Cheeze_Pavilion: Funny because to me, that's what everyone else is saying--'watch me use the Law of Large Numbers to come to a conclusion about...the smallest possible number!'
Simply using a large number isn't an application of the law of large numbers. The law of large numbers has to do with correlation between an expected average (from a probability calculation) and the actual observed value.
Cheeze_Pavilion: Not to mention that fact that the people who claim to have the right answer can't agree on the right method, which is...kinda troubling, don't you think?
*shrug* Discrete math isn't something that's taught effective in high school. Most people pick it up haphazardly.
Let's say RED represents a male puppy, and WHITE represents a female puppy. So a card with two RED sides represents a male pair, a card with one RED side and one WHITE side represents a mixed pair, and a card with two WHITE sides represents a female pair.
When we learn there is at least one male, that information allows us to remove the card with two WHITE sides, right? So our 'pool' is one card that is RED on both sides, and one card that is RED on one side, and WHITE on the other. What are the chances of picking either card? 50/50, right? So what are the chances of a male pair vs. a mixed pair? 50/50, just like the cards that represent those two possibilities.
That sound right to you? I think even if it doesn't it gets to the heart of this disagreement.
If you have one card of each type, that's correct. (However, one card of each type is an incorrect model for this particular.)
Now, what if I have X red cards, Y mixed cards, and Z white cards? What are the individual probabilities? How do they change when I take all the white cards away?
Alex_P: Okay, one big fundamental misconception I keep seeing referenced over and over but never directly addressed. It sounds kinda like this:
"Two females don't satisfy the 'at least one male' requirement, so any experiment that has the potential to produce the two females result is an inappropriate model for this scenario."
That's pretty much gambler's fallacy at work. You're saying "This fair coin was heads this time so it can never be tails!"
Untrue! It's still a fair coin.
And, yes, that means that if you simulate the problem with two fair coins, then the answer to "Is at least one male?" is gonna be "no" 25% of the time. That's how it works. That's okay.
If you do something like setting one coin to make sure that you can never get TT, then what you're actually doing is breaking the "fair coin" rule completely -- your model ceases to be representative of the problem.
Why are we simulating with two fair coins? We don't have two fair puppies!
Actually, I think I was wrong about insisting on throwing out the FFs. Then again, I don't think it matters if you throw out any FFs anymore, either. But I'm not certain about that.
I just still don't understand why we're using two fair coins to simulate one fair puppy and one unfair puppy.
I said there was a 50% chance. After all, in a female... well, anything, she has two X chromosomes with the potential of one of them being copied to her offspring. Either way the female is passing on an X. The male has both an X and a Y chromosome, meaning that there's a 1 in 2 chance that he passes on the opposite chromosome that the mate did.
Cheeze_Pavilion: Why are we simulating with two fair coins? We don't have two fair puppies!
Actually, I think I was wrong about insisting on throwing out the FFs. Then again, I don't think it matters if you throw out any FFs anymore, either. But I'm not certain about that.
I just still don't understand why we're using two fair coins to simulate one fair puppy and one unfair puppy.
Nope - the actual probability of getting one head and one tail is twice that of getting 2 heads.
EDIT: Not when you're only flipping one coin--see the original problem: we know one of the 'coins' is heads. We only ever flip one coin--the value of the non-flipped coin is always Heads/Male.
Try it - flip a pair of coins 100 times. You'll end up with approx. 25 H-H pairs, 25 T-T pairs, and 50 mixed pairs.
Heck, I could write you a program that would do thousands of tests if you'd like, it's pretty simple...
Think about it--if the odds are the same after one additional trial as after 100, then there's something seriously wrong:
Probability does have an impact upon small numbers of observations, however. It determines how likely a certain outcome is. You can't defend your probability based on an argument that probability actually isn't valid on a small scale. Even if this were actually true.
Alex_P: If you have one card of each type, that's correct. (However, one card of each type is an incorrect model for this scenario.)
Hmm, now I'm starting to see. I'm thinking the odds are:
Two male (.5 x .5) One of Each (.5 x .5) Two female (.5 x .5)
and are you saying the odds are:
Two male (.5 x .5) One of Each (1 x .5) Two female (.5 x .5)
Now, what if I have X red cards, Y mixed cards, and Z white cards? What are the individual probabilities? How do they change when I take all the white cards away?
