Alex_P: 2 is gambler's fallacy. You're saying "this particular pair of puppies contained a male, so I must create an arbitrary system for generating puppies that puts a male in every pair."
No it isn't--gamblers fallacy is that luck going one way will be balanced out by luck going the other way very soon. Gamblers fallacy would be that because we got one male, we need to construct a system that makes it more likely that we get a female for the next result--which is what you guys are going--rather than a system where we've got an equal shot at getting a male or a female next.
You can't describe order from information that we don't have. Saying "the next one" implies that you know the male dog is first. You can't place him first because you don't know who he is. I don't even think the rules of probability problems allow you to change ordering after a problem is already set up. Sparky and Othello have one set order. It doesn't matter if it's Sparky and Othello or Othello and Sparky. Consistency is important.
EDIT: Think about what P(1 Male and 1 Female) *really* is, and don't just fill in values from above, and remember: you don't add probabilites of two events together to find out the probability of them both occuring, you multiply them, otherwise the chances of two male dogs would be P(1 Male and 1 Other Male)=P(.5 + .5)=P(1) which is obviously wrong, you agree?
What are the chances of a male? .5 right? What are the chances of a female? .5 again, right? So shouldn't P(1 Male and 1 Female) actually be:
P(1 Male and 1 Female)=P(.5 x .5)=.25
and from there I think you'll see why the chances are actually 50/50.
No, I don't agree.
P(Anything) == 1 (i.e. One of the three has to be true, agreed?) P(2 Males) = P(2 Females) = 0.25 (we all agree that its 0.5x0.5, correct?)
P(Anything) = P(2 Males) + P(2 Females) + P(1 Male AND 1 Female) Hence: 1 = 0.25 + 0.25 + P(1 Male AND 1 Female) = 0.5 + P(1 Male AND 1 Female)
Therefore, P(1 Male AND 1 Female) = 1 - 0.5 = 0.5
IF P(1 Male AND 1 Female) was 0.25, the total would be 0.75
Doug: Let me put in another way... binary counting.
So, for two bits, we have possibly:
00 (P of 0.25) 01 (P of 0.25) 10 (P of 0.25) 11 (P of 0.25)
What is the probably of a randomly selected number containing a one and a zero is: P(01) + P(10) = P(one 1 and one 0) = 0.25 + 0.25 = 0.5
So, if we assume order isn't important, we get: P( two 0's ) = 0.25 P( one 1 and one 0 ) = 0.5 P( two 1's ) = 0.25
The maths isn't lying. You've just based your probabilities on false assumptions.
Ahh, I get it now--coins and dogs and children were confusing. The two mixed pairs really are two mixed pairs, not just the same way of expressing the same mixed pair possibility--that was the false assumption.
Doug: Let me put in another way... binary counting.
So, for two bits, we have possibly:
00 (P of 0.25) 01 (P of 0.25) 10 (P of 0.25) 11 (P of 0.25)
What is the probably of a randomly selected number containing a one and a zero is: P(01) + P(10) = P(one 1 and one 0) = 0.25 + 0.25 = 0.5
So, if we assume order isn't important, we get: P( two 0's ) = 0.25 P( one 1 and one 0 ) = 0.5 P( two 1's ) = 0.25
The maths isn't lying. You've just based your probabilities on false assumptions.
Ahh, I get it now--coins and dogs and children were confusing. The two mixed pairs really are two mixed pairs, not just the same way of expressing the same mixed pair possibility--that was the false assumption.
Now I get it! Thanks!
*smiles* Excellent - and no worries, its a confusing question at first.
What I'm about to say has no bearing on the actual math and may have already been mentioned, but, does the gender of the other dog matter? The guy said he wanted a male, and it was confirmed that at least one of them was a male, so why does he need to know if they both are? The only way it could possibly matter is if the shopkeeper had already promised a male to another person.
*smiles* Excellent - and no worries, its a confusing question at first.
You know what I think it is? I've never really trusted the fact that the flip of two coins is twice as likely to give a mixed pair than a double pair. Yet somehow, on/offs make sense.
