Seeing as this is currently breaking Facebook…

Try this: Before we know anything other than there being two dogs the probabilities are FF =1/4, MF = 1/2, MM = 1/4. MF is twice as likely as MM. Do you agree with that?

no i dont.  options are FF, MM, or MF/FM.  3 options.

convert, ill try 🙂

but as the majority of 50/50 comments state, you know the result of one hence, the next result is independent

No you don’t. As I’ve said previously if the dogs were called Rover and Fido and the question as was if Rover was male I would agree that the answer to the question is Fido male is independent and indeed 50/50. But we never names the dogs. We didn’t gender test them in a defined order. It really matters.

No you don’t.

you do.   the result is ‘one is male’, doesnt matter whether its rover, fido or noname.  ones male.  chances left are 2 males or male/female combo.  the sex of an individual dog, that dog there doesnt matter.

convert, ill try

I’m glad. But hang on….

no i dont.  options are FF, MM, or MF/FM.  3 options.

We’ve got some work to do first on this bit.

I’ve got some work to do; can you go play with a couple of coins for a bit and see if you still believe this and I’ll get back to you.

can you go play with a couple of coins for a bit and see if you still believe this

no, cos the coins may be different to the dogs, so im not going to confuddle my little head any more 🙂

no i dont.  options are FF, MM, or MF/FM.  3 options.

Ok, I think I see where the problem is. Do you also think that your chances of winning the lottery are 50/50? You either win or you don’t, 2 options 🙂

This thread is great. I voted to stay in the EU and understand how the thought process and probability of 1/3 answer works. Yet I still think 50/50 is the correct answer. Any other answer just seems daft when we already know ones a bloke.

Where does this leave me?

Does the bloke have his Beagles?

Are they to be kept as pets or used as working dogs?

Do the dogs know what gender they are and is one more likely to have a career as a nurse?

Yours sincerely,

Confused of Macclesfield.

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?

We know one dog is male. It doesn’t matter which one, we are only concerned with the sex of the other dog which can only be male or female.

Ok, I think I see where the problem is. Do you also think that your chances of winning the lottery are 50/50? You either win or you don’t, 2 options 🙂

poor, very poor.  there are 14,000,000 other people with a ticket, why would i think its 50/50?  must try harder.  again, RTFQ and see it for what it is.

Are any of the dogs pink?

We know one dog is male. It doesn’t matter which one, we are only concerned with the sex of the other dog which can only be male or female.

What is even worse is the fact that theoretical beagle buying man doesn’t give a shit and we’re at six pages 😀

Once he heard the words “Yep, ones a boy” he was done and happy and couldn’t care less about the other dogs gender.

MF is different from FM because the gender of one dog is independent of the gender of the other so there are four options available (MM, MF, FM, FF) but we are given an extra bit of information that one dog is male so we can discount the FF option. This means that both dogs being male is one option of three.

If I toss two coins one in each hand I have four possible outcomes: both heads; LH heads, RH tails; LH tails, RH heads; both tails. The probabilities of all four outcomes is 1/4. The probability of getting one coin heads and one tails is 2/4. If I toss the coins sequentially and it comes out heads that does not change those probabilities so the odds of the second coin being heads is 50% because it’s an independent event but the odds of both being heads is 1/3.

I got halfway through the first page and decided some people have way too much time on their hands!

Now, probably the tricky bit. The only thing we are told is that “there is at least one boy” and with that information all we can do is exclude option 1, it isn’t FF. MF is still twice as likely as MM so the odds are now MF = 2/3, MM = 1/3

It’s 50/50.

The position of the male puppy doesn’t matter because whichever position it is in, it exludes either MF or FM. Your MF at 50% overall likelihood doesn’t hold water.

You can’t say that both of those are still valid options just because you don’t know the position. You can only say one of them is valid, but we don’t know which. Either way, you are back to two options. 50/50. Anything else is a distraction.

The probability of a dog being male or famale is 0.5. It doesnt matter what the dog next to it is. Its a poorly worded question.

