Seeing as this is currently breaking Facebook…

your coins isn’t 1 in 3 you are taking tails out of the equation leaving 1 in 2

You can’t take the tails out to the equation.

The Monty Hall problem has three doors. We get a choice of doors. There are two options for each door. You choose one door and get a result. Should you stick or twist on you choice for your final and critical door? Heck maybe it should be a lesson on the wonder goats and the joy in abandoning mainstream materialism. But it has nothing to do with this. It is a red herring and there are a lot very clever/stupid/trolls on here.

Scrap that last post. I am getting it now. Convert is right. The head and tail combo will crop up 2/3 of the time if we discount the tail/tail combo. So if I ask if one is head then I am more likely to get a yes response, the tail/head or head/tail split is important.

After eliminating 2 tails, If I ask convert to confirm if one is a head and he says yes without confirming which one, and I can’t see them,

1/3 of the time the left coin will be a head and the right a tail.

1/3 of the time the right coin will be a head and the left a tail.

Only 1/3 of the time will both be heads.

your coins isn’t 1 in 3 you are taking tails out of the equation leaving 1 in 2

You toss two fair coins together, there are four possible outcomes, not three:

HH

HT

TH

TT

Probability of two heads is 1 in 4.  If you remove TT from the equation then 4-1=3 so it’s then 1 in 3.

This is the crux of the puzzle, people are conflating MF and FM as if they are the same thing.  They aren’t.  If you don’t believe me, sit down with a couple of coins and a notepad.

Try this. Take two coins. Toss one. Get the result. The toss the second what chances are there on the coins matching because the first is already chosen. The first coin cannot be retossed. That dog is out of Schrod8ngers box, go on, 2 coins. Try it it now.

Try this. Take two coins. Toss one. Get the result. The toss the second what chances are there on the coins matching because the first is already chosen. The first coin cannot be retossed. That dog is out of Schrod8ngers box, go on, 2 coins. Try it it now.

Dude, come back in the morning and give it another go. You’ll get there eventually.

Let’s really screw them over Cougar.

Its 2 in 3 that theres a boy and a girl.

So basically you should always swap for the cat as regardless of sex the shit will smell the same and you’re going to have to pick it up.

Try this. Take two coins.

Yup, you’re perfectly correct.  The chances of getting the same result on both coins is 50:50.  Same as getting a different result is 50:50.

Now, discard all the both-tails flips.  What are the chances now of getting both heads?

This is what the dog question asks.  We’re told that at least one dog is male, ergo there isn’t two females / tails.  Is it still 50:50?  How can you remove a permutation and still get the same odds?

You really did vote for Brexit didn’t you.

I would expect that there’s a strong correlation between fifty-percenters and fifty-two-percenters, TBH.  But let’s not derail the thread.

Scrap that last post. I am getting it now. Convert is right. The head and tail combo will crop up 2/3 of the time if we discount the tail/tail combo. So if I ask if one is head then I am more likely to get a yes response, the tail/head or head/tail split is important.

After eliminating 2 tails, If I ask convert to confirm if one is a head and he says yes without confirming which one, and I can’t see them,

1/3 of the time the left coin will be a head and the right a tail.

1/3 of the time the right coin will be a head and the left a tail.

Only 1/3 of the time will both be heads.

Excellent, my work here is done and I can go to bed with a warm fuzzy glow.

What would Jordan Peterson say?

Try this. Take two coins. Toss one. Get the result. The toss the second what chances are there on the coins matching because the first is already chosen. The first coin cannot be retossed. That dog is out of Schrod8ngers box, go on, 2 coins. Try it it now.

It sounds the same, it uses the same coins but you changed the method.

The head and tail combo will crop up 2/3 of the time if we discount the tail/tail combo.

Bingo.

My last post as it is late. Hope this helps. The shopkeepers wife is washing 2 dogs with gender neutral names, say Leslie and Lyndsey.

I will only buy these dogs if at least one of them is a boy. I am told that at least one of the dogs is a boy. I could purchase these dogs and have 1 of 3 equally likely scenarios knowing this info

1)Leslie is a boy and Lyndsey is a girl.

2)Lyndsey is a boy and Leslie is a girl.

