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but to me, if i am physically presented with these two dogs and am told one is a boy and the other is unknown i only need to physically check one of them to find out what sex both of them are…
How would you do that? You may need to check both to know the sex of both. Either the one you check is female, in which case you do know the sex of both, or the one you check is male, in which case you have no way of knowing the sex of the other without checking it.
You have one dog in each hand
Why?
You have a dog that is male and a dog that is male or female.
In the 1/3rd scenario there is a female dog which doesn't exist in the question. There is no female dog. There is a male dog, and a dog that is either male or female.
aaaargh! The ‘maths answer’ is the real world logical answer! Never play poker for money.
played last night and won £25 actually!
I say real world and maths answer, but to me, if i am physically presented with these two dogs and am told one is a boy and the other is unknown i only need to physically check one of them to find out what sex both of them are…
So if you check one and it is a boy, what sex is the other one?
copy text, click the speech marks in the reply box which indents the cursor an inch or so. paste. then hit return twice to get cursor to start your reply.
ah ok like that thanks spunk
if i am physically presented with these two dogs and am told one is a boy and the other is unknown i only need to physically check one of them to find out what sex both of them are…
This would only help you if you found it to be female...
Edit: missed a couple of posts there...
Why?
You have a dog that is male and a dog that is male or female.
Because doing it with hands, as I suggest, is a good way to make it clear when you are applying knowledge you do not have.
You know at least one dog is male, but you don't know which one. That is important.
There are two routes to a mixed pair and only one route to get two males. The mixed pair is twice as likely.
Why?
You have a dog that is male and a dog that is male or female.
In the 1/3rd scenario there is a female dog which doesn’t exist in the question. There is no female dog. There is a male dog, and a dog that is either male or female.
For the sake of the explanation, let's say the dogs are called Rover and Spot.
From the information we have, we know that Rover and Spot are not both female. They could both be male or they could be a male and a female.
Three possibilities therefore exist:
1) Rover male, Spot female
2) Rover female, Spot male
3) Rover male, Spot male.
Assuming a random dog has a 50/50 chance of being male or female, that means the three possibilities are each equally likely to be the true position, ergo the probability of it being option 3 is 1 in 3.
You know at least one dog is male, but you don’t know which one. That is important.
It's only important if coming from the false premise that one dog is female which you need to get your four outcomes and your 1/3 conclusion.
Neither dog is female.
Edit, dodgy forum double post
I say real world and maths answer, but to me, if i am physically presented with these two dogs and am told one is a boy and the other is unknown i only need to physically check one of them to find out what sex both of them are…
No no no no no no. Only if you are told which one is a boy. Anyway that not the question that the thread's about.
if i am physically presented with these two dogs and am told one is a boy and the other is unknown i only need to physically check one of them to find out what sex both of them are…
if you are physically presented with them, and told that one is a boy *but not told which one*, and you then look at the back end of one of the dogs and it's a boy dog, you still don't know for sure if there's one or two boys. You might have checked the same dog, you might have checked the other, but you don't know.
knowing which dog makes all the difference.
if the wife said "the first dog I checked is a boy, I didn't look at the other one" then it's 50:50 on the other.
BUT THAT'S NOT WHAT THE OP SAID,
Assuming a random dog has a 50/50 chance of being male or female
You're assuming that both dogs have a 50/50 chance of being either male or female and then discarding the F/F option to give you 1/3.
Trouble is the question states that the probability of one dog being male = 1.
1x0.5=0.5
0.5x1=0.5
No, probability of *at least one dog* being male = 1. Which is a different thing.
"one dog being male" can only happen one way
"at least one dog out of two (and maybe both) being male" can happen three ways.
You have a dog that is male and a dog that is male or female.
No, you don't. You have two dogs and you know that they aren't both female. You do not have one of anything at any point in this riddle.
You claim to know that "one dog is male" so if I were to show you a photo of the two dogs would you be able to point to the known-male one? You're including extra information which you don't have.
Neither dog is female.
This isn't quantum physics. Both dogs have a gender.
Given the 50:50 binary cisgendering there are only four possible equally-likely scenarios when you don't know any of the genders. At that point the chances that you are in the scenario with two male dogs is 1 in 4.
Once you find out that at least one of the dogs is male then you can narrow down which scenario you could be in by ruling out the all-female scenario. The chances that you are in the scenario with two males dogs increases to 1 in 3.
If that doesn't work for you then I'd urge you to look at the simulation on my spreadsheet where it empirically shows this to be true by simulating 1000 cases.
