Mathematics

Mathematics is the study of crisply specified formal objects — for example, [numbers](https://arbital.com/p/number) — and the ways of knowing their [properties](https://arbital.com/p/mathematical_properties) — such as [https://arbital.com/p/-proofs](https://arbital.com/p/-proofs).  We can see "[https://arbital.com/p/-258](https://arbital.com/p/-258)" as the study of "which conclusions follow with certainty from which premises".  Using this definition of logic, we can also see mathematics as the study of logical objects in logical universes — entities whose properties follow from specifications about them, rather than from observation of the real world.  The number 3 is a logical object because its behavior follows from axioms about addition and multiplication; Mount Everest is a physical object because we learn about it by physically measuring Mount Everest.

Tsvi BT
Eugene Dobry
Eliezer Yudkowsky
Zack M. Davis
Alexei Andreev
Eric Rogstad
Mark Chimes
Eric Bruylant
Nate Soares
Kevin Clancy
Jack Gallagher
Dylan Hendrickson
Travis Rivera
Chris Barnett
Team Arbital
Patrick Stevens


Bayesian update

A Bayesian *update* or *belief revision* is a change in [probabilistic](https://arbital.com/p/1rf) beliefs after gaining new knowledge.  For example, after observing a patient's test result, we might revise our probability that a patient has a certain disease.  If this belief revision obeys [Bayes's Rule](https://arbital.com/p/1lz), then it is called [Bayesian](https://arbital.com/p/1r8).

Bayesian belief updates have a number of other interesting properties, and exemplify many key principles of clear reasoning or [rationality](https://arbital.com/p/9l).  Mapping Bayes's Rule onto real-life problems of encountering new evidence allows us to reproduce many intuitive features that have been suggested for "how to revise beliefs in the face of new evidence".

- [](https://arbital.com/p/21v)
- The scientific virtues of [falsifiability, advance prediction, boldness, precision, and falsificationism](https://arbital.com/p/220) can be seen in a Bayesian light.

Eric Bruylant
Eliezer Yudkowsky
Nate Soares
Alexei Andreev

Summary:
*Bayesian updating* or *Bayesian belief revision* is a way of changing [probabilistic](https://arbital.com/p/1rf) beliefs, in response to evidence, in a way that obeys [https://arbital.com/p/-1bv](https://arbital.com/p/-1bv) and in particular [https://arbital.com/p/1lz](https://arbital.com/p/1lz).

When we observe a piece of evidence that is more likely to be seen if X is true than if X is false, we should assign more credence to X.  For example, if I take a random object from your kitchen, and then I tell you this object is sharp, you should assign more credence than you did previously that it's a knife (even though forks are also sharp, so you won't be certain).

For more on this subject see [the Arbital guide to Bayes's rule](https://arbital.com/p/1zq).


Bayes' rule

summary(Technical): Bayes' rule (aka Bayes' theorem) is the quantitative [law](https://arbital.com/p/1bv) governing how to [revise probabilistic beliefs in response to observing new evidence](https://arbital.com/p/1ly).  Suppose we previously thought that the probability of $h_1$, denoted $\mathbb {P}(h_1)$, was twice as great as $\mathbb {P}(h_2)$.  Now we see a new piece of evidence, $e_0$, such that the probability of our seeing $e_0$ if $h_1$ is true (denoted by $\mathbb {P}(e_0\mid h_1)$) is one-fourth as great as $\mathbb {P}(e_0\mid h_2)$ (the probability of seeing $e_0$ if $h_2$ is true).  After observing $e_0$, we should think that $h_1$ is now half as likely as $h_2$:

$$\frac{\mathbb {P}(h_1\mid e_0)}{\mathbb {P}(h_2\mid e_0)} = \frac{\mathbb {P}(h_1)}{\mathbb {P}(h_2)} \cdot \frac{\mathbb {P}(e_0\mid h_1)}{\mathbb {P}(e_0\mid h_2)}$$

[More generally](https://arbital.com/p/1zj), Bayes' rule states:  $\mathbb P(\mathbf{H}\mid e) \propto \operatorname{\mathbb {P}}(e\mid \mathbf{H}) \cdot \operatorname{\mathbb {P}}(\mathbf{H}).$

Bayes' rule (aka Bayes' theorem) is the quantitative law of [probability theory](https://arbital.com/p/1bv) governing how to [revise](https://arbital.com/p/1ly) probabilistic beliefs in response to observing new evidence.

You may want to start at the [Guide](https://arbital.com/p/1zq) or the [Fast Intro](https://arbital.com/p/693).

# The laws of reasoning

Imagine that, as part of a clinical study, you're being tested for a rare form of cancer, which affects 1 in 10,000 people. You have no reason to believe that you are more or less likely than average to have this form of cancer. You're administered a test which is 99% accurate, both in terms of [https://arbital.com/p/-specificity](https://arbital.com/p/-specificity) and [https://arbital.com/p/-sensitivity](https://arbital.com/p/-sensitivity): It _correctly_ detects the cancer (in patients who have it) 99% of the time, and it _incorrectly_ detects cancer (in patients who don't have it) only 1% of the  time. The test results come back positive. What's the chance that you have cancer?

Bayes' rule says that the answer is precisely a 1 in 102 chance, which is a probability a little below 1%. The remarkable thing about this is that there is only one answer: the odds of you having that type of cancer, given the above information, is _exactly_ 1 in 102; no more, no less.

%comment: (999,900 * 0.99 + 100 * 0.99) / (100 * 0.99) = (10098 / 99) = 102.  Please leave this comment here so the above paragraph is not edited to be wrong.%

This is one of the key insights of Bayes' rule: Given what you knew, and what you saw, the maximally accurate state of belief for you to be in is completely pinned down. While that belief state is quite difficult to find in practice, we know how to find it in principle. If you want your beliefs to become more accurate as you observe the world, Bayes' rule gives some hints about what you need to do.

# Learn Bayes' rule

- [__Bayes' rule: Odds form.__](https://arbital.com/p/1x5) Bayes' rule is simple, if you think in terms of relative odds.
- [__Bayes' rule: Proportional form.__](https://arbital.com/p/1zm) The fastest way to say something both convincing and true about belief-updating.
- [__Bayes' rule: Log-odds form.__](https://arbital.com/p/1zh) A simple transformation of Bayes' rule reveals tools for measuring degree of belief, and strength of evidence.
- [__Bayes' rule: Probabilistic form.__](https://arbital.com/p/554) The original formulation of Bayes' rule.
- [__Bayes' rule: Functional form.__](https://arbital.com/p/1zj) Bayes' rule for continuous variables.
- [__Bayes' rule: Vector form.__](https://arbital.com/p/1zg) For when you want to apply Bayes' rule to lots of evidence and lots of variables, all in one go.

