Poll: A little math problem

[Alex_P]

Err, that source says that it's 1/4 because it "matches up with long-run frequencies."

This problem is the shortest run possible.

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.

(Where they differ is that the Bayesian outlook is the One True Way while frequentism is for jerks.)

-- Alex

 

[Dragon_of_red]

You all are very very wrong.


0.999 does = 1

No, it dont, it aproximitaly equals one, not fully though, your rounding up.

 

[monalith]

um guys we know its generally 50/50 for humans but what is it for dogs? i mean salt water crocodiles and many species of birds require different tempratures for different sexes so maybe theirs a genetic predisposition for more males.females in beagels

 

[Alex_P]

um guys we know its generally 50/50 for humans but what is it for dogs? i mean salt water crocodiles and many species of birds require different tempratures for different sexes so maybe theirs a genetic predisposition for more males.females in beagels

You're meant to assume it's 50/50.

Which it is, pretty much. Sex selection in dogs is the same as sex selection in humans (X/Y).

-- Alex

 

[scape]

You all are very very wrong.


0.999 does = 1

No, it dont, it aproximitaly equals one, not fully though, your rounding up.

woooosh!

 

[minarri]

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.

 

[Alex_P]

Yes, but *why* should I use a Bayesian equation in this instance? The sources you gave me said of the correctness of 1/2, 1/3, and 1/4: "So far as the formal theory is concerned, they all are!"

I keep asking about short-run outcomes, and you keep talking about how I should use one long-term outcome method over another: if that's your only explanation for your answer, it means your answer isn't relevant to my question, THE question asked in the problem.

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.

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

 

[Alex_P]

That's not really an explanation as much as it is an order and a "trust me" statement.

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.

Okay, I agree, but we're talking about two coin flips here, not one.

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).

I don't care what the Bayesian perspective has to say unless you can convince me the Bayesian perspective applies and is better to use than any other perspective.

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.

No, because there's no way for 'mixed puppies' to be an event with only one puppy. You can't just mix and match like that and compare the chances of two independent events occurring with those of one event that is a result of two independent events itself, neither of which alone make it more or less likely.

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.

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

 

[Alex_P]

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

Okay. Two dealers, two rooms, two players. The game is Coin Flipping. This is Ocean's Fourteen or something.



Dealer A flips for Player A. Comes up heads.

Dealer B flips for Player B. Comes up tails

Probability that in the next flip:

Dealer A flips another heads = .25 (.5 x .5)

Player A flips another heads = .25 (.5 x .5)

Dealer B flips heads this time = .75 (1 - ((.5 x .5)) = (1 - .25)

Player B flips heads this time = .75 (1 - ((.5 x .5)) = (1 - .25)

Players switch rooms/dealers

Player A and Dealer B are both anticipating heads on the same coin flip with different odds: .25 for Player A, .75 for Dealer B.

Player B and Dealer A are both anticipating heads on the same coin flip with different odds: .75 for Player B, .25 for Dealer A.

How is that possible?

The participants aren't modifying their beliefs correctly.

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

 

[Alex_P]

No, the law of large numbers holds for large numbers, not small numbers. That's why they call it that.

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.

-- Alex

 

[Alex_P]

So we don't keep going round and round, here's what I'm looking for: something that explains why there's a different rule for when you get two bites at the apple ("at least one is male") than when you get one ("the first one is male" or "the second one is male and I checked both regardless of the sex of the first one").

That's where the explanation lies if there is one.

The information contained within the two statements is different.

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

 

[Jumplion]

FFFFFFUUUUUUUUCCCCCKKKKKKKKK!!!!!!

This thread is back from the dead!

And I still say it's 50%

 

[SerenadeOfAngels]

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)

 

[santaandy]

The wording of the question says "if they're male, female, or a pair." It makes it sound like a trick question. If they're male or female, then they're both the same thing. 6 of one, a half-dozen of the other. So if one's male, so is the other.

 

[Alex_P]

The wording of the question says "if they're male, female, or a pair." It makes it sound like a trick question. If they're male or female, then they're both the same thing.

... And if they're "a pair", that means one of each sex.

The wording sucks but it doesn't suck quite that badly.

-- Alex