Seeing as this is currently breaking Facebook…

It’s maths probability problem

Why’s it written in words and not numbers then? Stick to your own ways mathematicians!

One is basically “We have determined Dog A is male; what are the odds that Dog B is male?” Answer: 1/2.

If that is your interpretation then you have misinterpreted the problem.

@Pierre

Neither wording of the question says anything more than that one dog is male.

TIME FOR EXPERIMENTATION!!!

Take 2 coins (“two Beagle puppies”)

Tails is male, heads is female.

Toss both coins say 100 times, recording your results each time.

Strike out all results where you get 2 heads/females (“asks her if there’s at least one boy. She says yes”)

Work out the proportion of double tails/males versus the mixed results (heads/tails or tails/heads)

“What is the chance there are two boys?” – you have your answer.

Has anyone asked the dogs as what they identify as?

“We have determined Dog A is male; what are the odds that Dog B is male?” Answer: 1/2.

That isn’t quite right though is it? Dogs A and B are selected from a pool of two dogs so dog A becomes dog A only if it is male. If it isn’t then you effectively choose dog B to be dog A. That means dog B is less likely to be male as it is either the rejected female dog or a 50/50 random dog. You still get 1/3 for dog B being male. Sorry i probably haven’t written that very clearly.

One is basically “We have determined Dog A is male; what are the odds that Dog B is male?” Answer: 1/2.

The other is “We have determined that there are four equally possible outcomes and have eliminated exactly one. What is the probability that Dog A and B are both male?” Answer: 1/3

This.

Read my post above with the two explanations. The two interpretations give situations that are two different mathematical problems:

One is basically “We have determined Dog A is male; what are the odds that Dog B is male?” Answer: 1/2.

This is correct, and what we’ve all been saying for 15 pages.  However, nowhere in the puzzle does it suggest that we have identified one specific dog, at all.  The puzzle asks “is there at least one boy?”

Any solution that starts with “take one dog…” is inherently wrong because it’s assuming knowledge that we are not given.  Any solution that starts with “the wife checks the first dog and…” is inherently wrong because it’s assuming knowledge that we are not given.  50:50 is the totally correct answer to an entirely different question.

This has been explained in lots of different ways and at this point sbob and others have just chosen to ignore the fact that they are wrong. There is only one answer and that is 1/3.

The fact that those saying 1/2 can’t demonstrate it with any logic outside of “the wording” or “pick up dog a, then dog b” speaks volumes, especially since it’s already been stated a number of times, that any interpretation where the male dog is known leads to 1/2 probability.

The questions isn’t poorly worded, it’s very specifically worded.

Edit: Urrrghhh those formatting marks 🙁

However, nowhere in the puzzle does it suggest that we have identified one specific dog, at all.

And yet that is precisely what you have to do to arrive at the four freedoms, err outcomes. If you don’t do that then you can’t achieve M/F and F/M as the separate outcomes you need to reach the 1/3 conclusion.

Otherwise you return to the situation where either of the dogs is male, the other might be.

Right sbob, but the question takes place before we’ve checked which is male. We only know 1 is, with the information provided by a 3rd party.

As it could be either or both, we have to account for all outcomes.

This has been explained in lots of different ways and at this point sbob and others have just chosen to ignore the fact that they are wrong. There is only one answer and that is 1/3.

It has been explained in exactly the same way just with added levels of obfuscation. Stating someone is wrong is not proof that they are. I repeated several times that I fully understand the 1/3 maths but it is down to the interpretation which I have explained.

We only know 1 is, with the information provided by a 3rd party.

We know one is, but we also know the other might be.

The questions isn’t poorly worded, it’s very specifically worded.

I asked ages ago how those suggesting that it’s badly worded would reword it to make it clearer, but oddly no-one answered.  As I said, I tried myself to tidy up the wording before posting as I felt the version I’d started with was (deliberately?) misleading.

I think Drac sums it up well.

Not yet Drac, just five more pages.

Not yet Drac, just five more pages.

I’ve only two and a half hours til work, make ’em count.

It’s still going? Absolutely glorious 😀

Top work Cougar 😆

Top work Cougar

mattoutandabout would be proud!

I’ve only two and a half hours til work, make ’em count.

Didn’t stop you last night.

I asked ages ago how those suggesting that it’s badly worded would reword it to make it clearer

I’m not suggesting it was badly worded and it’s clear you did tidy up the wording from the WIki wording. However, if you wanted to make it clearer you could have explicitly stated that both Beagles were placed back in a pen together and were mixed up with no clue as to which Beagle had been identified as the Male.

But you’d *never* have got to 15 pages that way.

Ok, so in the interest of clarity: I have just tossed 2 coins 50 times.

The results are:

HH:  17

HT:   18

TT:   15

I mean what are the chances of that?

Wait, I’ve got a solution which satisfies both camps.

Assuming for the sake of argument that @sbob’s approach is valid, he’s asserting (paraphrasing so correct me if I’m wrong) “we know one dog is male, therefore the other is male or female, so the likelihood of the other being male is 50:50 / 1 in 2.”  Now.

we know one dog is male, therefore the other is male or female

This bit seems pretty inescapable, yeah?  But is it logical to then conclude,

so the likelihood of of the other being male is 50:50 / 1 in 2.