Cheeze_Pavilion: Why are we simulating with two fair coins? We don't have two fair puppies!
Actually, I think I was wrong about insisting on throwing out the FFs. Then again, I don't think it matters if you throw out any FFs anymore, either. But I'm not certain about that.
I just still don't understand why we're using two fair coins to simulate one fair puppy and one unfair puppy.
Alex_P: Okay, one big fundamental misconception I keep seeing referenced over and over but never directly addressed. It sounds kinda like this:
"Two females don't satisfy the 'at least one male' requirement, so any experiment that has the potential to produce the two females result is an inappropriate model for this scenario."
That's pretty much gambler's fallacy at work. You're saying "This fair coin was heads this time so it can never be tails!"
Untrue! It's still a fair coin.
And, yes, that means that if you simulate the problem with two fair coins, then the answer to "Is at least one male?" is gonna be "no" 25% of the time. That's how it works. That's okay.
If you do something like setting one coin to make sure that you can never get TT, then what you're actually doing is breaking the "fair coin" rule completely -- your model ceases to be representative of the problem.
Why are we simulating with two fair coins? We don't have two fair puppies!
Actually, I think I was wrong about insisting on throwing out the FFs. Then again, I don't think it matters if you throw out any FFs anymore, either. But I'm not certain about that.
I just still don't understand why we're using two fair coins to simulate one fair puppy and one unfair puppy.
Neither of them is an unfair puppy. If you have two coins, and one of them is heads on a particular flip, the coin is still fair.
Now that you've released the primary flaw in your argument, that any experiment where FF's appear is invalid, it should be pretty easy from here. Just perform it.
Two coins. At least one heads. If there are two tails, it doesn't matter, reflip. This probability should be correct for ALL pairs where at least one coin is heads. Therefore, if you have, out of 200 flips around:
50 tails tails, 100 1 heads, 1 tails, and 50 heads heads (this is the most likely distribution)
About 150 of these cases should have the same probability as in out problem. So if you use 50 percent, you end up with 75 pairs of heads heads, which is, of course, incorrect. Only 33 percent will return you the correct distribution here.
Cheeze_Pavilion: Why are we simulating with two fair coins? We don't have two fair puppies!
Actually, I think I was wrong about insisting on throwing out the FFs. Then again, I don't think it matters if you throw out any FFs anymore, either. But I'm not certain about that.
I just still don't understand why we're using two fair coins to simulate one fair puppy and one unfair puppy.
What makes one puppy "unfair"?
The fact that we know that it is male.
It doesn't affect each individual puppy. It affects both together, as a pair.
Each individual puppy still has a certain chance of being male, and a certain chance of being female. Neither is guaranteed to be male 100 percent of the time.
No. You're conflating the general probability distribution with an observed fact about this one particular instance.
That's "gambler's fallacy."
We're not looking to figure out general probability distribution, we're trying to figure out something about this one particular instance.
Like you said, That's "gambler's fallacy."
Gambler's fallacy is where you assume something about an event based on outcomes already observed. Like when you flip a coin, see that it's heads, and expect the second one to be more likely to be female. That's gambler's fallacy, it's quite simple.
I don't think either of you are correct when you fault it to the other man.
Also, one particular instance has the same probabilities as a million identical ones. The size of your experiment is irrelevant, in terms of probabilities. The size of your experiment just determines how accurate the results will be, in terms of outcomes.
No. You're conflating the general probability distribution with an observed fact about this one particular instance.
That's "gambler's fallacy."
We're not looking to figure out general probability distribution, we're trying to figure out something about this one particular instance.
Like you said, That's "gambler's fallacy."
No, the general probability distribution is EXACTLY what we're trying to figure out! Essentially, what they want to know is, if this scenario were to happen an infinite number of times, what percentage of the scenarios (that have at least one male) would have two males?
Hell, that right there is why the computer simulation works for a rough estimate. You test a bunch of times, throw away the tests that don't fit the scenario, and use the remaining results to calculate rough probabilities. It's not exact, but it's great for testing your math.
...
Sigh. I forgot how annoying internet arguments can be.
Where did the two puppies in the problem come from? How were they initially selected?
What do you mean "initially"?