Cougar….tut tut. I have now read the solution linked to at the bottom of page 1 and you changed the genders of the shop keeper and the washer of dogs when you posted in the OP. Gender stereotyping right there with your male shop owner and female skivvy dog washer.
I really haven't, that's how it was presented when I came across it. (I'd never have done that anyway, I actually considered removing all references to the gender of the dog-washer).
You’re assuming that both dogs have a 50/50 chance of being either male or female and then discarding the F/F option to give you 1/3.
Trouble is the question states that the probability of one dog being male = 1.
1×0.5=0.5
0.5×1=0.5
This is now a personal journey of discovery for you to establish why that is not the correct answer. I am afraid we are long past the point where you being right is a possibility.
Go flip some coins. Flip two together and discard any where you get two tails. Count how many valid flips (discounting the two tails) it takes you to get 10 double heads. I'll eat my hat and you'll win the internet if it does not take you somewhere between 27 and 33 attempts. At that point come back and ask us again to explain how it works.
I don't know how often I can explain the same thing in a slightly different way, but.
Consider these three situations (or read the previous pages):
1: I've just flipped two coins. What is the probability that both are heads?
2: I've just flipped two coins. Looking at them I can see that the one on the left is heads. What is the likelihood that both are heads?
2: I've just flipped two coins. Looking at them I can see that at least one of them is heads. What is the likelihood that both are heads?
In case 1 I don't think anyone is suggesting that the answer is anything other than 1 in 4 (and if they are they should probably step away from the thread).
In case 2 we prescribe a value to a specific coin, at which point it becomes irrelevant to the value of the other one. The other coin is either H or T, so the probability of two heads is 50:50.
In case 3 we know one is a head but not which one. With two coins, there are three permutations which could give us a tableau containing at least one head. So the chances of getting two heads is one in three.
Pretty much everyone who is saying the dogs answer is 50:50 is interpreting it as though it's the second scenario here. But there's nothing in the riddle to say that one specific dog is male, just that there's at least one somewhere in the mix. It's outside information which they're erroneously including. One of the dogs is black and the other is brown, can you tell me which is the male one? If you can't then it's demonstrably scenario 3 not 2.
You claim to know that “one dog is male” so if I were to show you a photo of the two dogs would you be able to point to the known-male one? You’re including extra information which you don’t have.
I do know that one dog is male.
I do not know which dog is male. You're ascribing extra information to my claim that I haven't made.
I'm more of as cat person really.
I’ll eat my hat and you’ll win the internet if it does not take you somewhere between 27 and 33 attempts.
Possibly a little precise to wager for pilleusphagia, but yes, it should be roughly around those figures.
That is very much what my speadsheet does: It flips two coins/genders/whatevers 1000 times and examines the results.
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?
I've just re read it and the answer is 1, is at no point on the question stated "1 dog is male" nor is it asked "what are the chances both dogs are male" it's just inference.
Clearly there is at least one boy, clearly there are two boys, i can see them from my desk. The question has nothing to do with dogs or maths it's an obscured logic problem with a very simple answer but very good distraction.
The dogs are in fact not beagles but red herring and we're all wrong (unless this has already come up in pages 2-6, in which case i apologise).
That is very much what my speadsheet does: It flips two coins/genders/whatevers 1000 times and examines the results.
[whisper] Yes, I know. I love your spreadsheet. I'm going to be thinking about as I go to sleep tonight in a naughty way. But - monkey see, monkey do, learning through doing etc I think we have more chance of him coming round if he does it himself [/whisper]
Gosh, aren't these threads irritating? A group of people that have the right answer (and are sure of it) and another group of people who are completely wrong but are equally sure they're right.
Have we mentioned Dunning-Kruger yet?
It's no wonder we've got a bloody Brexit on our hands.
Go flip some coins
Why is everyone insisting on using a model that is proven to give an impossible outcome?
All your coin tosses/spreadsheets do is prove that they are not applicable to the problem.
this has to be deliberate?
2: I’ve just flipped two coins. Looking at them I can see that the one on the left is heads. What is the likelihood that both are heads?
Bad cougar.
Knowing you've flipped two coins before revealing the result means the probability of the second is determined. So it's a 1 in 4 chance the rh coin is a head.
If you flip a coin get a head, then flip a second coin the chances of a head on the second coin are 1 in 2 as it's wholly independent of the first.
The whole reason the problem in the op is a problem is down to when the probability is locked in.
Why is everyone insisting on using a model that is proven to give an impossible outcome?
OK, go to the dog pound, ask them to pick two random dogs out for you and do the same experiment (check they have roughly 50:50 girls to boys first). Repeat it 100 times and see the results. Its easier with the coins but the result will be the same.
ask them to pick two random dogs
They're not random in the question.