Also, it seems elitheiceman was just waiting for me to get it to reward me with free xbox games and gear, so!
Well, the second gender does not rely on the first gender(of the dogs), so it's nothing more than a 50% chance. I also don't know what you mean by the pair. Also, this is not like the monty hall problem at all. The monty hall problem is conditional probablity. The probablity of one door depends on how many other doors, there are. The monty hall problem removes a door, so therefore changes the probability. They rely on each other, they are dependant to use the proper term. Meanwhile the gender of the two dogs is independent, I am ignoring the pair at the moment, because I don't know what you mean.
Then again, I'm bloody tired, and I may be remembering wrong.
Alex_P: 2 is gambler's fallacy. You're saying "this particular pair of puppies contained a male, so I must create an arbitrary system for generating puppies that puts a male in every pair."
No it isn't--gamblers fallacy is that luck going one way will be balanced out by luck going the other way very soon. Gamblers fallacy would be that because we got one male, we need to construct a system that makes it more likely that we get a female for the next result--which is what you guys are going--rather than a system where we've got an equal shot at getting a male or a female next.
I'm not doing anything to make the "second female" somehow more likely than it should be, though. All I'm saying is that you should start with each one being "fair" -- random with an equal probability of being male or female -- and construct sets from that. Otherwise you're not creating a probability distribution that's representative of the initial setup of the problem.
It's a mistake to say "well, in this problem there was a male so I will just make one a male every time" because what happened is that some fair coin ended up heads this particular time.
geizr: This thread proves how much people actually do misunderstand probabilities and how they can change with change in knowledge. Of the respondents, 65.5% got the wrong answer. Let me make an attempt to explain why 33% is the correct answer.
At the start of the problem, we have only know that there are two puppies, but we have no idea what their sexes are. So, the possible outcomes are the following: dog1 dog2 M M M F F M F F
There are 4 possible outcomes. This means that the probability of each outcome is 1/4 or 25%. If we ask what is the probability that both dogs are male, the answer is 25%. This is because only one of the four outcomes is M/M. But, this is the case that we don't know anything at all about the dogs other than the fact we have 2 of them.
Now, let's suppose we are told that dog 1 is a male. This means that we now know the F/M and F/F outcomes did not occur. So now we ask what are the possible outcomes. The possible outcomes are dog1 dog2 M M M F
There are only 2 possible outcomes. So the probability of each outcome is 1/2 or 50%. If we are told that dog2 is the one that is male, then we have again 2 possible outcomes: dog1 dog2 M M F M
But, look closely at what happened as a result of our change in knowledge. While the first possible outcome remains the same, the second possible outcome is a different configuration from the case when we knew dog1 was the male. Regardless, because there are only 2 possible outcomes now, the probability of each outcome is 50%.
Now, here is where things change. In the problem, we are told that at least one of the dogs is male. But, we don't know which one. Because we don't know which dog is male, we must consider all possible outcomes in which there is a male dog. There are 3 such outcomes: dog1 dog2 M M M F F M
Notice that the number of outcomes has changed to 3 because we don't know which dog is male, only that at least one is male. Because there are now only 3 possible outcomes, thanks to us knowing now that at least one dog is male, the probability of each outcome is 1/3 or 33%. The problem is asking us what is the probability that both dogs turn out to be male when we know that at least one dog is male. We see that only 1 of the three outcomes have both dogs as male. So, the probability that both dogs are male is 33%.
The KEY of this problem is realizing that the probabilities change as a result of our change in knowledge about the situation. Notice carefully how the probability of having 2 male dogs changed as a result in our change in knowledge about the situation. When we don't know anything at all about the two dogs, the probability of 2 males is 25%. When we know that dog1 or dog2 is the one that male, then the probability of 2 males is 50%. When we only know that at least one dog is male(but not which one), then the probability of 2 males is 33%.
One last try to give a thorough explanation why the correct answer is 33%(I quote my own post because I don't want to type all that again). Someone mentioned about conditional probabilities; that's exactly what's going on in this problem. The problem is asking what is the probability of 2 males given that you know at least 1 dog is male.