3)Lyndsey and Leslie are both boys

So it is 1/3 chance both are boys.

Got it in the end cougar

You are all correct but this isn’t mathematical logic or statistical conundrum, nor does it have anything to do with flipping coins or opening doors. This is simply a matter of interpretation of the English language and subsequent assumptions made.
From the OP, the relevant part is, and I quote, “asks her if there’s at least one boy. She says yes.”

We don’t know whether the wife has checked one animal or both.

The 33 percenters are assuming she only checked one and are assuming MF and FM as different options because of this; whereas the 50 percenters are assuming both animals are checked and MF & FM are the same thing.

Maybe the 33 percenters are fundamentally lazy types who only do the absolute minimum amount of work whereas the 50 percenters are hard-working but inefficient types who like to dot all the “i’s” and cross all the “t’s”…

We don’t know whether the wife has checked one animal or both.

We don’t need to know.  The information provided here is “they aren’t both female.”  Observation does not change the probability distribution.

What are the odds before the question was asked?

What are the odds after the question was asked?

Whilst I know categorically that I am a lazy toad and that I seldom dot or cross anything it has got very little to do with the answer being right. Cos I am.

Cougar: you’ve just proved exactly the point I’m making…you’re just as guilty as jumping to assumptions as everyone else.

Your applying a very tight definition of logic which no doubt works in your world  but isnt the same as the real world…

😉

Your applying a very tight definition of logic which no doubt works in your world  but isnt the same as the real world…

That’s the point it’s a probability question in the real world the question would not be asked or then replied  like that, you’re changing the question and reply to get you answer.

That’s the point it’s a probability question

Is it? Where does it say that in the OP?

Again, assumptions which I’m guessing (assuming!) the mathematicians amongst us are getting their knickers in a twist about…

😉

I suspect something got lost in the translation because these types of riddles relay on very careful phrasing.  What this one wants you to do is:

Man will buy dogs if :

Out of MF FM MM FF only FF is false.  Therefore 3/4 odds he walks out with 2 dogs.  Out of MF FM MM, what are the odds of MM?  Why 1/3 obvs.

However the riddle is not written in a way that forces both MF and FM to be a possiblity. If it asked instead what are the chances the wife washed a male dog followed by another male dog then that forces that choice.

As it is written man will buy 2 dogs if Mx is true. M is true therefore the question becomes what is the probability that MM is true.  That is 100*1/2. There is nothing that forces the inclusion of both MF and FM as being false

Equally the coin toss example is only useful if you define if order is important.

What are the chances of getting at least one tail in a 2 coin toss – one out of two twice – 1/2*1/2 or TT TH HT HH = 3/4? or both?

For some good explanations of the Monty Hall Problem (including some nice diagrams) read the Wikipedia page.

The “trick” I had to get my head around for the dog problem is that the wife is not giving you as much information as you think she is. The natural assumption when you first read the problem is that she is telling you that dog A is male, and dog B could be either gender, hence 50% chance of two males. In fact she is telling you that dog A and dog B are not both female, which is a very different piece of information. As others have pointed out, you don’t actually know the gender of either dog, hence the three possible outcomes.

But the way it is written it doesn’t matter what dog A or dog B is.  All that is necessary is that one of dog A and dog B is male and what are the odds that dog A and B are both male.  Since we know one of dog A and B is male then it’s 1/2 that the other of the pair is male.

I suspect something got lost in the translation because these types of riddles relay on very careful phrasing.

It does which is why I use to struggle as I have dyslexia so didn’t get the wording.

But the way it is written it doesn’t matter what dog A or dog B is.

It’s because it is not specific that it matters.

Has the man bought the beagles yet?

Which beagle?

Jeremy

Is he about?

It’s because it is not specific that it matters.

As I said, careful wording.  I looked it up and the original version of this that goes with the answer being proposed doesn’t say “what is the chance there are two boys” it says “what is the probability that the other one is male”.  Something got lost in the translation allowing for both interpretations to be correct.

.

I suspect something got lost in the translation because these types of riddles relay on very careful phrasing.  What this one wants you to do is:

Man will buy dogs if :

Out of MF FM MM FF only FF is false.  Therefore 3/4 odds he walks out with 2 dogs.  Out of MF FM MM, what are the odds of MM?  Why 1/3 obvs.