He means "in the beginning." Like, how were the two puppies for the problem selected? Randomly, right? They represent a random pair, where at least one happens to be male. If you have a random pair, there is a certain likelihood that at least one will be male. This falls in that category. But it's information that doesn't change the fact that they are random puppies.
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%.
We're not looking to figure out general probability distribution, we're trying to figure out something about this one particular instance.
Like you said, That's "gambler's fallacy."
No, the general probability distribution is EXACTLY what we're trying to figure out! Essentially, what they want to know is, if this scenario were to happen an infinite number of times, what percentage of the scenarios (that have at least one male) would have two males?
Then why didn't they just ask that? They asked about *this* instance, just like a gambler wants to know about the *next* bet, not the bets that he could make if he had an infinite bankroll.
Sigh. I forgot how annoying internet arguments can be.
Umm, I was discussing this. I'm not arguing with anyone.
We're not looking to figure out general probability distribution, we're trying to figure out something about this one particular instance.
Like you said, That's "gambler's fallacy."
No, the general probability distribution is EXACTLY what we're trying to figure out! Essentially, what they want to know is, if this scenario were to happen an infinite number of times, what percentage of the scenarios (that have at least one male) would have two males?
Then why didn't they just ask that? They asked about *this* instance, just like a gambler wants to know about the *next* bet, not the bets that he could make if he had an infinite bankroll.
Sigh. I forgot how annoying internet arguments can be.
Umm, I was discussing this. I'm not arguing with anyone.
How is the probability for the next bet different than the probabilities for the next million? If they're identical bets, they all have the same probabilities. That includes each individual instance. Doing it a million times just assures that your results will likely mirror the probabilities very closely.
Where did the two puppies in the problem come from? How were they initially selected?
What do you mean "initially"?
Initially. In the beginning. At the start of the scenario.
Before the phonecall, the puppy washer was handed a set of puppies. How was that set selected?
One puppy was picked from an infinite number of puppies, then another puppy was picked from an infinite number of puppies.
Or
One set of puppies was picked from either an infinite set of male pairs, an infinite set of mixed pairs, or an infinite set of female pairs.
Are you saying something like the infinite set of mixed pairs is bigger than the infinite set of either same-sex pairs? Like if you went to pair off mixed sets with same sex sets, it would be like pairing off the number of real numbers between 0 and 1 with natural numbers?
(that's about as far as my knowledge of infinite set theory goes)
"What is the probability that the other one is a male?"
That is exactly what they did ask.
No, to ask what you're talking about, they would have asked "if this scenario we've just described to you were to happen not once as it was presented to you, but were to happen an infinite number of times, what percentage of the scenarios (that have at least one male) would have two males?"
Where did the two puppies in the problem come from? How were they initially selected?
What do you mean "initially"?
Initially. In the beginning. At the start of the scenario.
Before the phonecall, the puppy washer was handed a set of puppies. How was that set selected?
One puppy was picked from an infinite number of puppies, then another puppy was picked from an infinite number of puppies.
Or
One set of puppies was picked from either an infinite set of male pairs, an infinite set of mixed pairs, or an infinite set of female pairs.
Are you saying something like the infinite set of mixed pairs is bigger than the infinite set of either same-sex pairs? Like if you went to pair off mixed sets with same sex sets, it would be like pairing off the number of real numbers between 0 and 1 with natural numbers?
(that's about as far as my knowledge of infinite set theory goes)
Hehe, wow, now this is really irrelevant.
I concede that the infinities can't be proven to be different sizes. Fortunately, this little math problem is relatively simple and secular, and doesn't concern such elusive concepts as infinity.
Back to the matter at hand, please?
The proportions remain the same.
1 (2 male pair): 2 (1 male 1 female pairs): 1 (2 female pair)
Again, this is the distribution for random pairs of puppies.
"What is the probability that the other one is a male?"
That is exactly what they did ask.
No, to ask what you're talking about, they would have asked "if this scenario we've just described to you were to happen not once as it was presented to you, but were to happen an infinite number of times, what percentage of the scenarios (that have at least one male) would have two males?"
Again, the probabilities remain the same for all scenarios in that infinite number of identical scenarios. Repeating the experiment many times just increases how accurately the physical results resemble the probailities.
Cheeze_Pavilion: One puppy was picked from an infinite number of puppies, then another puppy was picked from an infinite number of puppies.