# Implications of Bayes' rule

- [__A Bayesian view of scientific virtues.__](https://arbital.com/p/220) Why is it that science relies on bold, precise, and falsifiable predictions? Because of Bayes' rule, of course.
- [__Update by inches.__](https://arbital.com/p/update_by_inches) It's virtuous to change your mind in response to overwhelming evidence. It's even more virtuous to shift your beliefs a little bit at a time, in response to *all* evidence (no matter how small).
- [__Belief revision as probability elimination.__](https://arbital.com/p/1y6) Update your beliefs by throwing away large chunks of probability mass.
- [__Shift towards the hypothesis of least surprise.__](https://arbital.com/p/552) When you see new evidence, ask: which hypothesis is *least surprised?*
- [__Extraordinary claims require extraordinary evidence.__](https://arbital.com/p/21v) The people who adamantly claim they were abducted by aliens do provide *some* evidence for aliens. They just don't provide quantitatively *enough* evidence.
- [__Ideal reasoning via Bayes' rule.__](https://arbital.com/p/) Bayes' rule is to reasoning as the [Carnot cycle](https://arbital.com/p/https://en.wikipedia.org/wiki/Carnot_cycle) is to engines: Nobody can be a perfect Bayesian, but Bayesian reasoning is still the theoretical ideal.

# Related content

- [__Subjective probability.__](https://arbital.com/p/4vr) Probability is in the mind, not the world. If you don't know whether a tossed coin came up heads or tails, that's a fact about you, not a fact about the coin.
- [__Probability theory.__](https://arbital.com/p/1bv) The quantification and study of objects that represent uncertainty about the world, and methods for making those representations more accurate.
- [__Information theory.__](https://arbital.com/p/3qq) The quantification and study of information, communication, and what it means for one object to tell us about another.

# Other articles and introductions

- [Wikipedia](https://arbital.com/p/http://en.wikipedia.org/wiki/Bayes%27_rule)
- [Better Explained](https://arbital.com/p/http://betterexplained.com/articles/an-intuitive-and-short-explanation-of-bayes-theorem/)

Dan Davies
Eliezer Yudkowsky
Alexei Andreev
Eric Bruylant
Nate Soares
Sandi x

Summary:
When an observation is more likely given one state of the world than another, we should increase our credence that we're in the world that was more likely to have produced that observation.

Suppose that Professor Plum and Miss Scarlet are two suspects in a murder, and that we start out thinking that Professor Plum is twice as likely to have committed the murder as Miss Scarlet.  We then discover that the victim was poisoned.  We think that, on occasions where they do commit murders, Professor Plum is around one-fourth as likely to use poison as Miss Scarlet.  Then after observing the victim was poisoned, we should think Professor Plum is around half as likely to have committed the murder as Miss Scarlet: $2 \times \dfrac{1}{4} = \dfrac{1}{2}.$

The quantitative rule at work here, in its various forms, is known as Bayes' rule.


Math 2

Do you work with math on a fairly routine basis?  Do you have little trouble grasping new mathematical ideas that use language you already know?  Having this requisite would be typical of a computer programmer, a physical engineer, or someone else who routinely works with mathematically-structured ideas.  At this level, you start to see LaTeX formulas in passing, but they'll have explanations attached.

Eliezer Yudkowsky
Alexei Andreev
Eric Bruylant
Nate Soares
Patrick Stevens
Joe Zeng


Bayesian reasoning

According to its advocates, Bayesian reasoning is a way of seeing the world, and our beliefs about the world, in the light of [probability theory](https://arbital.com/p/1bv), in particular Bayes's Theorem or [Bayes's Rule](https://arbital.com/p/1lz).  This probability-theoretic way of seeing the world can apply to scientific issues, to tasks in machine learning, and to everyday life.

To [start learning](https://arbital.com/p/1zq), visit [Arbital's Guide to Bayes's Rule](https://arbital.com/p/1zq).

After that, consider visiting the [Bayesian update](https://arbital.com/p/1ly) page.

Nate Soares
Alexei Andreev
Eliezer Yudkowsky

Summary: Bayesian reasoning is a way of interpreting the world, and our beliefs about the world, in the light of [probability theory](https://arbital.com/p/1bv), in particular Bayes's Theorem or [Bayes's Rule](https://arbital.com/p/1lz).  Applications include scientific statistics, machine learning, and everyday life.


Odds

summary(Technical): Odds express relative chances.  If the odds for X versus Y are 2 : 3, this expresses that we think that X is 2/3 = 0.666... times as likely as Y.  Clearly, odds of 6 : 9 express the same idea; odds are invariant up to multiplication by a positive factor.  When an odds ratio [exhausts all the possibilities](https://arbital.com/p/1rd), then we can convert its components to probabilities by [normalizing](https://arbital.com/p/1rk) them so that they sum to 1.  In the example above, the probabilities would be $2:3 = \frac{2}{3+2}:\frac{3}{3+2} = 0.4:0.6.$

![](https://i.imgur.com/GVZnz2c.png?0)

Odds are a tool for expressing relative [chances](https://arbital.com/p/1rf).  If the odds of a tree in a forest being sick versus healthy are 2 : 3, this says that there are 2 sick trees for every 3 healthy trees.  (The probability of a tree being sick, in this case, is 2/5 or 40%.)

![](https://i.imgur.com/GVZnz2c.png?0)

Odds are expressed in the form "X to Y", e.g. "7 to 9 for X versus Y", more compactly written as $7:9$.

The representation of chances as odds is often used in gambling and [https://arbital.com/p/-statistics](https://arbital.com/p/-statistics).

Gregor Gerasev
Eliezer Yudkowsky
Emile Kroeger
Alexei Andreev
Eric Rogstad
Stephanie Zolayvar
Eric Bruylant
Nate Soares

Summary:
Odds are a tool for expressing *relative* chances.  If the odds of a tree in a forest being sick versus healthy are 2 : 3, this says that there are 2 sick trees for every 3 healthy trees.  (The *probability* of a tree being sick, in this case, is 2/5 or 40%.)

![](https://i.imgur.com/GVZnz2c.png?0)


Mutually exclusive and exhaustive

A set of propositions is "mutually exclusive and exhaustive" when exactly one of the propositions must be true.  For example, of the two propositions "The sky is blue" and "It is not the case that the sky is blue", exactly one of those must be the case.  Therefore, the [probabilities](https://arbital.com/p/1rf) of those propositions must sum to exactly 1.