OK, go to the dog pound, ask them to pick two random dogs out for you...Repeat it 100 times
And if you think that includes an "impossible outcome" then get them to only tell you the results that include at least one male dog.
Spoiler: you'll get roughly 25 both-male pairs and roughly 50 mixed pairs.
And if you think that includes an “impossible outcome” then get them to only tell you the results that include at least one male dog.
So you want to use a model that gives incorrect answers, but then to just disregard those to fit your own agenda?
Do you work for the government?
Spoiler: you’ll get roughly 25 both-male pairs and roughly 50 mixed pairs.
Only by incorrectly using a 50/50 scenario that doesn't exist and discarding the answers you don't like.
Knowing you’ve flipped two coins before revealing the result means the probability of the second is determined. So it’s a 1 in 4 chance the rh coin is a head.
I'm not sure as I follow what you mean?
go to the dog pound, ask them to pick two random dogs out for you and do the same experiment
About 33% of the time I got both heads and the RSPCA are on the phone.
I don't think this is a modified Monty Hall problem, because you aren't selecting a dog. This leaves the known M against an unknown dog.
I'm on the 1/3 side and the best way I think about it is that MF and FM are important and different because you don't know which dog the wife picked up first and if she needed to pick a second one up.
Thus of the 4 combinations there are, only 1 is discounted by the wife leaving 3 equally likely outcomes.
That is, M first, or F then M, or F then M.<span style="font-size: 0.8rem;">That means MM is 1/3.</span>
Or alternatively, it's not the same as saying there is Dog A and Dog B. Dog A is M, what is the likelihood that both are M? As this eliminates FM and FF from the 4 possibilities.
Gosh, aren’t these threads irritating?
Nope, they’re great. One of the threads with the most participation I’ve seen for ages. A bit of debate, a few laughs and irritating people who just have to involve brexit in every bloody thread 😉
Hate to say but I think sbob is lost to us.
Hate to say but I think sbob is lost to us.
I understand the maths you are using, I'm questioning whether or not you should be using it.
Your coin toss models give outcomes that do not fit with the question, yet you insist on using them. Do you disagree with that fact?
I fear that the time may have come to give up on this one.
Do you disagree with that fact?
Yes.
Before you have gained information you can be in any scenario. At that point all four outcomes are equally likely. As represented by the coin tosses, or spreadsheet results, or whatever.
Once you gain the information that they are not both girls then you can remove one of the possible outcomes as a possibility. But the remaining three outcomes are still equally likely and only one of them is "both boys".
Do you disagree with that fact?
Vociferously.
If you understand the maths used by the 1/3 gang then hopefully you can see that all the information in the phone call does is discount the two female option. That is exactly like tossing two coins and discarding the two tail option when it occurs. The scenarios are identical.
Keeping it to dogs, but ickle little models of dogs...I've got two models of dogs - they could be ickle boy dog models or ickle girl dog models. I have one in each hand out in front of me. I know what gender each is but you don't. You ask me one question "is one of your model dogs a boy?" - I chearly reply yes! Because it is true. You then ask me to show you.
Reality one....I open my left hand and show you a boy dog and open my right hand and show you a girl dog
Reality two....I open my right right hand and show you a boy dog and my left hand and show you a girl dog
Reality three...I open both my hands and both have got a boy dog.
It took 3 versions of events to show you the options of what could have been hidden in my hands. Ergo 1/3.
There are only still three outcomes if you start off with the wrong four.
We’ve been going about this all wrong people. How have we got this far and not used bike parts to explain it?
There are only still three outcomes if you start off with the wrong four.
So explain what you think the possible outcomes are when using convert's ickle little models of dogs?
Your coin toss models give outcomes that do not fit with the question
No. The coin toss model doesn't fit with your answer. Which is kinda the point.
There are only still three outcomes if you start off with the wrong four.
Huh?
they could be ickle boy dog models or ickle girl dog models.
That's the same error I've already pointed out.
It's not really the Monty Hall problem but it might be related in that it sounds intuitive.
For Monty Hall it'd be more like you and the dog washer know there are 3 dogs, and both know exactly 1 is male. You say "I'd like the one nearest the plughole", the dog washer knows which of the 3 is the male one, and says "the one furthest from the plughole is a bitch... do you want to stick with the first choice or perhaps change your mind and have the other one?"
Since you really want a male one, you're not going to choose the one you know is female, but you can double the chance of getting the male by correctly choosing from the other 2.
There must be a video on Numberphile youtube channel explaining this? They probably did it for the general case too with numbers other than 3.
