Poll: A little math problem
- Thread starter Fud
- Start date
Recommended Videos
Err, google "two coins probability" or something.Cheeze_Pavilion said:
This text [http://books.google.com/books?id=L6IWgaCuilwC&pg=PA14&lpg=PA14&dq=two+coins+probability&source=bl&ots=yzBvncg2JA&sig=agzX9UzzqBDkmHuhheVo7Tg40_Q&hl=en&ei=2tK8ScjRD8G0twfyseX2Cw&sa=X&oi=book_result&resnum=1&ct=result] may be helpful.
-- Alex
The frequentist and Bayesian outlooks produce identical mathematical results. If the long-running frequency of an outcome is 1/4, that means your Bayesian belief about a particular instance achieving that outcome is also true with P = 1/4.Cheeze_Pavilion said:
(Where they differ is that the Bayesian outlook is the One True Way while frequentism is for jerks.)
-- Alex
Unless beagles are much more likely to give birth to one sex than the other, then the probability of the other puppy being male must be 50%.
The only way I can think of the answer being other than 50% is if this is a trick question, such as taking into account the ages of the puppies and therefore making three possible outcomes:
Older puppy male, younger puppy male
Older puppy male, younger puppy female
Older puppy female, younger puppy male
Unless I'm mistaken.
Still, generally speaking the probability is 50% that if one puppy is male the other will be as well, since the sex of one puppy isn't dependent upon the sex of the other.
The only way I can think of the answer being other than 50% is if this is a trick question, such as taking into account the ages of the puppies and therefore making three possible outcomes:
Older puppy male, younger puppy male
Older puppy male, younger puppy female
Older puppy female, younger puppy male
Unless I'm mistaken.
Still, generally speaking the probability is 50% that if one puppy is male the other will be as well, since the sex of one puppy isn't dependent upon the sex of the other.
Okay, I shouldn't have pointed you to that book, because that book sucks. It's not saying the right thing in the right way.Cheeze_Pavilion said:
You should use the Bayes' theorem equation here because you should always use the Bayes' theorem equation. If you set up the priors correctly, Bayes' theorem is never wrong.
Since each coin flip is an independent event, the number of times you toss the coin is irrelevant. The probabilities for the first coin out of one, the first coin out of 10,000,000, and the 10,000,000th coin out of 10,000,000 are the same. There's no difference between "large-scale" and "small-scale" probabilities here.
From the Bayesian perspective:
I know someone has two puppies. At this point, I don't know anything about these specific puppies, so I assume that, for each, P(this puppy is male) = 1/2.
Given two independent events with a probability of 1/2, their joint probability is 1/4.
Now, do you agree so far?
-- Alex
Think of Bayes' theorem as a generalization of Aristotelian logic that accepts truth values other than 0 or 1. This page [http://en.wikipedia.org/wiki/Bayesian_probability] links to several formal justifications for why Bayesian inference is the only logically consistent system of inductive inference. The short form is that any other approach creates a Dutch book, which inherently contradicts part of your definition of probability.Cheeze_Pavilion said:
Two flips that are initially independent. It doesn't matter whether you flip one coin twice, or flip two coins at the same time, or flip one coin and then 10,000,000 coins that you just ignore and then another coin. That means you can safely apply reasoning based on a very large number of retries to this problem, if you need to (you don't, not really).Cheeze_Pavilion said:
All that means is "I am using probability to talk about the certainty of a particular belief about the puppies". You've consistently attacked frequentism as inappropriate because we're only working with one set of puppies, so you're already implicitly using probabilities to represent beliefs.Cheeze_Pavilion said:
You're jumping the gun here by trying to mush them together into a "mixed" category. This is where you keep tripping yourself up. For now, focus on the two events as independent events you are looking at together.Cheeze_Pavilion said:
You have one "X" that can be in state X or state ~X and no others, with P(X)=a and P(~X)=b.
You have another thing "Y" that can be in state Y or state ~Y and no others. P(Y)=c and P(~Y)=d.
Do you agree that a+b=c+d=1?
Since they're independent, you don't have to worry about contingent probabilities. P(X|Y) = P(X|Y') = P(X) and vice versa. So you can put those out of your mind for now.
Does this make sense only up to this point?
How do you find the probability of a combination of the two? Just multiply them together.
P(X, Y) = P(X) * P(Y)
Does you agree that P(X, Y) = P(X) * P(Y) for these things?
-- Alex
The participants aren't modifying their beliefs correctly.Cheeze_Pavilion said:
P (dealer flips two heads consecutively | nothing, I just started playing the game) = 1/4
P (dealer flips two heads consecutively | I just saw that the first one is heads) = 1/2
The second set of odds should be 1/2, 1/2, 1/2, 1/2.
-- Alex
The mathematical probability for a fair coin flip turning up heads is always 1/2. All that the law of large numbers tells you is that if you only test a few times, you might end up not observing the true range and mean -- you're undersampling, like someone who only looks out the window at night and concludes that there is no sun.Cheeze_Pavilion said:
-- Alex
The information contained within the two statements is different.Cheeze_Pavilion said:
P (both male | whatever we know) = P (1st is male | whatever we know) * P (2nd is male | whatever we know)
As we modify "whatever we know", the probabilities shift. Initially this is 1/2 * 1/2 = 1/4.
If I say "the first one is male", P (1st is male | 1st is male) is equal to 1. P (2nd is male | 1st is male) is still 1/2. So now P (both male | 1st is male) = 1/2, thanks to the new information we have. (Symmetrical for "#2 is male and I would've looked at both even if #1 were to be male".)
So, here's the difference for "at least one is male":
Well, now P (1st is male | at least one is male) isn't 1, obviously, since there is the possibility that the 1st still isn't male. But "at least one is male" still gives us useful information. It's just that quantifying it is a bit more challenging.
So, to be all (vaguely) formal here, let's use Bayes' theorem.
P (1st is male | at least one is male) = P (at least one is male | 1st is male) * P (1st is male) / P (at least one is male)
P (1st is male) = 1/2 -- this is the prior probability (i.e. initial, unmodified by our new condition).
P (at least one is male) = 1 - P (both are female) = 1 - P (1st is female) * P (2nd is female) = 1 - 1/4 = 3/4 -- against, this is the prior probability unmodified by our new condition (that we know one is male).
P (at least one is male | 1st is male) = 1.
So, P (1st is male | at least one male) = 1 * (1/2) / (3/4) = 2/3.
Hopefully that result makes some intuitive sense: it's less than 1 since you can't be certain that the 1st one is male, but it's more than it was before because if you *know* that *at least one* is male, it's a bit more likely that the *this particular one* is male, too.
-- Alex
how is zero an option theres always the probability the other one is male if its unkown (unknown would mean zero and one hundred wouldnt work)