❓

We’ve taken this as a valid conclusion for 15 pages, and it turns out, no, it’s an erroneous logical leap.

We know the other is male or female, sure.  But we also know that the probability of pairings are not evenly distributed.  From Graham’s spreadsheet we can see demonstrably that the likelihood of a M/F pairing is twice as likely as a M/M pairing.

So even using sbob’s logic, the answer is “we know one dog is male, therefore the other is male or female, with the given distribution the likelihood of the other being male is 33:67 / 1 in 3.”

*mic drop*

Didn’t stop you last night.

Boo to you, fun-sponge.

tl;dr

it is 50:50 right?

From Graham’s spreadsheet we can see demonstrably that the likelihood of a M/F pairing is twice as likely as a M/M pairing.

I wonder how that is even possible. 😂

Oh and it has been mentioned but it again it was ignored.

I asked ages ago how those suggesting that it’s badly worded

Apologies cougar, I didn’t see that.  Badly worded is probably unnecessarily perjorative.  The original wording goes out of it’s way to avoid using any language which suggests the dogs are 2 individuals and uses language which subtlety suggests regarding them as part of a pair.  Language which invites treating as individuals validates the 50% answer as legitimate.  Putting chances of two boys at the end does that.

It is also the reason why people are using arguments which seek to identify the “first dog” and the “second dog”.  The 1/3 ers which suggest naming the dogs or using individual identifiers are actually undermining their own position.

It is the difference between flip one coin twice, what are the odds of TT and flip a pair of coins, what are the odds of TT.

Or to put it another way.  If the dog the wife is bathing is male, what are the chances of the other being male vs. if one of the pair of dogs in the bath is male, what are the chances of the pair of dogs in the bath being male.

*mic drop*

That’s a terrible effort.

All you are saying is Y is wrong because X is right. You haven’t added anything. And we know X is wrong because it gives us impossible answers.

It is the difference between flip one coin twice, what are the odds of TT and flip a pair of coins, what are the odds of TT.

Are you sure about that?

We’ve taken this as a valid conclusion for 15 pages

Well I don’t think it is correct and tried to refute it and hour or so ago.

That’s a terrible effort.

Care to point out the flaw in my logic?

Actually, forget it, I give up.  You’re a lost cause.  You’re going to argue that black is white until you get run over on a zebra crossing.

nah black is only white for 1/6 of the cases (and white is black for 1/6) 😉

Care to point out the flaw in my logic?

Gladly.

You’re reverting back to a calculation that provides us with wrong answers.
Then instead of changing the calculation, you’re still just fudging the answers.

Actually, forget it, I give up. You’re a lost cause. You’re going to argue that black is white until you get run over on a zebra crossing.

You reap what you sow.

It’s not me that is the one with the entrenched way of thinking. 🙂

Can we get sbob on the original Monty Hall problem next?

Pigeons repeatedly exposed to the problem show that they rapidly learn always to switch, unlike humans

I asked ages ago how those suggesting that it’s badly worded would reword it to make it clearer, but oddly no-one answered.

In the explanation link posted on I think the first page it uses the word “also” which I think makes it much clearer. That would have led me to 0.25 and then I might have read it better, cant guarantee I’d have got to 0.33 though. I initially read it quickly saw that we had one dog as male and another dog so it must be 0.5. I’ll be honest i get easily bored by things I know are trying to trick me for some reason. I was thinking it was a question like babies if you have had 8 sons whats the chances of another son etc.

Are you sure about that?

Absolutely.  The final answer will be the same but the valid methodology to get there will be entirely different.

It’s a paradox… pair o’dogs!

Sorry.

In other words it depends whether you treat them as a pair of dogs, or two individual dogs. If they are a pair of dogs, you’ve got 4 possible combinations: FF, FM, MF, MM with equal probability. Eliminate one of those (FF) with the right question and you’re left with a 1/3 chance of MM being the case.

But if you treat each dog as an individual case, and the odds of any dog being male as 1/2, it doesn’t matter how many dogs you have already identified as one gender, you’ve excluded them from your analysis. If you’ve already removed 20 dogs that were all male and you still have one dog remaining, the odds of it also being male are 1/2.

To return to the coin tossing thing (sorry, I skipped those pages), it’s like arguing the probability of my next coin toss being heads – it’s still 1/2, it doesn’t matter how many times I’ve previously played those odds (assuming the coin is equally balanced. If I toss a coin 50 times and the first 49 times it comes up heads, that’s improbable. The odds of it being heads on the 50th toss are still 1/2.

Can we get sbob on the original Monty Hall problem next?

Always switch.
The plane takes off.
Remain.

HTH.

Language which invites treating as individuals validates the 50% answer as legitimate.

The 1/3 ers which suggest naming the dogs or using individual identifiers are actually undermining their own position.

Only if you know which individual is male. Which you don’t.