Yesterday I spoke of the Mind Projection Fallacy, giving the example of the alien monster who carries off a girl in a torn dress for intended ravishing—a mistake which I imputed to the artist's tendency to think that a woman's sexiness is a property of the woman herself, woman.sexiness, rather than something that exists in the mind of an observer, and probably wouldn't exist in an alien mind.
The term "Mind Projection Fallacy" was coined by the late great Bayesian Master, E. T. Jaynes, as part of his long and hard-fought battle against the accursèd frequentists. Jaynes was of the opinion that probabilities were in the mind, not in the environment—that probabilities express ignorance, states of partial information; and if I am ignorant of a phenomenon, that is a fact about my state of mind, not a fact about the phenomenon.
I cannot do justice to this ancient war in a few words—but the classic example of the argument runs thus:
You have a coin.
The coin is biased.
You don't know which way it's biased or how much it's biased. Someone just told you, "The coin is biased" and that's all they said.
This is all the information you have, and the only information you have.
You draw the coin forth, flip it, and slap it down.
Now—before you remove your hand and look at the result—are you willing to say that you assign a 0.5 probability to the coin having come up heads?
The frequentist says, "No. Saying 'probability 0.5' means that the coin has an inherent propensity to come up heads as often as tails, so that if we flipped the coin infinitely many times, the ratio of heads to tails would approach 1:1. But we know that the coin is biased, so it can have any probability of coming up heads except 0.5."
The Bayesian says, "Uncertainty exists in the map, not in the territory. In the real world, the coin has either come up heads, or come up tails. Any talk of 'probability' must refer to the information that I have about the coin—my state of partial ignorance and partial knowledge—not just the coin itself. Furthermore, I have all sorts of theorems showing that if I don't treat my partial knowledge a certain way, I'll make stupid bets. If I've got to plan, I'll plan for a 50/50 state of uncertainty, where I don't weigh outcomes conditional on heads any more heavily in my mind than outcomes conditional on tails. You can call that number whatever you like, but it has to obey the probability laws on pain of stupidity. So I don't have the slightest hesitation about calling my outcome-weighting a probability."
I side with the Bayesians. You may have noticed that about me.
Even before a fair coin is tossed, the notion that it has an inherent 50% probability of coming up heads may be just plain wrong. Maybe you're holding the coin in such a way that it's just about guaranteed to come up heads, or tails, given the force at which you flip it, and the air currents around you. But, if you don't know which way the coin is biased on this one occasion, so what?
I believe there was a lawsuit where someone alleged that the draft lottery was unfair, because the slips with names on them were not being mixed thoroughly enough; and the judge replied, "To whom is it unfair?"
To make the coinflip experiment repeatable, as frequentists are wont to demand, we could build an automated coinflipper, and verify that the results were 50% heads and 50% tails. But maybe a robot with extra-sensitive eyes and a good grasp of physics, watching the autoflipper prepare to flip, could predict the coin's fall in advance—not with certainty, but with 90% accuracy. Then what would the real probability be?
There is no "real probability". The robot has one state of partial information. You have a different state of partial information. The coin itself has no mind, and doesn't assign a probability to anything; it just flips into the air, rotates a few times, bounces off some air molecules, and lands either heads or tails.
So that is the Bayesian view of things, and I would now like to point out a couple of classic brainteasers that derive their brain-teasing ability from the tendency to think of probabilities as inherent properties of objects.
Let's take the old classic: You meet a mathematician on the street, and she happens to mention that she has given birth to two children on two separate occasions. You ask: "Is at least one of your children a boy?" The mathematician says, "Yes, he is."
What is the probability that she has two boys? If you assume that the prior probability of a child being a boy is 1/2, then the probability that she has two boys, on the information given, is 1/3. The prior probabilities were: 1/4 two boys, 1/2 one boy one girl, 1/4 two girls. The mathematician's "Yes" response has probability ~1 in the first two cases, and probability ~0 in the third. Renormalizing leaves us with a 1/3 probability of two boys, and a 2/3 probability of one boy one girl.