It’s a very carefully worded question, which is why the correct answer is not the “obvious” one.

There are two dogs and you know nothing specific about either one. The only thing you know it’s that both of them are not female. So the unknown dogs could be MM, MF or FM. That’s three options, only one of which is the outcome you want.

The probability of a random, new dog being born male is 0.5, but that’s not what is being asked here. The two dogs already exist, and all you know it’s that they are not both female.

So the unknown dogs could be MM, MF or FM.

no they cant, MF is the same as FM.  which dog is which doesnt matter.  crack this one and we’ll all agree 😀

Yay! The Boy/Girl Paradox. Sorry to be late to the party.

I posted this to the old old forum many many years ago and it went for twenty odd pages.

The correct answer is 1/3.

The easiest approach for people that don’t get it is to do a spreadsheet and prove it empirically.

Hang on.. I’ll knock on up…

If the first dog being checked was female what is the probability of both dogs being male?

You’ll answer, correctly, zero.

If we are only concerned with the gender of the second dog, as stated above, then the answer apparently is 50%!

@sadexpunk – look at my coins example which is the same problem.

The 3 possible outcomes of MM, MF, FM makes perfect sense though the way the question is worded means MF and FM is the same as one they both mean only one is male. Its a good one that makes a lot of people over think a question. We know one is male so the question is simply what are the odds of the other one being male. 50/50.

I just asked my wife….what is the probability we will be getting a divorce?

You can’t say that both of those are still valid options just because you don’t know the position. You can only say one of them is valid, but we don’t know which. Either way, you are back to two options. 50/50. Anything else is a distraction.

But both of those options are still valid. I know SP doesn’t like it but the coin toss analogy works well here. Toss two coins. HH 1/4, HT 1/2, TT 1/4. Now discount any results where you don’t have at least one head (ie one or more of them is a head, so all you can discount is TT) the results will now be HT 2/3, HH 1/3. Same for the dogs. You can try it with the coins, should only take 20 or so tosses to start seeing a trend

@mjsmke – the probability of the second dog being male is 50% but that isn’t what is being asked. MF & FM are valid and separate probabilities so have to be taken in to account since we don’t know which of the two dogs was checked for gender.

herein maybe lies the crux of the issue if we can bash this out. i say its not twice as likely, its the same.

But… well, it’s not.  I don’t know how else to explain this.

Take the second bit of the question out of the equation for a moment.  You have two dogs at random.  The likelihood that they’re the same gender is the same as the likelihood that they’re different gender, yes?  50:50, they’re either the same of different.

Now we remove one of those “same gender” options, we’re told that they aren’t both girls.  You’re arguing that the likelihood of them both being either the same gender or different is still 50:50?  How is that possible?

another way of wording it……whats the chances of the other one now being the same as that male one

That’s not another way of wording it, it’s a different question entirely.  You’re arbitrarily singling out a specific dog and then asking questions of the other one.  What if it is, in fact, the other dog which has been identified as male?  You’re completely ignoring this possibility, which is why your maths is wrong.

You can try it with the coins, should only take 20 or so tosses to start seeing a trend

Coin tosses are not a valid analogy, we are looking at a single pair of dogs and we know that one of them is male.

If the question posed was asking what the likelihood of a second dog being male in any given family randomly chosen from a dataset where we knew that at least one puppy was male then that would be valid. In this case, we are not. We are looking at a single data point, and we know that a randomly slected puppy is male.

Just had a thought – could we tweet this to Donald Trump? It’d keep him busy for years!!

And as others have said, the coins is a perfect analogy.  It’s the same question.  You’ve got a pair of things that can be one thing or another and are asked to discount one permutation.

Take two coins and flip them repeatedly.  Record the results of two heads, two tails, or one of each.  Now, cross out any results for whom the answer to “is there at least one head?” is “no”.  Would you expect to see a similar tally of two-heads to one-of-each?  If your answer is “yes” I strongly suggest that you actually try it as an exercise.