I just went out riding. It was ace.
no dogs, no coins, no (awesome) spreadsheets. It was *ace*
(this post is adding about as much to the substance of the thread as many other offerings, so i though it fair game to report).
That’s the same error I’ve already pointed out.
Explain how you think you pointed it out again? Humour us.
No.
If I toss two coins I could get HH or TT. This is not applicable to the scenario in the question.
The coin toss model doesn’t fit with your answer.
Because it is wrong.
Which is kinda the point.
Precis, as they say in Sweden.
Okay, how much do we actually agree on @sbob?
Do you agree that without any extra information about gender, the chances of randomly being in a male-male pair scenario is 25%?
For Monty Hall it’d be more like you and the dog washer know there are 3 dogs, and both know exactly 1 is male. You say “I’d like the one nearest the plughole”, the dog washer knows which of the 3 is the male one, and says “the one furthest from the plughole is a bitch… do you want to stick with the first choice or perhaps change your mind and have the other one?”
Where’s my giant goat gone? Did the nasty American lady shoot him? 😕
Do you agree that without any extra information about gender, the chances of randomly being in a male-male pair scenario is 25%?
Absolutely, but this isn't the case in the OP.
Absolutely, but this isn’t the case in the OP.
It's the case in the first half of the OP though, yes? Before we gain additional information?
sbob: If we exclude FF from the outset then there are still 3 possible outcomes.
I don't know how you can't see the difference between these two scenarios:
An owner gives you a dog and says this one is male. He gives you a second and says what is the probability that both are male? <span style="font-size: 0.8rem;">This is 50% because only 2 options are available to you at this point. MF or MM.</span>
An owner gives you two dogs and says at least one of these is male. He says what is the likelihood of both of them being male? This is 33% because 3 options are available to you. MF, MM and FM.
And then with the additional information, we rewrite the model. We don't just fudge the results of the old one that we now know is not true.
Actually I've just had a lightbulb moment in how sbob (and maybe others) have come to their answer. Still got it wrong mind, but it makes more sense to me now.
Imagine your were a film director and wanting to set up the scene. Two dogs in a bath, bloke on the phone, woman fondling the dogs. You know how the scene is going to play out and you know when the woman is asked about the dogs she is going to say at least one of them is a boy. So you lob in a male dog into the bath so she can guarantee answering the question truthfully and then find another dog at random and lob that in there too. Then you say action and watch the scene play out. From that point there are only two outcomes - the other dog being male or the other dog being female.
If that's how you are viewing it I don't actually know how to help you other than you can't 'cheat' and go into the scene from that point. You have to start from the beginning and work forwards.
3 options are available to you. MF, MM and FM.
There is no female dog, there is only a dog that is either male or female, we just don't know!
I cannot for the life of me understand why anyone would think that four minus one equals two.
The four has ceased to exist.
And then with the additional information, we rewrite the model.
Rewrite it to what?? The chances of being in any of the remaining scenarios remains equal.
We don’t just fudge the results of the old one that we now know is not true.
It's not a fudge, it is how you calculate odds with new information.
e.g. What are the chances of me rolling a 6 and then another 6 on a fair six-sided die? 1 in 36 right?
36 possible outcomes and only one is the winner.
But once we know that my first roll was successful then we can discard 30 of those possible outcomes and we are left with just six, where only one is the winner.
There is no female dog, there is only a dog that is either male or female, we just don’t know!
Jesus wept. There are two dogs which are either male or female, we just don't know. What we do know is that the number of male dogs altogether is non-zero.
You're doggedly (ho ho) clinging to the notion that one specific dog is known to be male. It isn't. The male dog could be the other one.
There is no female dog, there is only a dog that is either male or female, we just don’t know!
Is that dog A or B. The only personal people who know are the dog washer or the shopkeeper.
We don’t so it makes 3 options.
Jesus wept. There are two dogs which are either male or female, we just don’t know.
Incorrect. We know one of them is male and the other is either.
You're stuck in a 50/50 loop that no longer exists.
We don’t so it makes 3 options.
No, you have one perfectly possible option and two mutually exclusive options. That is effectively two options. I can't believe you guys are letting these guys win with this disingenuous attempt at statistical sleight of hand.
Is that dog A or B?
Who knows? It doesn't matter.
50% chance. She has one dog thats a boy its in her hand as she talks on the phone- the other dog we don't know so 50/50 chance.
This thread is pure popcorn.
Chapeau sbob.
We know one of them is male and the other is either.
Okay so pick up one of those dogs at random then.
If it is a female then the other one must be the male. (1 possible outcome)
If it is a male then the other one can be either male or female. (2 possible outcomes)
Only one of these 3 possible outcomes is male-male.