But suppose that instead you had asked, "Is your eldest child a boy?" and the mathematician had answered "Yes." Then the probability of the mathematician having two boys would be 1/2. Since the eldest child is a boy, and the younger child can be anything it pleases.
Likewise if you'd asked "Is your youngest child a boy?" The probability of their being both boys would, again, be 1/2.
Now, if at least one child is a boy, it must be either the oldest child who is a boy, or the youngest child who is a boy. So how can the answer in the first case be different from the answer in the latter two?
Or here's a very similar problem: Let's say I have four cards, the ace of hearts, the ace of spades, the two of hearts, and the two of spades. I draw two cards at random. You ask me, "Are you holding at least one ace?" and I reply "Yes." What is the probability that I am holding a pair of aces? It is 1/5. There are six possible combinations of two cards, with equal prior probability, and you have just eliminated the possibility that I am holding a pair of twos. Of the five remaining combinations, only one combination is a pair of aces. So 1/5.
Now suppose that instead you asked me, "Are you holding the ace of spades?" If I reply "Yes", the probability that the other card is the ace of hearts is 1/3. (You know I'm holding the ace of spades, and there are three possibilities for the other card, only one of which is the ace of hearts.) Likewise, if you ask me "Are you holding the ace of hearts?" and I reply "Yes", the probability I'm holding a pair of aces is 1/3.
But then how can it be that if you ask me, "Are you holding at least one ace?" and I say "Yes", the probability I have a pair is 1/5? Either I must be holding the ace of spades or the ace of hearts, as you know; and either way, the probability that I'm holding a pair of aces is 1/3.
How can this be? Have I miscalculated one or more of these probabilities?
If you want to figure it out for yourself, do so now, because I'm about to reveal...
That all stated calculations are correct.
As for the paradox, there isn't one. The appearance of paradox comes from thinking that the probabilities must be properties of the cards themselves. The ace I'm holding has to be either hearts or spades; but that doesn't mean that your knowledge about my cards must be the same as if you knew I was holding hearts, or knew I was holding spades.
It may help to think of Bayes's Theorem:
P(H|E) = P(E|H)P(H) / P(E)
That last term, where you divide by P(E), is the part where you throw out all the possibilities that have been eliminated, and renormalize your probabilities over what remains.
Now let's say that you ask me, "Are you holding at least one ace?" Before I answer, your probability that I say "Yes" should be 5/6.
But if you ask me "Are you holding the ace of spades?", your prior probability that I say "Yes" is just 1/2.
So right away you can see that you're learning something very different in the two cases. You're going to be eliminating some different possibilities, and renormalizing using a different P(E). If you learn two different items of evidence, you shouldn't be surprised at ending up in two different states of partial information.
Similarly, if I ask the mathematician, "Is at least one of your two children a boy?" I expect to hear "Yes" with probability 3/4, but if I ask "Is your eldest child a boy?" I expect to hear "Yes" with probability 1/2. So it shouldn't be surprising that I end up in a different state of partial knowledge, depending on which of the two questions I ask.
The only reason for seeing a "paradox" is thinking as though the probability of holding a pair of aces is a property of cards that have at least one ace, or a property of cards that happen to contain the ace of spades. In which case, it would be paradoxical for card-sets containing at least one ace to have an inherent pair-probability of 1/5, while card-sets containing the ace of spades had an inherent pair-probability of 1/3, and card-sets containing the ace of hearts had an inherent pair-probability of 1/3.
Similarly, if you think a 1/3 probability of being both boys is an inherent property of child-sets that include at least one boy, then that is not consistent with child-sets of which the eldest is male having an inherent probability of 1/2 of being both boys, and child-sets of which the youngest is male having an inherent 1/2 probability of being both boys. It would be like saying, "All green apples weigh a pound, and all red apples weigh a pound, and all apples that are green or red weigh half a pound."