If a set $X$ of propositions is "mutually exclusive", this states that for every two distinct propositions, the probability that both of them will be true simultaneously is zero:

$\forall i: \forall j: i \neq j \implies \mathbb{P}(X_i \wedge X_j) = 0.$

This implies that for every two distinct propositions, the probability of their union equals the sum of their probabilities:

$\mathbb{P}(X_i \vee X_j) = \mathbb{P}(X_i) + \mathbb{P}(X_j) - \mathbb{P}(X_j \wedge X_j) = \mathbb{P}(X_i) + \mathbb{P}(X_j).$

The "exhaustivity" condition states that the union of all propositions in $X,$ has probability $1$ (the probability of at least one $X_i$ happening is $1$):

$\mathbb{P}(X_1 \vee X_2 \vee \dots \vee X_N) = 1.$

Therefore mutual exclusivity and exhaustivity imply that the probabilities of the propositions sum to 1:

$\displaystyle \sum_i \mathbb{P}(X_i) = 1.$

Eric Bruylant
Eliezer Yudkowsky
Fedor Belolutskiy


Probability

*Probabilities* are the central subject of the discipline of [https://arbital.com/p/1bv](https://arbital.com/p/1bv).  $\mathbb{P}(X)$ denotes our level of belief, or someone's level of belief, that the proposition $X$ is true.  In the classical and canonical representation of probability, 0 expresses absolute incredulity, and 1 expresses absolute credulity. %%knows-requisite([https://arbital.com/p/1mh](https://arbital.com/p/1mh)):  Furthermore, [mutually exclusive](https://arbital.com/p/1rd) events have additive classical probabilities: $\mathbb{P}(X \wedge Y) = 0 \implies \mathbb{P}(X \vee Y) = \mathbb{P}(X) + \mathbb{P}(Y).$%%

For the standard probability axioms, see https://en.wikipedia.org/wiki/Probability_axioms.

# Notation

$\mathbb{P}(X)$ is the probability that X is true.

$\mathbb{P}(\neg X) = 1  - \mathbb{P}(X)$ is the probability that X is false.

$\mathbb{P}(X \wedge Y)$ is the probability that both X and Y are true.

$\mathbb{P}(X \vee Y)$ is the probability that X or Y or both are true.

$\mathbb{P}(X|Y) := \frac{\mathbb{P}(X \wedge Y}{\mathbb{P}(Y)}$ is the **[conditional probability](https://arbital.com/p/1rj) of X given Y.**  That is, $\mathbb{P}(X|Y)$ is **the degree to which we would believe X, assuming Y to be true.**  $\mathbb{P}(yellow|banana)$ expresses "The probability that a banana is yellow." $\mathbb{P}(banana|yellow)$ expresses "The probability that a yellow thing is a banana".

# Centrality of the classical representation

While there are other ways of expressing quantitative degrees of belief, such as [odds ratios](https://arbital.com/p/1rb), there are several especially useful properties or roles of classical probabilities that give them a central  / convergent / canonical status among possible ways of representing credence.

[Odds ratios](https://arbital.com/p/1rb) are isomorphic to probabilities - we can readily go back and forth between a probability of 20%, and odds of 1:4.  But unlike odds ratios, probabilities have the further appealing property of being able to add the probabilities of two [mutually exclusive](https://arbital.com/p/1rd) possibilities to arrive at the probability that one of them occurs.  The 1/6 probability of a six-sided die turning up 1, plus the 1/6 probability of a die turning up 2, equals the 1/3 probability that the die turns up 1 or 2.  The odds ratios 1:5, 1:5, and 1:2 don't have this direct relation (though we could convert to probabilities, add, and then convert back to odds ratios).

Thus, classical probabilities are uniquely the quantities that must appear in the [expected utilities](https://arbital.com/p/18v) to weigh how much we proportionally care about the uncertain consequences of our decisions.  When an outcome has classical probability 1/3, we multiply the degree to which we care by a factor of 1/3, not by, e.g., the odds ratio 1:2.

If the amount you'd pay for a lottery ticket that paid out on 1 or 2 was more or less than twice the price you paid for a lottery ticket that only paid out on 1, or a lottery ticket that paid out on 2, then I could buy from you and sell to you a combination of lottery tickets such that you would end up with a certain loss.  This is an example of a [Dutch book argument](https://arbital.com/p/dutch_book), which is one kind of coherence theorem that underpins classical probability and its role in choice.  (If we were dealing with actual betting and gambling, you might reply that you'd just refuse to bet on disadvantageous combinations; but in the much larger gamble that is life, "doing nothing" is just one more choice with an uncertain, probabilistic payoff.)

The combination of several such coherence theorems, most notably including the [Dutch Book arguments](http://plato.stanford.edu/entries/dutch-book/), [Cox's Theorem](https://en.wikipedia.org/wiki/Cox%27s_theorem) and its variations for probability theory, and the [Von Neumann-Morgenstern theorem](https://en.wikipedia.org/wiki/Von_Neumann%E2%80%93Morgenstern_utility_theorem) (VNM) and its variations for expected utility, together give the classical probabilities between 0 and 1 a *central* status in the theory of [epistemic and instrumental rationality](https://arbital.com/p/9l).  Other ways of representing scalar probabilities, or alternatives to scalar probability, would need to be converted or munged back into classical probabilities in order to animate [agents](https://arbital.com/p/agents) making coherent choices.

This also suggests that bounded agents which *approximate* coherence, or at least manage to avoid blatantly self-destructive violations of coherence, might have internal mental states which can be *approximately* viewed as corresponding to classical probabilities.  Perhaps not in terms of such agents necessarily containing floating-point numbers that directly represent those probabilities internally, but at least in terms of our being able to look over the agent's behavior and deduce that they were "behaving as if" they had assigned some coherent classical probability.

Eliezer Yudkowsky
Alexei Andreev
Eric Rogstad
Eric Bruylant
Nate Soares
Alan Liddell


Joint probability

To write "the chance that both X and Y are true" using standard [https://arbital.com/p/1rf](https://arbital.com/p/1rf) notation, we write $\mathbb{P}(X \wedge Y)$ or $\mathbb{P}(X, Y)$.


Conditional probability

summary:  $\mathbb{P}(X\mid Y)$ means "The [probability](https://arbital.com/p/1rf) that X is true, assuming Y is true."

- $\mathbb{P}(yellow\mid banana)$ is "the chance that something is yellow, given that we know it is a banana" or equivalently "the chance that a banana is yellow".
- $\mathbb{P}(banana\mid yellow)$ expresses "how much we should think a random object is a banana, after being told that it was yellow" or "the chance that a yellow thing is a banana".

To calculate a conditional probability $\mathbb{P}(X\mid Y)$, we consider only the cases where Y is true, and ask about the cases where X is also true.

Suppose a barrel contains 15 round green marbles, 5 round blue marbles, 70 square green marbles, and 10 square blue marbles.  "The probability that a marble is blue, after we've been told that it's round" or "The probability a round marble is blue" is calculated by restricting our attention to only the 20 round marbles, and asking about the 5 marbles that are both blue and round.

