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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 [b]any[/b] 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, [u]that's[/u] 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.
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 [u]other[/u] one male?" or "is the other dog [u]also[/u] 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 [u]this[/u] 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...