However the riddle is not written in a way that forces both MF and FM to be a possiblity. If it asked instead what are the chances the wife washed a male dog followed by another male dog then that forces that choice.

As it is written man will buy 2 dogs if Mx is true. M is true therefore the question becomes what is the probability that MM is true.  That is 100*1/2. There is nothing that forces the inclusion of both MF and FM as being false

Equally the coin toss example is only useful if you define if order is important.

What are the chances of getting at least one tail in a 2 coin toss – one out of two twice – 1/2*1/2 or TT TH HT HH = 3/4? or both?

As I said, careful wording.  I looked it up and the original version of this that goes with the answer being proposed doesn’t say “what is the chance there are two boys” it says “what is the probability that the other one is male”.  Something got lost in the translation allowing for both interpretations to be correct.

There would be some merit in anything you typed if the question had said that the wife was asked to look at one of the beagles and one of the beagles only and tell her husband that beagle was male. But that’s not what was said. So your premise is gash.

From the blog post with the “”answer”

A shopkeeper says she has two new baby beagles to show you, but she doesn’t know whether they’re both male, both female, or one of each. You tell her that you want only a male, and she telephones the fellow who’s giving them a bath. “Is at least one a male?” she asks him. She receives a reply. “Yes!” she informs you with a smile. What is the probability that the other one is a male?

Reduced to its essence, the problem is asking this: “There are two dogs. One of them is male. What is the probability that the other one is also male?

I dont doubt the maths but the first says 1 dog is male, whats the probability another dog is male. Thats 0.5. If you have a baby thats male the probability of the next baby being male is also 0.5, the probability of having 2 male babies is 0.25.

In his second statement where he talks about the “essence” he adds the word “also” which utterly changes the meaning.

I dont doubt his maths but do doubt his language.

I dont doubt the maths but the first says 1 dog is male, whats the probability another dog is male. Thats 0.5. If you have a baby thats male the probability of the next baby being male is also 0.5, the probability of having 2 male babies is 0.25.

In his second statement where he talks about the “essence” he adds the word “also” which utterly changes the meaning.

I dont doubt his maths but do doubt his language.

You can grumble all you like (and paraphase to suit – nowhere does he say “1 dog is male, whats the probability another dog is male. ‘) but nothing will cover up that your answer on page one was wrong and your are now looking for excuses 😉

As I said, in the original wording of the problem the phrase used was “the other one” which introduces the necessary uncertainty to allow for the 3rd possibility.  I suspect you’ve read the solution but don’t really understand how it relates to the question because everything you’ve said doesn’t apply to the riddle as posed in the first post but does apply to the original phrasing of the riddle.

Although I would argue that even the correct solution is correct logically but wrong mathematically.  If we label them as known and unknown and the first is the known male then “the other” has a 1/2 chance of being female.  If the first is the known male then the “the other” has 1/2 chance of being male.  If the first is the unknown then “the other” has a 1/1 chance of being male.

1/2*1/2*1/1= the probability of “the other” being male is 3/4.  1/3 doesn’t account for the differential weighting of “the other” being the one we know 100% is male

Ah phrasing riddle and probabilities. Such fun. Oh how the brain and other people’s brain works .

I hope the buyer checks when he picks them up , as I had a workmate recently who got a cat called <span style=”font-size: 0.8rem;”>tinkerbell. </span>she recently discovered it’s actually a tom so had to change the name .. it’s now one confused cat

I suspect you’ve read the solution but don’t really understand how it relates to the question because everything you’ve said doesn’t apply to the riddle as posed in the first post but does apply to the original phrasing of the riddle.

Let you into a tiny secret – some of us havn’t needed to look at the solution. Scary isn’t it!

 some of us havn’t needed to look at the solution. Scary isn’t it!

Not scary at all.  If you think your solution is the answer to the riddle as posed is post 1 then it’s just wrong.  It is, however, the answer to the original riddle.  Right answer, wrong question not at all uncommon.

A woman hides behind a black door. She can appear at the door with a deal or no deal. What are the odds she will appear with a deal?