Letting $\mathbb{P}(blue \wedge round)$ denote "the probability a marble is both blue and round":

$\mathbb{P}(blue\mid round) := \frac{\mathbb{P}(blue \wedge round)}{\mathbb{P}(round)} = \frac{\text{5% blue and round marbles}}{\text{20% round marbles}} = \frac{5}{20} = 0.25.$

In general, $\mathbb{P}(X\mid Y) := \frac{\mathbb{P}(X \wedge Y)}{\mathbb{P}(Y)}.$

summary(Technical):  $\mathbb{P}(X\mid Y) := \frac{\mathbb{P}(X \wedge Y)}{\mathbb{P}(Y)}$ is the answer to the question, "Assuming $Y$ to be true, what is the probability of $X$?" or "Constraining our attention to only possibilities where $Y$ is true, what is the probability of $X \wedge Y$ inside those cases?"  (Where $X \wedge Y$ denotes "X and Y" or "Both X and Y are true".)

Thus, $\mathbb P(observation\mid hypothesis)$ would denote the [likelihood](https://arbital.com/p/56v) of seeing some observation, *if* a hypothesis is true.   $\mathbb P(hypothesis\mid observation)$ would denote the [revised](https://arbital.com/p/1ly) probability we ought to assign to a hypothesis, after learning that the observation was true.

The conditional probability $\mathbb{P}(X\mid Y)$ means "The [probability](https://arbital.com/p/1rf) of $X$ given $Y$."  That is, $\mathbb P(left\mid right)$ means "The probability that $left$ is true, assuming that $right$ is true."

$\mathbb P(yellow\mid banana)$ is the probability that a banana is yellow - if we know something to be a banana, what is the probability that it is yellow?

$\mathbb P(banana\mid yellow)$ is the probability that a yellow thing is a banana - if the right side is known to be $yellow$, then we ask the question on the left, what is the probability that this is a $banana$?

# Definition

To obtain the probability $\mathbb P(left \mid right),$ we constrain our attention to only cases where $right$ is true, and ask about cases within $right$ where $left$ is also true.

Let $X \wedge Y$ denote "$X$ and $Y$" or "$X$ and $Y$ are both true".  Then:

$$\mathbb P(left \mid right) = \dfrac{\mathbb P(left \wedge right)}{\mathbb P(right)}.$$

We can see this as a kind of "zooming in" on only the cases where $right$ is true, and asking, *within* this universe, for the cases where $right$ *and* $left$ are true.

# Example 1

Suppose you have a bag containing objects that are either red or blue, and either square or round, where the number of each is given by the following table:

$$\begin{array}{l\mid r\mid r}
& Red & Blue \\
\hline
Square & 1 & 2 \\
\hline
Round & 3 & 4
\end{array}$$

If you reach in and feel a round object, the conditional probability that it is red is given in by zooming in on only the round objects, and asking about the frequency of objects that are round *and* red inside this zoomed-in view:

$$\mathbb P(red\mid round) = \dfrac{\mathbb P(red \wedge round)}{\mathbb P(round)} = \dfrac{3}{3 + 4} = \dfrac{3}{7}$$

If you look at the object nearest the top, and can see that it's blue, but not see the shape, then the conditional probability that it's a square is:

$$\mathbb P(square\mid blue) = \dfrac{\mathbb P(square \wedge blue)}{\mathbb P(blue)} = \dfrac{2}{2 + 4} = \dfrac{1}{3}$$

![conditional probabilities bag](https://i.imgur.com/zscEdLj.png?0)

# Example 2

Suppose you're Sherlock Holmes investigating a case in which a red hair was left at the scene of the crime.

The Scotland Yard detective says, "Aha!  Then it's Miss Scarlet.  She has red hair, so if she was the murderer she almost certainly would have left a red hair there.  $\mathbb P(red hair\mid Scarlet) = 99\%,$ let's say, which is a near-certain conviction, so we're done."

"But no," replies Sherlock Holmes.  "You see, but you do not correctly track the meaning of the conditional probabilities, detective.  The knowledge we require for a conviction is not $\mathbb P(redhair\mid Scarlet),$ the chance that Miss Scarlet would leave a red hair, but rather $\mathbb P(Scarlet\mid redhair),$ the chance that this red hair was left by Scarlet.  There are other people in this city who have red hair."

"So you're saying..." the detective said slowly, "that $\mathbb P(redhair\mid Scarlet)$ is actually much lower than $1$?"

"No, detective.  I am saying that just because $\mathbb P(redhair\mid Scarlet)$ is high does not imply that $\mathbb P(Scarlet\mid redhair)$ is high.  It is the latter probability in which we are interested - the degree to which, *knowing* that a red hair was left at the scene, we *infer* that Miss Scarlet was the murderer.  This is not the same quantity as the degree to which, *assuming* Miss Scarlet was the murderer, we would *guess* that she might leave a red hair."

"But surely," said the detective, "these two probabilities cannot be entirely unrelated?"

"Ah, well, for that, you must read up on [Bayes' rule](https://arbital.com/p/1lz)."

# Example 3

> "Even if most Dark Wizards are from Slytherin, very few Slytherins are Dark Wizards. There aren't all that many Dark Wizards, so not all Slytherins can be one."

> "So yeh're saying, that most Dark Wizards are Slytherins... but..."

> "But most Slytherins are not Dark Wizards."

— Harry Potter and the Methods of Rationality, Ch. 100

Eliezer Yudkowsky
Nate Soares
Alexei Andreev

Summary:
$\mathbb{P}(X\mid Y)$ means "The [probability](https://arbital.com/p/1rf) that X is true, assuming Y is true."

$\mathbb{P}(yellow\mid banana)$ is "the chance that something is yellow, given that we know it is a banana" or "the chance that a banana is yellow".

Conversely, $\mathbb{P}(banana\mid yellow)$ expresses "how much we should think a random object is a banana, after being told that it was yellow."


Normalization (probability)

"Normalization" is an arithmetical procedure carried out to obtain a set of [probabilities](https://arbital.com/p/1rf) summing to exactly 1, in cases where we believe that [exactly one of the corresponding possibilities is true](https://arbital.com/p/1rd), and we already know the [relative probabilities](https://arbital.com/p/1rb).

For example, suppose that the [odds](https://arbital.com/p/1rb) of Alexander Hamilton winning a presidential election are 3 : 2.  But Alexander Hamilton must either win or not win, so the *probabilities* of him winning *or* not winning should sum to 1.  If we just add 3 and 2, however, we get 5, which is an unreasonably large probability.

If we rewrite the odds as 0.6 : 0.4, we've preserved the same proportions, but made the terms sum to 1.  We therefore calculate that Hamilton has a 60% probability of winning the election.

