chances of randomly being in a male-male pair scenario is 25%?

Absolutely, but this isn't the case in the OP.

Absolutely, but this isn’t the case in the OP.

It's the case in the first half of the OP though, yes?  Before we gain additional information?

sbob: If we exclude FF from the outset then there are still 3 possible outcomes.

I don't know how you can't see the difference between these two scenarios:

An owner gives you a dog and says this one is male. He gives you a second and says what is the probability that both are male? <span style="font-size: 0.8rem;">This is 50% because only 2 options are available to you at this point. MF or MM.</span>

An owner gives you two dogs and says at least one of these is male. He says what is the likelihood of both of them being male? This is 33% because 3 options are available to you. MF, MM and FM.

And then with the additional information, we rewrite the model. We don't just fudge the results of the old one that we now know is not true.

Actually I've just had a lightbulb moment in how sbob (and maybe others) have come to their answer. Still got it wrong mind, but it makes more sense to me now.

Imagine your were a film director and wanting to set up the scene. Two dogs in a bath, bloke on the phone, woman fondling the dogs. You know how the scene is going to play out and you know when the woman is asked about the dogs she is going to say at least one of them is a boy. So you lob in a male dog into the bath so she can guarantee answering the question truthfully and then find another dog at random and lob that in there too. Then you say action and watch the scene play out. From that point there are only two outcomes - the other dog being male or the other dog being female.

If that's how you are viewing it I don't actually know how to help you other than you can't 'cheat' and go into the scene from that point. You have to start from the beginning and work forwards.

3 options are available to you. MF, MM and FM.

There is no female dog, there is only a dog that is either male or female, we just don't know!

I cannot for the life of me understand why anyone would think that four minus one equals two.

The four has ceased to exist.

And then with the additional information, we rewrite the model.

Rewrite it to what?? The chances of being in any of the remaining scenarios remains equal.

We don’t just fudge the results of the old one that we now know is not true.

It's not a fudge, it is how you calculate odds with new information.

e.g. What are the chances of me rolling a 6 and then another 6 on a fair six-sided die? 1 in 36 right?
36 possible outcomes and only one is the winner.

But once we know that my first roll was successful then we can discard 30 of those possible outcomes and we are left with just six, where only one is the winner.

There is no female dog, there is only a dog that is either male or female, we just don’t know!

Jesus wept.  There are two dogs which are either male or female, we just don't know.  What we do know is that the number of male dogs altogether is non-zero.

You're doggedly (ho ho) clinging to the notion that one specific dog is known to be male.  It isn't.  The male dog could be the other one.

There is no female dog, there is only a dog that is either male or female, we just don’t know!

Is that dog A or B. The only personal people who know are the dog washer or the shopkeeper.

We don’t so it makes 3 options.

Jesus wept.  There are two dogs which are either male or female, we just don’t know.

Incorrect. We know one of them is male and the other is either.

You're stuck in a 50/50 loop that no longer exists.

We don’t so it makes 3 options.

No, you have one perfectly possible option and two mutually exclusive options. That is effectively two options. I can't believe you guys are letting these guys win with this disingenuous attempt at statistical sleight of hand.

Is that dog A or B?

Who knows? It doesn't matter.

50% chance.  She has one dog thats a boy its in her hand as she talks on the phone- the other dog we don't know so 50/50 chance.

This thread is pure popcorn.

Chapeau sbob.

We know one of them is male and the other is either.

Okay so pick up one of those dogs at random then.

If it is a female then the other one must be the male. (1 possible outcome)

If it is a male then the other one can be either male or female. (2 possible outcomes)

Only one of these 3 possible outcomes is male-male.

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

Mea culpa, i think...

Incorrect. We know one of them is male and the other is either.

Which one?

Who knows? It doesn’t matter.

Yes, yes it does.  If you know that one specific dog is male rather than one out of two then it changes the probability.

Let me put it another way.  You say one dog is male and the other is either.  Fine, that then gives us a probability.  But then there's also the possibility that the other dog is male and your one dog is in fact female.  You're ignoring this probability which is why you're getting a higher figure.

Imagine the dogs being washed are a Great Dane, and a Yorkshire terrier. Can you tell them apart?

are you telling me that a male yorkie and a female Great Dane is *exactly the same* as a female Yorkshire terrier and a male Great Dane?

The dog washer didn't tell us if she's found a male yorkie or a male Great Dane, just that she found a male. We didn't ask, she didn't tell us

so we have a male dog, either big or small, could be either.

now we have THREE OPTIONS! Male yorkie female GD, two male dogs, or female tiny pooch and male enormous one.

still sure that the MF and FM cases are the same?

Okay so pick up one of those dogs at random then.

But they are not random.

Actually I’ve just had a lightbulb moment in how sbob (and maybe others) have come to their answer. Still got it wrong mind, but it makes more sense to me now.

Imagine your were a film director and wanting to set up the scene. Two dogs in a bath, bloke on the phone, woman fondling the dogs. You know how the scene is going to play out and you know when the woman is asked about the dogs she is going to say at least one of them is a boy. So you lob in a male dog into the bath so she can guarantee answering the question truthfully and then find another dog at random and lob that in there too. Then you say action and watch the scene play out. From that point there are only two outcomes – the other dog being male or the other dog being female.

If that’s how you are viewing it I don’t actually know how to help you other than you can’t ‘cheat’ and go into the scene from that point. You have to start from the beginning and work forwards.