Doug: Let me put in another way... binary counting.
So, for two bits, we have possibly:
00 (P of 0.25) 01 (P of 0.25) 10 (P of 0.25) 11 (P of 0.25)
What is the probably of a randomly selected number containing a one and a zero is: P(01) + P(10) = P(one 1 and one 0) = 0.25 + 0.25 = 0.5
So, if we assume order isn't important, we get: P( two 0's ) = 0.25 P( one 1 and one 0 ) = 0.5 P( two 1's ) = 0.25
The maths isn't lying. You've just based your probabilities on false assumptions.
Ahh, I get it now--coins and dogs and children were confusing. The two mixed pairs really are two mixed pairs, not just the same way of expressing the same mixed pair possibility--that was the false assumption.
Now I get it! Thanks!
*smiles* Excellent - and no worries, its a confusing question at first.
Cheeze gets it? Yay! All hail Doug, master of logic and explanation! Sir, I bow before your superior teaching ability.
It's a mistake to say "well, in this problem there was a male so I will just make one a male every time" because what happened is that some fair coin ended up heads this particular time.
The maths isn't lying. You've just based your probabilities on false assumptions.
Cheeze gets it? Yay! All hail Doug, master of logic and explanation! Sir, I bow before your superior teaching ability.
You know what it is? What I think confuses people is not this particular problem. I think what confuses people is how counter-intuitive probability itself is, that it's natural to have "based your probabilities on false assumptions" as Doug pointed out.
I think what makes it confusing is that it doesn't seem like permutations should have anything to do with the probability of different results in a two-event, on/off outcome in a system without memory. What's counter intuitive isn't what happens when the new information comes along, it's the initial probabilities. What's fishy is that the probabilities involved in two independent random trials is the exact same as two picks from a system with memory.
I thought this was never going to end. Since I bid this "adieu", I've been coming back every day just to see how many pages there were. Each day, I was increasingly perplexed that it was still going so strong. In fact, I was increasingly suspicious that Cheeze_Pavilion was taking advantage of this and doing an incredible job of that. Honestly Cheeze, I mean that with the utmost respect; I couldn't fathom the purported explanation that you sincerely didn't get it, and it seemed to make much more sense if you were doing it on purpose.
Still, I have a mild suspicion that this is all just one of these...
Vallen00: What I'm about to say has no bearing on the actual math and may have already been mentioned, but, does the gender of the other dog matter? The guy said he wanted a male, and it was confirmed that at least one of them was a male, so why does he need to know if they both are? The only way it could possibly matter is if the shopkeeper had already promised a male to another person.
I've pointed this out numerous times, but nobody seems to care. (ps., the answers zero!)
Geoffrey42: Honestly Cheeze, I mean that with the utmost respect; I couldn't fathom the purported explanation that you sincerely didn't get it, and it seemed to make much more sense if you were doing it on purpose.
Heh, didn't get it until Doug's post made me think 'well, let's try something different--let's try homogeneous results vs. heterogeneous results: now why *wouldn't* they be 50/50?'
Cheeze_Pavilion: SNIP You know what it is? What I think confuses people is not this particular problem. I think what confuses people is how counter-intuitive probability itself is, that it's natural to have "based your probabilities on false assumptions" as Doug pointed out.
I think what makes it confusing is that it doesn't seem like permutations should have anything to do with the probability of different results in a two-event, on/off outcome in a system without memory. What's counter intuitive isn't what happens when the new information comes along, it's the initial probabilities. What's fishy is that the probabilities involved in two independent random trials is the exact same as two picks from a system with memory.
As I said long ago in this thread, I think the difficult part is that your mind seizes on what you know and stops. If the question had been the chance that, say, three other dogs out of six were male, you'd work the problem.
Cheeze_Pavilion: SNIP You know what it is? What I think confuses people is not this particular problem. I think what confuses people is how counter-intuitive probability itself is, that it's natural to have "based your probabilities on false assumptions" as Doug pointed out.