We normalized those odds by dividing each of the terms by the sum of terms, i.e., went from 3 : 2 to $\frac{3}{3+2} : \frac{2}{3+2} = 0.6 : 0.4.$

In converting the odds $m : n$ to $\frac{m}{m+n} : \frac{n}{m+n},$ the factor $\frac{1}{m+n}$ by which we multiply all elements of the ratio is called a [normalizing constant](https://arbital.com/p/https://en.wikipedia.org/wiki/Normalizing_constant).

More generally, if we have a relative-odds function $\mathbb{O}(H)$ where $H$ has many components, and we want to convert this to a probability function $\mathbb{P}(H)$ that sums to 1, we divide every element of $\mathbb{O}(H)$ by the sum of all elements in $\mathbb{O}(H).$  That is:

$\mathbb{P}(H_i) = \frac{\mathbb{O}(H_i)}{\sum_i \mathbb{O}(H_i)}$

Analogously, if $\mathbb{O}(x)$ is a continuous distribution on $X$, we would normalize it (create a proportional probability function $\mathbb{P}(x)$ whose integral is equal to 1) by dividing $\mathbb{O}(x)$ by its own integral:

$\mathbb{P}(x) = \frac{\mathbb{O}(x)}{\int \mathbb{O}(x) \operatorname{d}x}$

In general, whenever a probability function on a variable is *proportional* to some other function, we can obtain the probability function by *normalizing* that function:

$\mathbb{P}(H) \propto \mathbb{O}(H) \implies \mathbb{P}(H) = \frac{\mathbb{O}(H)}{\sum \mathbb{O}(H)}$

Summary:
"Normalization" obtains a set of [probabilities](https://arbital.com/p/1rf) summing to 1, in [cases where they ought to sum to 1](https://arbital.com/p/1rd).  We do this by dividing each pre-normalized number by the sum of all pre-normalized numbers.

Suppose the [odds](https://arbital.com/p/1rb) of Alexander Hamilton winning an election are 3 : 2.  We think the proportions are right (Alexander is 1.5 times as likely to win as not win) but we want *probabilities*.  To say that Hamilton has probability 3 of winning the election would be very strange indeed.  But if we divide each of the terms by the sum of all the terms, they'll end up summing to one:  $3:2 \cong \frac{3}{3+2} : \frac{2}{3+2} = 0.6 : 0.4.$  Thus, the probability that Hamilton wins is 60%.


Prior probability

"Prior [probability](https://arbital.com/p/1rf)", "prior [odds](https://arbital.com/p/1rb)", or just "prior" refers to a state of belief that obtained before seeing a piece of new evidence.  Suppose there are two suspects in a murder, Colonel Mustard and Miss Scarlet.  After determining that the victim was poisoned, you think Mustard and Scarlet are respectively 25% and 75% likely to have committed the murder.  *Before* determining that the victim was poisoned, perhaps, you thought Mustard and Scarlet were equally likely to have committed the murder (50% and 50%).  In this case, your "prior probability" of Miss Scarlet committing the murder was 50%, and your "posterior probability" after seeing the evidence was 75%.

The prior probability of a hypothesis $H$ is often being written with the unconditioned notation $\mathbb P(H)$, while the posterior after seeing the evidence $e$ is often being denoted by the [conditional probability](https://arbital.com/p/1rj) $\mathbb P(H\mid e).$%%note: [E. T. Jaynes](http://bayes.wustl.edu/) was known to insist on using the explicit notation $\mathbb P (H\mid I_0)$ to denote the prior probability of $H$, with $I_0$ denoting the prior, and never trying to write any entirely unconditional probability $\mathbb P(X)$.  Since, said Jaynes, we always have *some* prior information.%% %%knows-requisite([https://arbital.com/p/1r6](https://arbital.com/p/1r6)): This however is a heuristic rather than a law, and might be false inside some complicated problems.  If we've already seen $e_0$ and are now updating on $e_1$, then in this new problem the new prior will be $\mathbb P(H\mid e_0)$ and the new posterior will be $\mathbb P(H\mid e_1 \wedge e_0).$ %%

For questions about how priors are "ultimately" determined, see [https://arbital.com/p/11w](https://arbital.com/p/11w).

Cuyler Brehaut
Eliezer Yudkowsky
Alexei Andreev
Eric Bruylant
Nate Soares


Interest in mathematical foundations in Bayesianism

Eric Bruylant
Eliezer Yudkowsky
Alexei Andreev

Summary: "Want" this [requisite](https://arbital.com/p/1ln) if you prefer to see extra information about the mathematical foundations in [Bayesian reasoning](https://arbital.com/p/1r8).


Posterior probability

"Posterior [probability](https://arbital.com/p/1rf)" or "posterior [odds](https://arbital.com/p/1rb)" refers our state of belief *after* seeing a piece of new evidence and doing a [Bayesian update](https://arbital.com/p/1ly).  Suppose there are two suspects in a murder, Colonel Mustard and Miss Scarlet.  Before determining the victim's cause of death, perhaps you thought Mustard and Scarlet were equally likely to have committed the murder (50% and 50%).  After determining that the victim was poisoned, you now think that Mustard and Scarlet are respectively 25% and 75% likely to have committed the murder.  In this case, your "[prior probability](https://arbital.com/p/1rm)" of Miss Scarlet committing the murder was 50%, and your "posterior probability" *after* seeing the evidence was 75%.  The posterior probability of a hypothesis $H$ after seeing the evidence $e$ is often denoted using the [conditional probability notation](https://arbital.com/p/1rj) $\mathbb P(H\mid e).$

Eliezer Yudkowsky
Alexei Andreev
Eric Rogstad
Eric Bruylant
Nate Soares


Relative likelihood

Relative likelihoods express how *relatively* more likely an observation is, comparing one hypothesis to another.  For example, suppose we're investigating the murder of Mr. Boddy, and we find that he was killed by poison. The suspects are Miss Scarlett and Colonel Mustard. Now, suppose that the [probability](https://arbital.com/p/-1rf) that Miss Scarlett would use poison, if she _were_ the murderer, is 20%. And suppose that the probability that Colonel Mustard would use poison, if he were the murderer, is 10%. Then, Miss Scarlett is *twice as likely* to use poison as a murder weapon as Colonel Mustard. Thus, the "Mr. Boddy was poisoned" evidence supports the "Scarlett" hypothesis twice as much as the "Mustard" hypothesis, for relative likelihoods of $(2 : 1).$

These likelihoods are called "relative" because it wouldn't matter if the respective probabilities were 4% and 2%, or 40% and 20% — what matters is the _relative proportion_.