Yup, exactly! Except from the point the question is being asked the scene has been played out. We know one dog is male, we are simply being asked what the probability is that the other is as well.

FWIW I agree with the 1/3 working but this is just getting too Orwellian for me as I seem to be double thinking it. I did take a knock to the head yesterday (not an actual joke, smashed helmet and everything) and can see how both outcomes seem likely. I blame the language and the assumptions used (isn't that the whole point?) in the setting of the problem.

1/2ers:

Say a dog has 2 puppies one after the other. There is only 1 way the puppies can both be male. The first puppy has to be male (P=0.5), and the second puppy also has to be male (P=0.5). The overall probability is the sum of the two probabilities (0.5 x 0.5 = 0.25). There is also only 1 way the puppies can both be female. The first puppy has to be female (P=0.5), and the second puppy again has to be female (P=0.5). And again the overall probability is the sum of the two probabilities, (0.5 x 0.5 = 0.25).

We have accounted for so far a probability total of 0.25 + 0.25 = 0.5. The only remaining situation (1 male and 1 female) must account for the remaining 0.5 probabilty (since the total must be 1.0). This makes sense mathematically, and it makes sense conceptually because there are 2 ways of having a male and a female puppy and not 1 as was the case for MM and FF. The dog can have the female first and then the male, or have the male first and then the female.

Chapeau sbob.

Cheers!
I'd say I'm here all night but I've got to go to work.

Options are

MM

MF

FM

but surely if you include MF and FM you should also have MM and another MM

But they are not random.

I'm reformulating the model with the new information, exactly as you suggested doing.

Your population consists of two dogs and we know that there is at least one male.

Pick up a dog.

If it is a female then the other must be the male. (1 possible outcome)

If is a male then the other could be male or a female. (2 possible outcomes)

Only one of those 3 outcomes is male-male.

Imagine the dogs being washed are a Great Dane, and a Yorkshire terrier

For good reason, the OP clearly states they are both Beagles.

but surely if you include MF and FM you should also have MM and another MM

No. A perfect distribution would give:

MM: 25%
MF: 25%
FM: 25%
FF: 25%

Chance of a mixed-pair: 50%

Chance of single sex pair: 50%

Pick up a dog.

But we haven't picked up any dogs. You're still ascribing the old model.

We have one dog that is male, we'll call him Ishmael, and we have one dog that is either, we'll call them Leslie.

I think where you are going wrong is trying to split the new information into little bits you consider one at a time. That doesn't happen. You gain all the new info all at once.

It doesn't matter if Ishmael is the puppy on the left or the puppy on the right, the other puppy is Leslie.

I can see why you'd do it mind.

How do you know Leslie isn't male and Ishmael is female?  Why have you ruled that out as an option?

You're trolling us, aren't you.

we know one is a male, this is M* so

M* M

M M*

M* F

F M*

50% chance

To present the same question in a different context:

You are a medieval king.

Your wife is pregnant with (non-identical) twins. You are desperate to have a male heir to continue your ruling line, and not start a bloody war.

(you know it is twins, and that they are fraternal, not identical)

The day before she gives birth, you calculate the change of you getting at least one heir is 75%, as there are 4 possibilties MM, MF, FM, FF. Which one of these is correct was decided 9 months ago, but is unknown.

If there is a male heir, the midwife is to send up a signal of white smoke. If there is no heir, black smoke. Ergo - white smoke 75%, black smoke 25%.

Immediately after the birth, you see white smoke. [you have the answer to your very specific question, which requires a Y/N answer. While there is more information, smoke signals cannot communicate this].

At this point after the smoke is white, there are now 3 possibilities. MM, MF, FM, where the first M or F represents the first born.

You are happy that you have an heir, but a new problem has presented itself:

At this point in time (after the smoke, and before you actually meet your children) what is the chance you have 2 boys who will be forever fighting over who is first born, and thus entitled to the throne?

Answer - one in three of the remaining legitimate possibilities, which have an equal probability of occurring, as it was decided months before.

There is no maths answer v common sense answer - the only non-real-world occurrence is that the pet shop owners wife who can easily find out the sex of both dogs, and then explain this in a full sentence through the wonders of telephone; instead chooses to correctly answer the exact, precise question asked with a yes/no one word answer, and venture no more information on the topic. Which is unlike any woman I have ever met.

We have one dog that is male, we’ll call him Ishmael, and we have one dog that is either, we’ll call them Leslie.

And by doing that you have made it a specific dog. That is information you do not have.

That is exactly why I originally explained this in terms of hands. Stating that Ishamael is the male is exactly the same mistake as saying the one in your left hand is male. You do not know that.

If you want to phrase it with names then we can do that.
But you don't know the gender of either Ismael or Leslie, you just know there is at least one is a male.
So either:

1) Ismael is male and Leslie is male
2) Ismael is male and Leslie is female
3) Ismael is female and Leslie is male

Surely this is only a question about one dog.

two dogs are advertised

one is revealed to be male.

the only question of 'chance' is whether the other one is male.

How do you know Leslie isn’t male and Ishmael is female? Why have you ruled that out as an option?

It doesn't matter, one of them is, one of them might be.

Note to self: must find a way to monetise this.

It doesn’t matter, one of them is, one of them might be.

Exactly,

So either:

1) Ismael is male and Leslie is male
2) Ismael is male and Leslie is female
3) Ismael is female and Leslie is male

What happened to Ishmael?