Seeing as this is currently breaking Facebook…

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?

Can we stick to M & F ffs 😆

“Call me Ismael.” 😉

GrahamS – is this more fun than arguing with flat earthers?

Can I interest you in a sexy spreadsheet @ctk?

GrahamS – is this more fun than arguing with flat earthers?

It is remarkably similar 😂

Your spreadsheet has GG as an outcome.  We KNOW one is a boy.

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

M* M

M M*

M* F

F M*

50% chance

*sigh*

*sigh*

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

100 door monty hall, with a small stake to win a reasonable prize? And hope everyone’s a 50/50er. And doesn’t watch the vids.

is this more fun than arguing with flat earthers?

oh and the latest Numberphile vid is indeed a map projection that demonstrates a flat earth 🙂

Yes, @ctk GG (or FF) is in the initial population.

It is a possible outcome at the start, before we receive the additional information.

As you can see, in that initial population of 1000 fairly assigned pairs, we have around 250 MM, 500 Mixed, 250 FF.

Once we find out that our pair is definitely not one of the FF pairs then we are left knowing that our pair comes from the remaining population of 250 MM and 500 Mixed pairs.

😂

ctk – Do you.not realise that by ascribing a * to one of them, you are assigning that a specific dog of the 2 is known to be male, thus making it a different problem to the OP.

It’s already been noted a number of.times that if someone point to one of the 2 dogs and says that one is male, then the probability becomes 50%

The issue is that what you are presenting is not correct..

I can’t believe this is still going, excellent 😀

When I read it first thing this morning I was in the 1/2 camp, but after the 2nd page and I think the sitting dog explanation it clicked and I got the 1/3.

Page 10, well done STW. Chances this is still going the same time tomorrow?

Peanut Butter M&Ms?

Would Revels not be better for this game?

I saw the light after a slow start. It’s definitely 1/3

Say my wife is expecting twins. I tell the midwife that I will take both babies home if at least one of them is a boy.

I am in the waiting room, the midwife comes out to say I can take the babies home as at least one of them is a boy. One of three possible and equally likely scenarios has occurred

1) My wife has given birth to a boy then a girl. Outcome is I go home with 2 children of different sexes. This will happen 1/3 of the tim

2) My wife has given birth to a girl then a boy. Outcome is I go home with 2 children of different sexes. This will happen 1/3 of the time

3) My wife has given birth to a boy then another boy. Outcome is I go home with 2 children that are both boys. This will happen 1/3 of the time.

Importantly, outcome 1 and outcome 2 CANNOT be lumped into one. They are separate outcomes although on the face of it they appear the same.

Only 1/3 of the time will I be heading away with 2 boys.

For it to be 50/50 I would need to fix another fact to a particular child, such as the blond baby is a boy, what is the chance that both babies are boys. This insinuates that the OTHER child is not blond, hence the chance that the OTHER child is male or female is 50%. You can’t do this with the OP because the problem does not give you enough info to pin down facts about a particular dog.

Hope this has cleared it up.

Probability 1 in 3

Chances 1 in 2

Probability is unable to realise that for the questioner m-f and f-m are one and the same.

this is somewhat marvellous.  And I have to agree with The Cougar that the 52% think it is 50%

well done all

Do you not realise that by ascribing a * to one of them, you are assigning that a specific dog of the 2 is known to be male

Also it’s like counting two ways to roll a double six: 6 6* or 6* 6

Probability 1 in 3

Chances 1 in 2

How exactly are you defining the difference between “Probability” and “Chances”??

Funnily enough we are going to the dog home for a look on Saturday.  Dont like beagles ;-0

I genuinely cannot fathom the logic here.  We have four options.  We’re then told that one of those options isn’t valid.  We’re then left with two options.  What?

I’m suddenly reminded of this.

Yeah that’s probably a better way to think of it.

I’ve probably only got one other way of describing the problem and that’s no.foubt been covered already in the thread but here goes.

Dog A and Dog B

FF

FM

MF

MM

If I say A is a male, you can eliminate 2 possibilities FF and FM leaving 2 equally likely options.