I’m not sure as I follow what you mean?
Mea culpa, i think...
Incorrect. We know one of them is male and the other is either.
Which one?
Who knows? It doesn’t matter.
Yes, yes it does. If you know that one specific dog is male rather than one out of two then it changes the probability.
Let me put it another way. You say one dog is male and the other is either. Fine, that then gives us a probability. But then there's also the possibility that the other dog is male and your one dog is in fact female. You're ignoring this probability which is why you're getting a higher figure.
Imagine the dogs being washed are a Great Dane, and a Yorkshire terrier. Can you tell them apart?
are you telling me that a male yorkie and a female Great Dane is *exactly the same* as a female Yorkshire terrier and a male Great Dane?
The dog washer didn't tell us if she's found a male yorkie or a male Great Dane, just that she found a male. We didn't ask, she didn't tell us
so we have a male dog, either big or small, could be either.
now we have THREE OPTIONS! Male yorkie female GD, two male dogs, or female tiny pooch and male enormous one.
still sure that the MF and FM cases are the same?
Okay so pick up one of those dogs at random then.
But they are not random.
Actually I’ve just had a lightbulb moment in how sbob (and maybe others) have come to their answer. Still got it wrong mind, but it makes more sense to me now.
Imagine your were a film director and wanting to set up the scene. Two dogs in a bath, bloke on the phone, woman fondling the dogs. You know how the scene is going to play out and you know when the woman is asked about the dogs she is going to say at least one of them is a boy. So you lob in a male dog into the bath so she can guarantee answering the question truthfully and then find another dog at random and lob that in there too. Then you say action and watch the scene play out. From that point there are only two outcomes – the other dog being male or the other dog being female.
If that’s how you are viewing it I don’t actually know how to help you other than you can’t ‘cheat’ and go into the scene from that point. You have to start from the beginning and work forwards.
Yup, exactly! Except from the point the question is being asked the scene has been played out. We know one dog is male, we are simply being asked what the probability is that the other is as well.
FWIW I agree with the 1/3 working but this is just getting too Orwellian for me as I seem to be double thinking it. I did take a knock to the head yesterday (not an actual joke, smashed helmet and everything) and can see how both outcomes seem likely. I blame the language and the assumptions used (isn't that the whole point?) in the setting of the problem.
1/2ers:
Say a dog has 2 puppies one after the other. There is only 1 way the puppies can both be male. The first puppy has to be male (P=0.5), and the second puppy also has to be male (P=0.5). The overall probability is the sum of the two probabilities (0.5 x 0.5 = 0.25). There is also only 1 way the puppies can both be female. The first puppy has to be female (P=0.5), and the second puppy again has to be female (P=0.5). And again the overall probability is the sum of the two probabilities, (0.5 x 0.5 = 0.25).
We have accounted for so far a probability total of 0.25 + 0.25 = 0.5. The only remaining situation (1 male and 1 female) must account for the remaining 0.5 probabilty (since the total must be 1.0). This makes sense mathematically, and it makes sense conceptually because there are 2 ways of having a male and a female puppy and not 1 as was the case for MM and FF. The dog can have the female first and then the male, or have the male first and then the female.
Chapeau sbob.
Cheers!
I'd say I'm here all night but I've got to go to work.
Options are
MM
MF
FM
but surely if you include MF and FM you should also have MM and another MM
But they are not random.
I'm reformulating the model with the new information, exactly as you suggested doing.
Your population consists of two dogs and we know that there is at least one male.
Pick up a dog.
If it is a female then the other must be the male. (1 possible outcome)
If is a male then the other could be male or a female. (2 possible outcomes)
Only one of those 3 outcomes is male-male.
Imagine the dogs being washed are a Great Dane, and a Yorkshire terrier
For good reason, the OP clearly states they are both Beagles.
but surely if you include MF and FM you should also have MM and another MM
No. A perfect distribution would give:
MM: 25%
MF: 25%
FM: 25%
FF: 25%
Chance of a mixed-pair: 50%
Chance of single sex pair: 50%
Pick up a dog.
But we haven't picked up any dogs. You're still ascribing the old model.
We have one dog that is male, we'll call him Ishmael, and we have one dog that is either, we'll call them Leslie.
I think where you are going wrong is trying to split the new information into little bits you consider one at a time. That doesn't happen. You gain all the new info all at once.
It doesn't matter if Ishmael is the puppy on the left or the puppy on the right, the other puppy is Leslie.
I can see why you'd do it mind.
How do you know Leslie isn't male and Ishmael is female? Why have you ruled that out as an option?
You're trolling us, aren't you.