I think what makes it confusing is that it doesn't seem like permutations should have anything to do with the probability of different results in a two-event, on/off outcome in a system without memory. What's counter intuitive isn't what happens when the new information comes along, it's the initial probabilities. What's fishy is that the probabilities involved in two independent random trials is the exact same as two picks from a system with memory.
As I said long ago in this thread, I think the difficult part is that your mind seizes on what you know and stops. If the question had been the chance that, say, three other dogs out of six were male, you'd work the problem.
My mind seized on why an event that was 50% likely at the start could become 66% likely--and not even 75% likely, or possibly 62.5% likely--when a piece of information was revealed that is one and only one of two necessary conditions for that event to come true.
gerrymander61: It's 50%, or slightly higher given that the dogs could be identical twins. Anyone who says otherwise should take a class in basic stats or probability to find out just how STUPID they are.
EDIT: Wow, just checked the poll results. 34.6% of people here are stupid.
Take discrete math, please. It's actually like the best don't-have-to-be-a-math-major college math class: useful and light on memorization.
-- Alex
Um, excuse me? The fact that one dog is male has NO BEARING WHATSOEVER on the other dog's gender, if anything, it makes it slightly more likely that it is male due to the possibility, albeit unlikely, of the dogs being twins, putting the second puppy's chance at being male slightly over 50%. What you're arguing for is what we Statistics and probability people call "The Gambler's Fallacy." See? It's even got a name because there are so many idiots like YOU out there in the world.
gerrymander61: It's 50%, or slightly higher given that the dogs could be identical twins. Anyone who says otherwise should take a class in basic stats or probability to find out just how STUPID they are.
EDIT: Wow, just checked the poll results. 34.6% of people here are stupid.
Take discrete math, please. It's actually like the best don't-have-to-be-a-math-major college math class: useful and light on memorization.
-- Alex
Um, excuse me? The fact that one dog is male has NO BEARING WHATSOEVER on the other dog's gender, if anything, it makes it slightly more likely that it is male due to the possibility, albeit unlikely, of the dogs being twins, putting the second puppy's chance at being male slightly over 50%. What you're arguing for is what we Statistics and probability people call "The Gambler's Fallacy." See? It's even got a name because there are so many idiots like YOU out there in the world.
gerrymander61: It's 50%, or slightly higher given that the dogs could be identical twins. Anyone who says otherwise should take a class in basic stats or probability to find out just how STUPID they are.
EDIT: Wow, just checked the poll results. 34.6% of people here are stupid.
Take discrete math, please. It's actually like the best don't-have-to-be-a-math-major college math class: useful and light on memorization.
-- Alex
Um, excuse me? The fact that one dog is male has NO BEARING WHATSOEVER on the other dog's gender, if anything, it makes it slightly more likely that it is male due to the possibility, albeit unlikely, of the dogs being twins, putting the second puppy's chance at being male slightly over 50%. What you're arguing for is what we Statistics and probability people call "The Gambler's Fallacy." See? It's even got a name because there are so many idiots like YOU out there in the world.
Do a little reading, and then maybe instead of taking a probability course, pay a-f**king-ttention in YOUR discrete course. I've taken it.
The fact that one dog in a random set is male has no bearing on the sex of the other dog, yes. (I don't know why everyone keeps saying "gender" when this is all about looking at nards rather than the dogs' self-identity or which color the adorable bowties around their heads happen to be.)
However -- and this is something that you have completely missed -- "at least one dog is male" tells you something specific about the set, but is not a definite fact about a specific dog. Discrete math is about using logic systematically, not applying pithy truisms at random even when they don't fit the problem.
Rude AND wrong - it's the new killer combination everyone wants!
Killer combination maybe, but new? Blind Punk Riot tried that 15 pages ago. Although to be fair, that was mainly sarcastic rude, this is more look-at-me-laugh-at-the-stupids rude.
Rude AND wrong - it's the new killer combination everyone wants!