Relative likelihoods may be given between many different hypotheses at once. Given the evidence $e_p$ = "Mr. Boddy was poisoned", it might be the case that Miss Scarlett, Colonel Mustard, and Mrs. White have the respective probabilities 20%, 10%, and 1% of using poison any time they commit a murder. In this case, we have three hypotheses — $H_S$ = "Scarlett did it", $H_M$ = "Mustard did it", and $H_W$ = "White did it". The relative likelihoods between them may be written $(20 : 10 : 1).$

In general, given a list of hypotheses $H_1, H_2, \ldots, H_n,$ the relative likelihoods on the evidence $e$ can be written as a [scale-invariant list](https://arbital.com/p/) of the likelihoods $\mathbb P(e \mid H_i)$ for each $i$ from 1 to $n.$ In other words, the relative likelihoods are

$$ \alpha \mathbb P(e \mid H_1) : \alpha \mathbb P(e \mid H_2) : \ldots : \alpha \mathbb P(e \mid H_n) $$

where the choice of $\alpha > 0$ does not change the value denoted by the list (i.e., the list is [scale-invariant](https://arbital.com/p/scale_invariant_list)). For example, the relative likelihood list $(20 : 10 : 1)$ above denotes the same thing as the relative likelihood list $(4 : 2 : 0.20)$ denotes the same thing as the relative likelihood list $(60 : 30 : 3).$ This is why we call them "relative likelihoods" — all that matters is the ratio between each term, not the absolute values.

Any two terms in a list of relative likelihoods can be used to generate a [https://arbital.com/p/-56t](https://arbital.com/p/-56t) between two hypotheses. For example, above, the likelihood ratio $H_S$ to $H_M$ is 2/1, and the likelihood ratio of $H_S$ to $H_W$ is 20/1. This means that the evidence $e_p$ supports the "Scarlett" hypothesis 2x more than it supports the "Mustard" hypothesis, and 20x more than it supports the "White" hypothesis.

Relative likelihoods summarize the [strength of the evidence](https://arbital.com/p/22x) represented by the observation that Mr. Boddy was poisoned — under [Bayes' rule](https://arbital.com/p/1lz), the evidence points to Miss Scarlett to the same degree whether the absolute probabilities are 20% vs. 10%, or 4% vs. 2%.

By Bayes' rule, the way to update your beliefs in the face of evidence is to take your [prior](https://arbital.com/p/1rm) [https://arbital.com/p/-1rb](https://arbital.com/p/-1rb) and simply multiply them by the corresponding relative likelihood list, to obtain your [posterior](https://arbital.com/p/1rp) odds. See also [https://arbital.com/p/1x5](https://arbital.com/p/1x5).

Eliezer Yudkowsky
Alexei Andreev
Eric Rogstad
Eric Bruylant
Nate Soares
Al Prihodko


Bayes' rule examples

This page and its tabs store exemplar problems for Bayes' rule.  You can suggest additional example problems by leaving a comment on the appropriate tab.

Problem types by tab:

- [Introductory](https://arbital.com/p/22w).  Meant as a small set of problems for people who haven't heard of Bayes' rule before, with examples that will be discussed in the introductory pages for Bayes' rule.
- [Homework](https://arbital.com/p/).  Problems for people who already know Bayes' rule and want to test their grasp of it in straightforward ways.
- [Clever](https://arbital.com/p/).  Problems for people who wish to be mathematically entertained, or that exhibit some surprising facet or gotcha of Bayes' rule.
- [Realistic](https://arbital.com/p/1x4).  Instances of Bayesian reasoning that have arisen in real life.

Eric Bruylant
Nate Soares
Eliezer Yudkowsky


Waterfall diagram

Waterfall diagrams, like [frequency diagrams](https://arbital.com/p/560), provide a way of visualizing [Bayes' Rule](https://arbital.com/p/1lz). Recall that Bayes' rule (in the [odds form](https://arbital.com/p/1x5)) says that the posterior odds between any two hypotheses is equal to the prior odds times the relative likelihoods.

For example, in the [Diseasitis](https://arbital.com/p/22s) problem, a patient is 20% likely to be sick (and 80% likely to be healthy) a priori, and they take a test that is 3x more likely to come back positive if they are sick. The odds of a patient being sick _given_ a positive test are thus $(1 : 4) \times (3 : 1) = (3 : 4).$

Using waterfall diagrams, we can visualize the prior odds as two separate streams of water at the top of a waterfall, and the relative likelihoods as the proportion of water from each stream that makes it to the shared pool at the bottom. The posterior odds can then be visualized as the proportion of water in the shared pool that came from each different prior stream.

![Waterfall diagram](https://i.imgur.com/CXsoZhA.png?0)

See [https://arbital.com/p/1x1](https://arbital.com/p/1x1) for a walkthrough of the diagram.

Waterfall diagrams make it clear that, when calculating the posterior odds, what matters is the _relative proportion_ between how much each gallon of red water vs each gallon of blue water makes it into the shared pool. If 45% of red water and 15% of blue water made it to the bottom, that would give the same _relative proportion_ of red and blue water in the shared pool at the bottom as 90% and 30%.

![Same relative proportions](https://i.imgur.com/6FOndjc.png?0)

Thus, it is only the relative likelihoods (and not the absolute likelihoods) that matter when calculating posterior odds.

Similarly, changing the water flows at the top of the waterfall from (20 gallons/sec red water : 80 gallons/sec blue water) to (40 gallons/sec red water : 160 gallons/sec blue water) would double the total water at bottom, but not change the relative proportions of blue and red water.  So only the *relative* prior odds matter to the *relative* posterior odds.

Eliezer Yudkowsky
Alexei Andreev
Eric Rogstad
Eric Bruylant
Nate Soares
Salil Kalghatgi

Summary:
Waterfall diagrams, like [frequency diagrams](https://arbital.com/p/560), provide a way of visualizing [Bayes' Rule](https://arbital.com/p/1lz).  For example, if 20% of the patients in the screening patient are sick (red) and 80% are healthy (blue); and 90% of the sick patients get positive test results; and 30% of the healthy patients get positive test results, we could visualize the probability flows using the following diagram:

![Waterfall diagram](https://i.imgur.com/CXsoZhA.png?0)


Waterfall diagrams and relative odds

Imagine a waterfall with two streams of water at the top, a red stream and a blue stream.  These streams separately approach the top of the waterfall, with some of the water from both streams being diverted along the way, and the remaining water falling into a shared pool below.

![unlabeled waterfall](https://i.imgur.com/D8EhY65.png?0)

Suppose that:

- At the top of the waterfall, 20 gallons/second of red water are flowing down, and 80 gallons/second of blue water are coming down.
- 90% of the red water makes it to the bottom.
- 30% of the blue water makes it to the bottom.

Of the purplish water that makes it to the bottom of the pool, how much was originally from the red stream and how much was originally from the blue stream?

