Poll: A little math problem
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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.
-- Alex
"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.
-- Alex
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 post=18.73797.826674 said:
*shrug* Discrete math isn't something that's taught effective in high school. Most people pick it up haphazardly.Cheeze_Pavilion post=18.73797.826674 said:
If you have one card of each type, that's correct. (However, one card of each type is an incorrect model for this particular.)Cheeze_Pavilion post=18.73797.826674 said:
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
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.
So it's a 50-50 chance.
So it's a 50-50 chance.
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.Cheeze_Pavilion post=18.73797.826387 said:
And again, this isn't the Gambler's Fallacy.
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.Cheeze_Pavilion post=18.73797.826685 said:
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.
No. You're conflating the general probability distribution with an observed fact about this one particular instance.Cheeze_Pavilion post=18.73797.826728 said:
That's a "gambler's fallacy" (sometimes called "inverse gambler's fallacy").
-- Alex
It doesn't affect each individual puppy. It affects both together, as a pair.Cheeze_Pavilion post=18.73797.826728 said:
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.
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.Cheeze_Pavilion post=18.73797.826761 said:
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, 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?Cheeze_Pavilion post=18.73797.826761 said:
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.
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.Cheeze_Pavilion post=18.73797.826802 said:
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%.
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%.
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.Cheeze_Pavilion post=18.73797.826840 said:
"What is the probability that the other one is a male?"Cheeze_Pavilion post=18.73797.826840 said:
That is exactly what they did ask.
Hehe, wow, now this is really irrelevant.Cheeze_Pavilion post=18.73797.826880 said:
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.
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 post=18.73797.826902 said:
So, that's analogous to "two random puppies," right?Cheeze_Pavilion post=18.73797.826880 said:
So, what's the matrix of permutations?
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 post=18.73797.826880 said:
This would be more like just taking a limit.
-- Alex