Killer combination maybe, but new? Blind Punk Riot tried that 15 pages ago. Although to be fair, that was mainly sarcastic rude, this is more look-at-me-laugh-at-the-stupids rude.
gerrymander61: It's 50%, or slightly higher given that the dogs could be identical twins. Anyone who says otherwise should take a class in basic stats or probability to find out just how STUPID they are.
EDIT: Wow, just checked the poll results. 34.6% of people here are stupid.
Take discrete math, please. It's actually like the best don't-have-to-be-a-math-major college math class: useful and light on memorization.
-- Alex
Um, excuse me? The fact that one dog is male has NO BEARING WHATSOEVER on the other dog's gender, if anything, it makes it slightly more likely that it is male due to the possibility, albeit unlikely, of the dogs being twins, putting the second puppy's chance at being male slightly over 50%. What you're arguing for is what we Statistics and probability people call "The Gambler's Fallacy." See? It's even got a name because there are so many idiots like YOU out there in the world.
Do a little reading, and then maybe instead of taking a probability course, pay a-f**king-ttention in YOUR discrete course. I've taken it.
Rude AND wrong - it's the new killer combination everyone wants!
They're just making themselves look stupid. If you're going to be wrong, you can at least be quiet about it. Of course, what's the point of them looking stupid if they don't even know they look stupid.
Congrats, Cheeze. I was actually starting to think you might be carrying on intentionally. It's nice to know that sometimes 19 pages of argument isn't wasted time.
It's ambiguous. We don't know if Shopkeeper Woman and Puppy Washing Man are telling us something about the set, or about one of the puppies AND the set. We need to know why he's telling us "Yes!": is it because he's looking at/thinking about puppy nads on one of them, or is it because he knows the puppies came from Puppy Breeder Person who guarantees puppy pairs with at least one male in them?
Rude AND wrong - it's the new killer combination everyone wants!
Killer combination maybe, but new? Blind Punk Riot tried that 15 pages ago. Although to be fair, that was mainly sarcastic rude, this is more look-at-me-laugh-at-the-stupids rude.
At the risk of opening all this up again, I want to see exactly how this logic plays out in other settings, so, let's say there's a dice roll. There are various possibilities that are the result of a number of permutations, and therefore different chances of different things happening, from getting snake eyes--1/36--to getting a seven--1/6--to getting boxcars which, again, are 1/36.
Someone tells you that there is at least one 6. Or at least one 1, or 2, or whatever you want.
Now what are the probabilities and what are the outcomes?
EDIT: looked it up myself--seems people disagree about this the same way they disagree about coins or males or anything. Some people say the answer is 2/11, as are all the other outcomes but 8 (4 was the known number/number known to be in the pair in the author's example), which is 1/11. Others say 1/6.
I'm inclined to think now after looking at the different arguments for the dice problem* that the crux of the question is the probability that the Puppy Washing Man answered "Yes!" because he got his puppies from a Puppy Breeding Person that guarantees pairs with at least one male puppy, or that when the Shopkeeper Woman called, he looked for a male puppy and found one, and answered on the basis of that search.
So yeah, I'd say it's at best ambiguous, and probably the 50% people have the stronger case that they interpreted the word problem correctly.
Why would the question ask "What is the probability that the other one is a male?" if there is no other 'one', if the Puppy Washing Man came to his knowledge by knowing the pair was ordered from Guaranteed Puppy Breeding Person? Implicit in the final question is the additional fact that the Puppy Washing Man is referring to one of the puppies when he says "Yes!" otherwise what does the question mean by "the other one is male?"
As I see it now, to remain true to the question, the matrix has to look like this:
Barack Obama (i.e., "that one")/ The Other One M/M M/F F/M F/F
clearly the bottom two drop out, and we're left with 50%, right? Anyone got any other way of setting up the matrix when we're being asked about 'the other one'?
