Seeing as this is currently breaking Facebook…

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, <span class=”bbcode-underline”>that’s</span> improbable. The odds of it being heads on the 50th toss are still 1/2.

It’s not the same as this scenario though! You already have some information about both tosses because you know they’re not both Tails. If the scenario was wife washing one dog upstairs whilst the other slept downstairs and she said “The one I’m washing is male but i don’t know about the one downstairs” you would get your 0.5.

For the wife to make her statement she has to have done either of the following:

– check dog 1 confirming it’s male

or

– check dog 1, see it’s female and check dog 2 confirming it’s female

this means you’ve changed the uncertainty in the gender of dog 2.

Back to the coins:

Toss a 50p and a £1 100 times each,note the outcomes of each in 2 separate columns, columns A and B

Taking the original question you can discount all rows with 2 tails, statistically this will leave 75 rows, this is what the wifes information allows us to filter to. 25 of those rows will be HH so the probability is 25/75 (0.3333)

Taking my 1 dog is downstairs version of the question is akin to just looking at the column for the tosses of the 50p, the wife says the dog she has is male so discount all the T from the 50p column. This will filter to 50 rows of those 50 rows 25 will correspond to a H on the £1. So the HH probability is 25/50 (0.5) – this is not the information given by the wife though!

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…..wait……. I know this one.

It is 50/50 right?

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

Not exactly.  As soon as you treat them as individuals that changes the outcome from if you treat them as each 1/2 of a pair.  It doesn’t matter if you know which individual is which as long as they are individuals.  Pierre’s example above is perfect.  Flip a coin, do fifty trials.  The odds on any given trial are bit dependant in the outcome of any other trial or condition. You don’t need to know if you are on the first trial or the 5 hundreth.  Flip a pair of coins any number of times and you will never get a trial where the outcome is one where the result of one coin can be considered independant of the other – this is true whether you know which coin is which or not

Always switch.
The plane takes off.
Remain.

This man talks sense.

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

…but, depending on the language of the question, you do. If the second question is “is the other one male?” or “is the other dog also male?” we have one definitely male dog set to one side and we’re only dealing with the sex of one dog.

If he only asks one question, “is one male?” means we have the sexes of two dogs to consider. And then it’s the simple punnet square thing and 1/3.

The 50/50ers are undermining their position by reading plain english words, and answering a different question.

Now the one about the dog being born on Tuesday, what is the probability that the other dog is male?  I’ll give you that one as one that is clearly not intuitive and is brain exploding (when using English words rather than conditional probability equations)

Assuming for the sake of argument that  @sbob’s approach is valid,

But it’s not valid, so the rest of your argument seems to be redundant.

I have two dogs.

The dogs are not both female.

What is the probability I have two male dogs?

That’s the original problem, without the window dressing of pet shop, etc.

That’s the original problem

No it isn’t.

I have two dogs.

Do you?  Or do you have a pair?  Makes a difference. That is how the riddle works.

Time to open the bar, I think

The original problem:

A man sees a sign in a window advertising two Beagle puppies for sale. He goes in and tells the shopkeeper he will only take the puppies if there’s at least one boy.

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?

The answer to this is 1/3, but sbob, I and many others have also been answering the other-worded versions of the problem to try and disambiguate. (is that a word?)

The answer to this is 1/3, but sbob and many others have been answering the other-worded versions of the problem to try and disambiguate. (is that a word?)

Other way round because remove all the preamble and go to the question

What are the chances there are two boys?

Vs

What are the chances the other one is male (also).

So, what is it?

A beagle

And another beagle

It’s a big building with patients, but that’s not important right now.,

Forget Monty Hall, this is the Monty Python argument clinic.

In all probability.

No it isn’t.

Do you?

Yes. Two dogs.

Yes. Two dogs

But are they 2 dogs or a pair of dogs?  Have you asked them?

All I know is that there are two dogs and they are not both female. That’s it. There’s nothing else.

There is also a lady washing them. I wonder what dog shampoo she is using? Is she cleaning them in the family bathroom or does she have an en suite?

Cromolyolly there is no word trickery, if you ignore the preceding comments for the wife you are asking a different question. In some instances the wife has to be telling you something about the second coin toss in order to give you that answer so it’s not the same as saying the result of 1 coin toss is influencing the result of the subsequent coin toss.

would you agree with the following:

to make the statement “one of the dogs is male” She must have either:

– checked one dog and seen it was male (if you got told this was the case then the chance of 2 males would be 50%)

– checked one dog and seen it was female so she checked the second dog which was male

Can you set up polls on here?

Now that this thing appears to be into its death throes it would be interesting to know how many are in each camp.

I would predict 2/3 to 1/3 in favour of 1/3!

It would also be interesting to know how many minds were changed one way or the other during the course of it.

It would also be interesting to know how many minds were changed one way or the other during the course of it.

… and in which direction.

word trickery may be an overstatement.  There is an art to writing these things, they must be vague but not ambiguous, they must hint at the valid solution without stating it openly enough that it is obvious.

I think you are missing the case where she checks both dogs just because it seems likea good idea and tells you one is male because that was the precondition.

But that really doesn’t help you solve the problem.  That is part of the art of writing these things too. Details which distract from the true information needed to arrive at the solution.