Coin tosses are not a valid analogy, we are looking at a single pair of dogs and we know that one of them is male.

Coin tosses are a valid analogy, we are looking at a single pair of coins and we know that one of them is heads.

Coin tosses are a valid analogy, we are looking at a single pair of coins and we know that one of them is heads.

And you are now suggesting that the probability of throwing another head is 1/3?

Okay here’s my spreadsheet…

https://docs.google.com/spreadsheets/d/1T7BqW-IdB-5xZp3CE0cql6qaWHLtLGSFCriMCpoYRsQ/edit?usp=sharing

It generates 1000 scenarios of two dogs, where each dog has a random 50:50 chance of being male or female.

You can see for yourself that empirically, in actual results, the probability of both dogs being male consistently comes out at around 1/3

No mathematical trickery involved.

A man sees a sign in a window advertising two recently-tossed coins mounted in a display case. He goes in and tells the shopkeeper he will only take the coins if there’s at least one showing heads.

The shopkeeper phones his wife who is washing the case and asks her if there’s at least one facing heads-up. She says yes.

What is the chance there are two heads-up coins?

That’s not another way of wording it, it’s a different question entirely.  You’re arbitrarily singling out a specific dog and then asking questions of the other one.  What if it is, in fact, the other dog which has been identified as male?

if it was the other dog that had been identified as male then we’d have the other dog that would be male or female.

 I know SP doesn’t like it but the coin toss analogy works well here. Toss two coins. HH 1/4, HT 1/2, TT 1/4. Now discount any results where you don’t have at least one head (ie one or more of them is a head, so all you can discount is TT) the results will now be HT 2/3, HH 1/3.

that just messes my head up trying to think that through, so im sticking to what we’ve been told about these 2 dogs.  ones male, whats the chances of them both being male then.  50/50

And you are now suggesting that the probability of throwing another head is 1/3?

Why are you throwing one of them again?  They’re already tossed, you’re then simply observing the results.

Observing the coins doesn’t change the probability, putting one arbitrarily aside and re-tossing the second demonstrably does.

LOOK AT THE SPREADSHEET. 

It will all become clear.

Hopefully.

Is it any wonder that bookies always win (sigh)

It generates 1000 scenarios of two dogs, where each dog has a random 50:50 chance of being male or female.

You can see for yourself that empirically, in actual results, the probability of both dogs being male consistently comes out at around 1/3

nice spreadsheet, but we’ve got a bit more information now, we’re told that one of the dogs is male, thats a certainty.  100% male.  so whats the chances of the second dog being male and hence having 2 male dogs?  50/50.

i can see what youre doing/saying, but all your maths is from a point where we know nowt about them poor wet pooches, we now know that at least one is male.  you dont need a spreadsheet for that.

LOOK AT THE SPREADSHEET. 

That’s got my geek gene a purring.

we’ve got a bit more information now,

Yes but that information only lets you narrow down which scenario you are in. It doesn’t change the original odds.

The original odds are that if you do this 1000 times you’ll see “both boys” about 250 times and “mixed or both girls” about 750 times. Correct?

The information that at least one is a boy means you can ditch the “both girls” scenarios and you now know you are either in one of the 250 scenarios where they are both boys or one of the 500 scenarios where they are mixed.

Do the 50%ers still think the chance of all 3 being male is 50%?

Well I dont. But as I said my problem was with the wording, which was imo different from what he wrote in the explanation of why the answer wasnt 0.5.

It might make a little more sense if you think about the odds improving, not getting worse.

Before you have any information there is a 1/4 chance that they are both boys.

Once you know that at least one is a boy then you know there is a 1/3 chance they are both boys.

Yes but that information only lets you narrow down which scenario you are in. It doesn’t change the original odds.

but why do you need to keep the original odds the same?  the odds have now changed, as we now have more info.

The original odds are that if you do this 1000 times you’ll see “both boys” about 250 times and “mixed or both girls” about 750 times. Correct?

yes.