*what is interesting is that if 7 in the dice problem represents the mixed pair in the puppy/coin/child problem, it doesn't become any more likely than any of the other combinations but 8, which is what tipped me off to what is going on here. In the puppy problem, it's not that the odds of the mixed pair go up, it's the the odds of all remaining pairs go up equally *except* the 'double' pair where we know at least one half of the double is true. And the really interesting thing, the thing that jibes with my intuitions, is that the middle/mixed pair that is originally the most likely outcome turns into one of five equally likely outcomes, which can include the least likely outcome when two dice were being thrown, the outcome that was only 1/6th as likely, when a die is thrown, becomes just as likely, as long as it wasn't one of the outcomes that relied on two dice showing the same face.
In other words, the middle outcome doesn't actually become more likely (which satisfies my mind, in that if an outcome is the result of two events one of which is bound to come true, that event coming true cannot make the outcome more likely unless it's illogical); it's that some outcomes become non-at-all likely, and all the other remaining outcomes get an even bigger boost, which we can't see in a two-trial situation like coins or puppies.
The tricky part here is that the most probable outcome is the outcome that can never be generated with a double result, no matter how many permutations there are. That I think it what makes this so difficult. One of the results is not like any of the other results in that it is neither made more likely nor possibly ruled out by any one result.
Congrats, Cheeze. I was actually starting to think you might be carrying on intentionally. It's nice to know that sometimes 19 pages of argument isn't wasted time.
It certainly isn't--now I know exactly why the answer is 50% in the most sensible reading of the question: anyone who thinks it is 33% didn't really read the question that was asked: what is the sex of the *other* puppy, meaning the information in the problem wasn't about the pair, it was about *one* of the puppies.
Once I saw Doug's answer, I saw a situation where it would be 33%, because 10=/=01. On the other hand, male/female=female/male where the extra information *isn't* referring to the pair of puppies, but to one of the puppies--otherwise the phrase "the other one" makes no sense.
gerrymander61: It's 50%, or slightly higher given that the dogs could be identical twins. Anyone who says otherwise should take a class in basic stats or probability to find out just how STUPID they are.
EDIT: Wow, just checked the poll results. 34.6% of people here are stupid.
Take discrete math, please. It's actually like the best don't-have-to-be-a-math-major college math class: useful and light on memorization.
-- Alex
Um, excuse me? The fact that one dog is male has NO BEARING WHATSOEVER on the other dog's gender, if anything, it makes it slightly more likely that it is male due to the possibility, albeit unlikely, of the dogs being twins, putting the second puppy's chance at being male slightly over 50%. What you're arguing for is what we Statistics and probability people call "The Gambler's Fallacy." See? It's even got a name because there are so many idiots like YOU out there in the world.
Do a little reading, and then maybe instead of taking a probability course, pay a-f**king-ttention in YOUR discrete course. I've taken it.
The fact that one dog in a random set is male has no bearing on the sex of the other dog, yes. (I don't know why everyone keeps saying "gender" when this is all about looking at nards rather than the dogs' self-identity or which color the adorable bowties around their heads happen to be.)
However -- and this is something that you have completely missed -- "at least one dog is male" tells you something specific about the set, but is not a definite fact about a specific dog.
Actually, it does, when it occurs in the same word problem that ends with the question: "What is the probability that the other one is a male?"
If we haven't been given a definite fact about a specific dog, how can the question ask us about the "other" dog? On what basis are we to tell the "other" dog from the, uh, non-other dog if we have no specific facts about either?
*smiles* Excellent - and no worries, its a confusing question at first.
I hate to do this, but I have to take that back--all the arguments for 33% are based on the idea that we only know something about the set. However, if we look at the end of the question, it asks us "What is the probability that the other one is a male?"
So the facts we have in the question have to pertain to one of the dogs specifically, otherwise, the question we are being asked makes no sense--how can there be an "other" dog if we can't tell one dog from another based on the info we've been given?
If it's any consolation, it was your example that got me thinking about the deep logic of the 33% answer, and why it's answering a question that wasn't asked, using less/different information than was actually given.
You can't describe order from information that we don't have. Saying "the next one" implies that you know the male dog is first. You can't place him first because you don't know who he is. I don't even think the rules of probability problems allow you to change ordering after a problem is already set up. Sparky and Othello have one set order. It doesn't matter if it's Sparky and Othello or Othello and Sparky. Consistency is important.