So, that's analogous to "two random puppies," right?
So, what's the matrix of permutations?
Cheeze_Pavilion: Are you saying something like the infinite set of mixed pairs is bigger than the infinite set of either same-sex pairs? Like if you went to pair off mixed sets with same sex sets, it would be like pairing off the number of real numbers between 0 and 1 with natural numbers?
Not quite. Natural numbers vs. real numbers is countable infinity vs. uncountable infinity. In informal language, one of those infinities is like "infinity times infinity" (basically the idea is that you can squeeze infinite real numbers between any two rational numbers).
Cheeze_Pavilion: One puppy was picked from an infinite number of puppies, then another puppy was picked from an infinite number of puppies.
So, that's analogous to "two random puppies," right?
So, what's the matrix of permutations?
Our problem isn't with the initial matrix of permutations. We've all agreed on that. M/M, M/F, F/M, F/F. The question isn't about permutations, it's about trying to figure out the probability of three mutually exclusive sets of combinations.
Not quite. Natural numbers vs. real numbers is countable infinity vs. uncountable infinity. In informal language, one of those infinities is like "infinity times infinity" (basically the idea is that you can squeeze infinite real numbers between any two rational numbers).
Cheeze_Pavilion: Our problem isn't with the initial matrix of permutations. We've all agreed on that. M/M, M/F, F/M, F/F. The question isn't about permutations, it's about trying to figure out the probability of three mutually exclusive sets of combinations.
Well, this is separate from that. It speaks to how to test the problem empirically if you want.
Here's what I'm driving at...
The problem scenario goes like this: 1. Someone makes a pair out of two random puppies. 2. You ask a question about that pair and find out the answer is yes, allowing you to use this new knowledge to adjust your probabilistic guess. 3. You are asked to make a guess about the pair.
In order to "test" something like this, you should copy that general flow. First, make a pair of random puppies. Then ask "Is at least one male?" Then look at whether there are two males in this set.
If you just try any old way to create a pair of puppies, then you're not effectively simulating the problem anymore.
Cheeze_Pavilion: One puppy was picked from an infinite number of puppies, then another puppy was picked from an infinite number of puppies.
So, that's analogous to "two random puppies," right?
So, what's the matrix of permutations?
Our problem isn't with the initial matrix of permutations. We've all agreed on that. M/M, M/F, F/M, F/F. The question isn't about permutations, it's about trying to figure out the probability of three mutually exclusive sets of combinations.
Not quite. Natural numbers vs. real numbers is countable infinity vs. uncountable infinity. In informal language, one of those infinities is like "infinity times infinity" (basically the idea is that you can squeeze infinite real numbers between any two rational numbers).
This would be more like just taking a limit.
Ahh, okay. My bad.
You said it yourself. There are three mutually exclusive situations. They are all equally probable, are they not?
Cheeze_Pavilion: Our problem isn't with the initial matrix of permutations. We've all agreed on that. M/M, M/F, F/M, F/F. The question isn't about permutations, it's about trying to figure out the probability of three mutually exclusive sets of combinations.
Well, this is separate from that. It speaks to how to test the problem empirically if you want.
Here's what I'm driving at...
The problem scenario goes like this: 1. Someone makes a pair out of two random puppies. 2. You ask a question about that pair and find out the answer is yes, allowing you to use this new knowledge to adjust your probabilistic guess. 3. You are asked to make a guess about the pair.
In order to "test" something like this, you should copy that general flow. First, make a pair of random puppies. Then ask "Is at least one male?" Then look at whether there are two males in this set.
If you just try any old way to create a pair of puppies, then you're not effectively simulating the problem anymore.
-- Alex
"1. Someone makes a pair out of two random puppies."
Two unknown puppies put in a pair.
"2. You ask a question about that pair and find out the answer is yes, allowing you to use this new knowledge to adjust your probabilistic guess."
One puppy is male, the other, we don't know.
"3. You are asked to make a guess about the pair."
Well, the male puppy is guaranteed to be male, and the unknown puppy is either male or female. And puppies are as likely to be male as they are to be female, so there's a .5 chance the unknown puppy is male, and a .5 chance the unknown puppy is female. So the chances of the pair being two males is 50%, and one of each, 50%
Now where, as the LOLCats would say, am I doin' it wrong?