%%if-after([https://arbital.com/p/55z](https://arbital.com/p/55z)):
This is structurally identical to the [Diseasitis](https://arbital.com/p/22s) problem from [before](https://arbital.com/p/55z):

- 20% of the patients in the screening population start out with Diseasitis.
- Among patients with Diseasitis, 90% turn the tongue depressor black.
- 30% of the patients without Diseasitis will also turn the tongue depressor black.
%%

%%!if-after([https://arbital.com/p/55z](https://arbital.com/p/55z)):
This is structurally similar to the following problem, such as medical students might encounter:

You are a nurse screening 100 patients for Diseasitis, using a tongue depressor which usually turns black for patients who have the sickness.

- 20% of the patients in the screening population start out with Diseasitis.
- Among patients with Diseasitis, 90% turn the tongue depressor black (true positives).
- However, 30% of the patients without Diseasitis will also turn the tongue depressor black (false positives).

What is the chance that a patient with a blackened tongue depressor has Diseasitis?
%%

The 20% of sick patients are analogous to the 20 gallons/second of red water; the 80% of healthy patients are analogous to the 80 gallons/second of blue water:

![top labeled waterfall](https://i.imgur.com/eQh2qUt.png?0)

The 90% of the sick patients turning the tongue depressor black is analogous to 90% of the red water making it to the bottom of the waterfall.  30% of the healthy patients turning the tongue depressor black is analogous to 30% of the blue water making it to the bottom pool.

![middle labeled waterfall](https://i.imgur.com/6GBBYO5.png?0)

Therefore, the question "what portion of water in the final pool came from the red stream?" has the same answer as the question "what portion of patients that turn the tongue depressor black are sick with Diseasitis?"

%%if-after([https://arbital.com/p/55z](https://arbital.com/p/55z)):
Now for the faster way of answering that question.
%%

We start with *4 times as much blue water as red water* at the top of the waterfall.

Then each molecule of red water is 90% likely to make it to the shared pool, and each molecule of blue water is 30% likely to make it to the pool.  (90% of red water and 30% of blue water make it to the bottom.)  So each molecule of red water is *3 times as likely* (0.90 / 0.30 = 3) as a molecule of blue water to make it to the bottom.

So we multiply prior proportions of $1 : 4$ for red vs. blue by relative likelihoods of $3 :  1$ and end up with final proportions of $(1 \cdot 3) : (4 \cdot 1) = 3 : 4$, meaning that the bottom pool has 3 parts of red water to 4 parts of blue water.

![labeled waterfall](https://i.imgur.com/QIrtuVU.png?0)

To convert these *relative* proportions into an *absolute* probability that a random water molecule at the bottom is red, we calculate 3 / (3 + 4) to see that 3/7ths (roughly 43%) of the water in the shared pool came from the red stream.

This proportion is the same as the 18 : 24 sick patients with positive results, versus healthy patients with positive test results, that we would get by [thinking about 100 patients](https://arbital.com/p/560).

That is, to solve the Diseasitis problem in your head, you could convert this word problem:

> 20% of the patients in a screening population have Diseasitis.  90% of the patients with Diseasitis turn the tongue depressor black, and 30% of the patients without Diseasitis turn the tongue depressor black. Given that a patient turned their tongue depressor black, what is the probability that they have Diseasitis?

Into this calculation:

> Okay, so the initial odds are (20% : 80%) = (1 : 4), and the likelihoods are (90% : 30%) = (3 : 1). Multiplying those ratios gives final odds of (3 : 4), which converts to a probability of 3/7ths.

(You might not be able to convert 3/7 to 43% in your head, but you might be able to eyeball that it was a chunk less than 50%.)

You can try doing a similar calculation for this problem:

- 90% of widgets are good and 10% are bad.
- 12% of bad widgets emit sparks.
- Only 4% of good widgets emit sparks.

What percentage of sparking widgets are bad? If you are sufficiently comfortable with the setup, try doing this problem entirely in your head.

(You might try visualizing a waterfall with good and bad widgets at the top, and only sparking widgets making it to the bottom pool.)
%todo: Have a picture of a waterfall here, with no numbers, but with the parts labeled, that can be expanded if the user wants to expand it.%

%%hidden(Show answer):
- There's (1 : 9) bad vs. good widgets.
- Bad vs. good widgets have a (12 : 4) relative likelihood to spark.
- This simplifies to (1 : 9) x (3 : 1) = (3 : 9) = (1 : 3), 1 bad sparking widget for every 3 good sparking widgets.
- Which converts to a probability of 1/(1+3) = 1/4 = 25%; that is, 25% of sparking widgets are bad.

Seeing sparks didn't make us "believe the widget is bad"; the probability only went to 25%, which is less than 50/50.  But this doesn't mean we say, "I still believe this widget is good!" and toss out the evidence and ignore it.  A bad widget is *relatively more likely* to emit sparks, and therefore seeing this evidence should cause us to think it *relatively more likely* that the widget is a bad one, even if the probability hasn't yet gone over 50%.  We increase our probability from 10% to 25%.%%

%%if-before([https://arbital.com/p/1x8](https://arbital.com/p/1x8)):
Waterfalls are one way of visualizing the "odds form" of "Bayes' rule", which states that **the prior odds times the likelihood ratio equals the posterior odds.**  In turn, this rule can be seen as formalizing the notion of "the strength of evidence" or "how much a piece of evidence should make us update our beliefs".  We'll take a look at this more general form next.
%%

%%!if-before([https://arbital.com/p/1x8](https://arbital.com/p/1x8)):
Waterfalls are one way of visualizing the [odds form](https://arbital.com/p/1x5) of [Bayes' rule](https://arbital.com/p/1lz), which states that **the prior odds times the likelihood ratio equals the posterior odds**.
%%

Malo Bourgon
Eliezer Yudkowsky
Alexei Andreev
Eric Bruylant
Nate Soares
Robert Eidschun
Khana Santamaria
Adom Hartell

Summary:
Waterfall diagrams, like [frequency diagrams](https://arbital.com/p/560), provide a way of visualizing [Bayes' Rule](https://arbital.com/p/1lz).  For example, if 20% of the patients in the screening patient are sick (red) and 80% are healthy (blue); and 90% of the sick patients get positive test results; and 30% of the healthy patients get positive test results, we could visualize the probability flows using the following diagram:

![Waterfall diagram](https://i.imgur.com/CXsoZhA.png?0)

This diagram helps show that only the *relative* ratios matter to the final answer.  Twice as much water flowing from both streams at the top, or half as much water from each stream making it to the bottom, wouldn't change the relative proportions in the end result.


Humans doing Bayes

Tag for examples of human beings applying explicitly Bayesian reasoning in everyday life.  Parent for rules, heuristics, and concepts that are specifically about the human, cognitive use of explicit Bayesian reasoning.


Explicit Bayes as a counter for 'worrying'

One of the possible [human uses of explicit Bayesian reasoning](https://arbital.com/p/1x2) is that making up explicit probabilities and doing the Bayesian calculation lets us summarize the relevant considerations in one place.  This can counter a mental loop of 'worrying' by bouncing back and forth between focusing on individual considerations.