That's what happens when you start thinking as if probabilities are in things, rather than probabilities being states of partial information about things.
Probabilities express uncertainty, and it is only agents who can be uncertain. A blank map does not correspond to a blank territory. Ignorance is in the mind.

My first post, so be gentle. :)
I disagree that there is a difference between "Bayesian" and "Frequentist;" or at least, that it has anything to do with what is mentioned in this article. The field of Probability has the unfortunate property of appearing to be a very simple, well defined topic. But it actually is complex enough to be indefinable. Those labels are used by people who want to argue in favor of one definition - of the indefinable - over another. The only difference I see is where they fail to completely address a problem.
Take the biased coin problem as an example. If either label applies to me, it is Frequentist, but my answer is that the one Eliezer_Yudkowsky says is the Bayesian's. He gets the wrong Frequentist solution because he only allows the Frequentist to acknowledge one uncertainty - one random variable - in the problem. Whether the coin came up heads or tails. If a Frequentist says the question is unanswerable, (s)he is wrong because (s)he is using an incomplete solution. The bias b - of a coin already selected - is just as much a random variable as the side s that came up in a coin already flipped. If you claim the answer must be based on the actual value of b for the coins, it must also be based on the actual value of s for this flip. That means the probability is either 0 or 1, which is absurd. (Technically, this error is one of confusing an outcome and an event. An outcome is the specific result of a specific trial, and has no probability. An event is a set of possible outcomes, and is what a probability is assigned to. Eliezer_Yudkowsky's Frequentist is treating the choice of a coin as an outcome, and the result of the flip as an event.)
We can answer the question without knowing anything more about b, than that it is not 1/2. For any 0<=b1<1/2, since we have no other information, b=b1 and b=1-b1 must be treated as equally likely. Regardless of what the distribution of b1 is, this makes the probability the coin landed on heads 1/2.
The classic Two Child Problem has a similar issue, but Eliezer_Yudkowsky did not ask the classic one. I find it best to explain this one in the manner Joseph Bertrand used for his famous Box Paradox. I have two children. What is the probability they share the same gender? That's easy: 1/2. Now I secretly write one gender on a note card. I then show the card to you, and tell you one of my children has that gender. If it says "boy," does the answer change to 1/3? What if it says "girl"? The answers can't be different for the two words you might see; but whatever that answer is, it has to be the same as the answer to the original question (proof by Bayes Theorem). So if the answer does change, we have a paradox.
Yet if presented with the information all at once, "I have two, and one is a boy," Frequentist and Bayesian alike will usually answer "1/3." And they usually will say that anybody who answers 1/2 is addressing the "I have two, and one specific child, by age, is a boy" version Eliezer_Yudkowsky mentioned. But that is not how I get 1/2. There are three random variables, not two: the older child's gender, the younger child's gender, and which gender I will mention if I have the choice of two. Allowing all three to be split 50/50 between "boy" and "girl" makes the answer 1/2, and there is no paradox.
Ironically, my reasoning is what the same mathematicians will use for either the Monty Hall Problem, or the identical Three Prisoners Problem. Two cases that were originally equally likely remain possible. But they are no longer equally possible, because the provider of information had a choice of two in one case, but no choice in the other. Bayesians may claim the difference is a property of the information, and Frequentists (if they use a complete solution) will say there is an additional, implicit random variable. Both work out the same, just by different methods. It is ironic, because while Bertrand's Box Paradox is often compared to these two problems because it is mathematically equivalent to them. The Two Child Problem is closer to being logically equivalent because of the way the information is provided, yet never gets compared. In fact, it is identical if you add a fourth box.
I can't speak for the rest of your post, but
is pretty clearly wrong. (In fact, it looks a lot like you're establishing a prior distribution, and that's uniquely a Bayesian feature.) The probability of an event (the result of the flip is surely an event,... (read more)