Cheeze_Pavilion: Our problem isn't with the initial matrix of permutations. We've all agreed on that. M/M, M/F, F/M, F/F. The question isn't about permutations, it's about trying to figure out the probability of three mutually exclusive sets of combinations.
Well, this is separate from that. It speaks to how to test the problem empirically if you want.
Here's what I'm driving at...
The problem scenario goes like this: 1. Someone makes a pair out of two random puppies. 2. You ask a question about that pair and find out the answer is yes, allowing you to use this new knowledge to adjust your probabilistic guess. 3. You are asked to make a guess about the pair.
In order to "test" something like this, you should copy that general flow. First, make a pair of random puppies. Then ask "Is at least one male?" Then look at whether there are two males in this set.
If you just try any old way to create a pair of puppies, then you're not effectively simulating the problem anymore.
-- Alex
"1. Someone makes a pair out of two random puppies."
Two unknown puppies put in a pair.
"2. You ask a question about that pair and find out the answer is yes, allowing you to use this new knowledge to adjust your probabilistic guess."
One puppy is male, the other, we don't know.
"3. You are asked to make a guess about the pair."
Well, the male puppy is guaranteed to be male, and the unknown puppy is either male or female. And puppies are as likely to be male as they are to be female, so there's a .5 chance the unknown puppy is male, and a .5 chance the unknown puppy is female. So the chances of the pair being two males is 50%, and one of each, 50%
Now where, as the LOLCats would say, am I doin' it wrong?
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."
What you should be doing is starting with a random pair -- because that's how the pair in the problem is actually constructed. Then for each pair you say "Does this pair contain one or more male puppies?" and "Does this pair contain two male puppies?" and you look at how the distributions are aligned.
Otherwise any "experiment" is flawed because you're replacing the original probability distribution of the sets with one of your own devising.
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.
Cheeze_Pavilion: Our problem isn't with the initial matrix of permutations. We've all agreed on that. M/M, M/F, F/M, F/F. The question isn't about permutations, it's about trying to figure out the probability of three mutually exclusive sets of combinations.
Well, this is separate from that. It speaks to how to test the problem empirically if you want.
Here's what I'm driving at...
The problem scenario goes like this: 1. Someone makes a pair out of two random puppies. 2. You ask a question about that pair and find out the answer is yes, allowing you to use this new knowledge to adjust your probabilistic guess. 3. You are asked to make a guess about the pair.
In order to "test" something like this, you should copy that general flow. First, make a pair of random puppies. Then ask "Is at least one male?" Then look at whether there are two males in this set.
If you just try any old way to create a pair of puppies, then you're not effectively simulating the problem anymore.
-- Alex
"1. Someone makes a pair out of two random puppies."
Two unknown puppies put in a pair.
"2. You ask a question about that pair and find out the answer is yes, allowing you to use this new knowledge to adjust your probabilistic guess."
One puppy is male, the other, we don't know.
"3. You are asked to make a guess about the pair."
Well, the male puppy is guaranteed to be male, and the unknown puppy is either male or female. And puppies are as likely to be male as they are to be female, so there's a .5 chance the unknown puppy is male, and a .5 chance the unknown puppy is female. So the chances of the pair being two males is 50%, and one of each, 50%
Now where, as the LOLCats would say, am I doin' it wrong?
You are, indeed, doin' it wrong. You are assuming that you know which puppy is the known puppy, and which is unknown. Indeed, if the washer told you, "yes, Sparky's a male," the answer would be 50 percent.
However, since you don't know, these three possibilities are all equally likely;
Sparky is a male, Othello is a male Sparky is a male, Othello is a female Sparky is a female, Othello is a male
You see how knowing one is male, and knowing which one is male are different? These three solutions are all exclusive possibilities, and all are equally likely. With what do you disagree?
Simply using a large number isn't an application of the law of large numbers. The law of large numbers has to do with correlation between an expected average (from a probability calculation) and the actual observed value.
*shrug* Discrete math isn't something that's taught effective in high school. Most people pick it up haphazardly.
If you have one card of each type, that's correct. (However, one card of each type is an incorrect model for this particular.)
Now, what if I have X red cards, Y mixed cards, and Z white cards? What are the individual probabilities? How do they change when I take all the white cards away?
-- Alex