If I say at least 9ne of the dogs is a male then you can only eliminate 1 possibility FF leaving 3 equally likely options. This is the problem in the OP.

This works.if you run it with 3 sets of 2 dogs (AB,CD and EF). 12 options. Knowing one of each pair is M leaves 9 options with MM for all 3 being a third. If I said A, D and E were male then it leaves on 6 options with MM being a half.

How exactly are you defining the difference between “Probability” and “Chances”??

Well to calculate the chance of there being two male dogs when you already know that there is at least one  male dog you can reduce m-f or f-m to one option but probability can’t do that.

This boils down to whether MF and FM are the same or different but the key phrase here is the ‘at least one‘ condition and the wife’s positive response and I think the one-thirders are conveniently ignoring this….

Scenario 1 MM – only needs to check first dog before providing positive answer
Scenario 2 MF – only needs to check first dog before providing positive answer
Scenario 3 FM – needs to check SECOND dog before providing POSITIVE answer
Scenario 4 FF – needs to check second dog before providing negative answer

So, although there are three different scenarios where a positive answer is provide, the difference between scenarios 2 & 3 is actually irrelevant meaning there are only two outcomes – both males or one of each.

So the REAL scenarios are

Scenario 1 – MM

Scenario 2 – One of Each

Scenario 3 – FF

We know FF does not apply, so 50%…

Sorry to spoil the fun…

I imagine her with one dog out of the bath having just had its sex checked and one dog left in the bath that is yet to be checked.  What I need to imagine is her pulling both dogs out of the bath at the same time, checking and confirming at least one is male and then putting them both back in 🙂

Probability 1 in 3

Chance 1 in 2

Probability 0.33333recurring

Chance 1 in 2

FTFY

Except the chance is 1 in 3 also…

when you already know that there is at least one male dog you can reduce m-f or f-m to one option

Sure, but, that one option is still twice as likely as the other one.

Scenario 1 MM – only needs to check first dog before providing positive answer
Scenario 2 MF – only needs to check first dog before providing positive answer
Scenario 3 FM – needs to check SECOND dog before providing POSITIVE answer
Scenario 4 FF – needs to check second dog before providing negative answer

Nowhere in the puzzle does it mention that she’s checking dogs at all, you’re adding / assuming information which isn’t provided.

For all you know she happened to notice a few days ago that one had a willy and can’t even remember now which one it was. In any case, even if she did know she doesn’t then convey that information to the shopkeeper so it’s lost knowledge at that point.

So the REAL scenarios are

Scenario 1 – MM

Scenario 2 – One of Each

Scenario 3 – FF

As above:

Scenario 1 – MM – 25% chance.

Scenario 2 – One of Each – 50% chance.

Scenario 3 – FF – 25% chance

Scenario three is reduced to 0% when we discover that they aren’t both female, but the other two’s ratio to each other does not and indeed cannot change.

It’s one in three, I’m a professional mathematician and have been doing this sort of calculation for decades but most people are too simple-minded to understand it, sorry. The correct answers (there have been many) are correct.

Bah, we’ve had enough of experts.

The important thing to note is that the prior probability of boy and girl (in any order) is double the prior odds of two boys. If this isn’t obvious you can check this yourself by tossing a pair of coins, repeatedly, and counting up how many times you get HH, HT, TH, and TT. Of the times that you got at least one H, in 1/3 of them you got HH, 1/3 was HT, and 1/3 was TH.

Nowhere in the puzzle does it mention that she’s checking dogs at all, you’re adding / assuming information which isn’t provided.

For all you know she happened to notice a few days ago that one had a willy and can’t even remember now which one it was. In any case, even if she did know she doesn’t then convey that information to the shopkeeper so it’s lost knowledge at that point.

Ok, point taken but you’re obfuscating…

At no point in the OP was ”’probability” mentioned either, and probability introduces a time component (ie the experiment is repeated). But for any one instance, this is a binary choice and the second dog is either male or female so for this one instance I don’t agree with your assertion that “one of each” is twice as likely to occur as MM…

Saying “there are two possibilities, therefore they are equally likely” is a simple fallacy. Tomorrow, the sun may rise, or it may not. They are not equally likely.