I think you are missing the case where she checks both dogs just because it seems likea good idea and tells you one is male because that was the precondition.

It still works if she always checks both dogs, would you agree she can’t simultaneously check the sex of both dogs at once? (Even if it’s just a shift of glance)

We have 4 options:

look at dog 1: male, look at dog 2: male – 25% chance

look at dog 1: male, look at dog 2: female – 25% chance

look at dog 1: female, look at dog 2: male – 25% chance

look at dog 1: female, look at dog 2: female – 25% chance – though we know this case cannot be true because of the information the wife has given

this leaves us with 3 equal probability outcomes – only one of which is 2 male dogs giving you 1/3 chance. From the wife’s perspective yes, as she glances from one dog to the next, the chance of that second dog being male is 50%. But this is not our reference point. The information we get is the partially filtered information obtained by looking at the gender of both dogs which allows us to arrive at a modified probability.

Sorry, but I just got caught up on all this and wanted to give it one more shot:

For the sake of the 50%ers, like sbob, who seem to be saying that the information doesn’t change the original odds then how would they cope with a much more straightforward example of a puzzle with developing information.

1) You are given FIVE cards, face down, from a normal 52 card deck.

2) The dealer tells you that ONE of them is definitely the Ace of Spades.

3) You turn over FOUR of them and don’t find the ace.

What are the odds that the last card is the Ace of Spades?

1/52 ?!?!?! 😂😂😂😂😂

..

It would also be interesting to know how many minds were changed one way or the other during the course of it.

… and in which direction.

My mind was changed, but I just didnt read it properly to start with.

For the sake of the 50%ers, like sbob, who seem to be saying

I’ve been very clear about my position, just that too many people are too busy thinking they are clever to try and understand what I have been saying.

If you pick up a dog and it is male, then the other dog is either male or female.
If you pick up a dog and it is female then the other dog is male.
MM, MF, FM, three options, 1/3rd. Never disputed this, the maths using this methodology is simple.

My position was whether or not this is using the information correctly as it is given in the conundrum. Using all instantaneous knowledge of the dogs from the wife’s “yes” it is arguably possible to come to a different conclusion as I have explained many times, rather than the step by step logic above which is not how we gain the information.

What I was baiting for was for someone to try and prove or disprove what I had said by expressing it mathematically, which would have led us down a much more interesting road, had anyone been able to.

There is no BODMAS for this question…

I heard a rumour on Facebook saying the shopkeeper had a vasectomy after his first boy was born. Effectively there’s zero chance of two boys. There’s two puppies but no one cares about them other than the shopkeepers wife. Social services took away their first son (recognizable by trademark skinhead, facial scar, clenched fists and pink high top sneakers).

Using all instantaneous knowledge of the dogs from the wife’s “yes” it is arguably possible to come to a different conclusion as I have explained many times, rather than the step by step logic above which is not how we gain the information.

So explain how that is different to the example I just gave with the cards that uses “step by step logic”

You are not applying the new information correctly. As demonstrated by Bayes Theorem and carefully explained by multiple stats experts.

I don’t really understand how you can comprehend Monty Hall but not this puzzle. There is a distinct overlap.

What I was baiting for..

i think the term is “trolling”

So explain how that is different to the example I just gave with the cards that uses “step by step logic”

You are not applying the new information correctly. As demonstrated by Bayes Theorem and carefully explained by multiple stats experts.

I ‘think’ what sbob is saying is the scenario to the point where the wife says yes is contrived/staged. Like a film director has setup the scene to end as it does by guaranteeing one of the dogs is male. This is done by throwing one male dog into the bath and one other dog of indeterminate gender so that when ‘action’ is called the wife will always answer yes.If you view it like this the answer will be 50%. He is ‘rigging the deck’ to the tune of 1/6th (difference between 1/3 and 1/2) to ensure the same result in the scenario every time it plays out. You and I view this as a scenario that is playing out as we observe it where the wife could have said no but didn’t. It goes without saying that we are right and he is wrong 😉

Still going huh? Yes if sbob snuck in into the pet shop and rigged the selection of dogs it’s 50%. I think it’s fair to assume he didn’t in the question as worded.

If someone rigged the section of dogs the probability is 100% cos they made sure it was a pair of boys at the start.

Prove me wrong

You are not applying the new information correctly. As demonstrated by Bayes Theorem and carefully explained by multiple stats experts.

Conversely, you are not applying Bayes Theorem correctly and I’m really not sure what “stats experts” we have here or what relevance they have to my point, though referring to Bayes is proof that you don’t

I can accept no-one is able to get my point and am perfectly happy with that but it does not mean it is invalid. 🙂

I think the term is “trolling”

That title goes to Cougar for deliberately posting a thread that he hoped would be contentious.

Are we all finally agreed on 1/3 now? If not here is why I changed from thinking 1/2:

Forget coin tosses, imagine 100 wives and baths (ahem). One dog is assigned to each bath randomly = 50M, 50F.

Second random dog gets wet, so now = 25MM, 25MF and 25FM, 25FF (100M, 100F)

Phone call confirms it isn’t one of the last 25 baths, so must be one of the remaining 75 baths.

Only a third of these are both dog dogs.

The end.