One example comes from a woman who was a test subject for an early version of the Bayes intro at the same time she was dating on OKCupid, and a 96% OKCupid match canceled their first date for coffee without explanation.  After bouncing back and forth mentally between 'maybe there was a good reason he canceled' versus 'that doesn't seem like a good sign', she decided to try Bayes.  She estimated that a man like this one had [prior](https://arbital.com/p/1rm) [odds](https://arbital.com/p/1rb) of 2 : 5 for desirability vs. undesirability, based on his OKCupid profile and her past experience with 96% matches.  She then estimated a 1 : 3 [likelihood ratio](https://arbital.com/p/1rq) for desirable vs. undesirable men flaking on the first date.  This worked out to 2 : 15 [posterior](https://arbital.com/p/1rp) odds for the man being undesirable, which she decided was unfavorable enough to not pursue him further.

The point of this mental exercise wasn't that the numbers she made up were exact.  Rather, by laying out all the key factors in one place, she stopped her mind from bouncing back and forth between %%knows-requisite([https://arbital.com/p/1rj](https://arbital.com/p/1rj)): visualizing $\mathbb P(\text{cancel}|\text{desirable})$ and visualizing $\mathbb P(\text{cancel}|\text{undesirable})$, which were pulling her back and forth as arguments pointing in different directions.%% %%!knows-requisite([https://arbital.com/p/1rj](https://arbital.com/p/1rj)): switching between imagining reasons why a good prospect might've canceled, versus imagining reasons why a bad prospect might've canceled.%%  By combining both ideas into a likelihood ratio, and moving on to posterior odds, she summarized the considerations and was able to stop unproductively 'worrying'.

Worrying isn't always unproductive.  Paying more attention to an individual consideration like "Maybe there were reasons a good dating prospect might've canceled anyway?" might cause you to adjust your estimate of this consideration, thereby changing your posterior odds.  But you can get the *illusion* of progress by switching your attention from one already-known consideration to another, since it feels like these considerations are pulling on your posterior intuition each time you focus on them.  It feels like your beliefs are changing and cognitive work is being performed, but actually you're just going in circles.  This is the 'unproductive worrying' process that you can interrupt by doing an explicitly Bayesian calculation that summarizes what you already know into a single answer.

(She did send him a rejection notice spelling out her numerical reasoning; since, if he wrote back in a way indicating that he actually understood that, he might've been worth a second chance.)

(He didn't.)

Summary: By making up plausible Bayesian probabilities, we can summarize all of our knowledge about a problem in the same place at the same time, rather than 'worrying' by bouncing back and forth between different parts of it.  To use one real-life example, if an OKCupid match cancels your first date without explanation, then you might be tempted to 'worry' by bouncing your focus back and forth between possible valid reasons to cancel the date, versus possible ways it was a bad sign.  Alternatively, you could estimate  [prior](https://arbital.com/p/1rm) [odds](https://arbital.com/p/1rb) of 2 : 5 for desirability vs. undesirability of a 96% OKCupid match, a 1 : 3 [likelihood ratio](https://arbital.com/p/1rq) for desirable vs. undesirable men flaking on the first date, decide that the [posterior](https://arbital.com/p/1rp) odds of 2 : 15 don't warrant further pursuit, and then be done.  The point isn't that the made-up numbers are exact, but that they summarize the relevant weights into a single place and calculation, terminating the 'worry' cycle of alternating attention on particular pieces, and letting us be done.


Realistic (Math 1)

### What's the chance that a potentially good partner would flake on the first date?

From a test subject for an early version of the Bayes intro:

A 96% OKCupid match canceled their first date for coffee without providing an explanation.

She estimated that a man like this one had [prior](https://arbital.com/p/1rm) [odds](https://arbital.com/p/1rb) of 2 : 5 for desirability vs. undesirability, based on his OKCupid profile and her past experience with 96% matches.  She then estimated a 1 : 3 [likelihood ratio](https://arbital.com/p/1rq) for desirable vs. undesirable men flaking on the first date.  This worked out to 2 : 15 [posterior](https://arbital.com/p/1rp) odds for the man being undesirable, which she decided was unfavorable enough to not pursue him further.

She [used this explicitly Bayesian calculation to interrupt a 'worrying' cycle](https://arbital.com/p/1x3) wherein she was focusing on one consideration, then a different consideration, arguing for pursuing further / not pursuing further.  Making up numbers and doing the Bayesian calculation [terminated this cycle](https://arbital.com/p/1x3).


Bayes' rule: Odds form

One of the more convenient forms of [Bayes' rule](https://arbital.com/p/1lz) uses [relative odds](https://arbital.com/p/1rb). Bayes' rule says that, when you observe a piece of evidence $e,$ your [posterior](https://arbital.com/p/1rp) odds $\mathbb O(\boldsymbol H \mid e)$ for your hypothesis [https://arbital.com/p/-vector](https://arbital.com/p/-vector) $\boldsymbol H$ given $e$ is just your [prior](https://arbital.com/p/1rm) odds $\mathbb O(\boldsymbol H)$ on $\boldsymbol H$ times the [https://arbital.com/p/-56s](https://arbital.com/p/-56s) $\mathcal L_e(\boldsymbol H).$

For example, suppose we're trying to solve a mysterious murder, and we start out thinking the odds of Professor Plum vs. Miss Scarlet committing the murder are 1 : 2, that is, Scarlet is twice as likely as Plum to have committed the murder [a priori](https://arbital.com/p/1rm).  We then observe that the victim was bludgeoned with a lead pipe.  If we think that Plum, *if* he commits a murder, is around 60% likely to use a lead pipe, and that Scarlet, *if* she commits a murder, would be around 6% likely to us a lead pipe, this implies [relative likelihoods](https://arbital.com/p/1rq) of 10 : 1 for Plum vs. Scarlet using the pipe.  The [posterior](https://arbital.com/p/1rp) odds for Plum vs. Scarlet, after observing the victim to have been murdered by a pipe, are $(1 : 2) \times (10 : 1) = (10 : 2) = (5 : 1)$.  We now think Plum is around five times as likely as Scarlet to have committed the murder.

# Odds functions

Let $\boldsymbol H$ denote a [https://arbital.com/p/-vector](https://arbital.com/p/-vector) of hypotheses. An odds function $\mathbb O$ is a function that maps $\boldsymbol H$ to a set of [https://arbital.com/p/-1rb](https://arbital.com/p/-1rb). For example, if $\boldsymbol H = (H_1, H_2, H_3),$ then $\mathbb O(\boldsymbol H)$ might be $(6 : 2 : 1),$ which says that $H_1$ is 3x as likely as $H_2$ and 6x as likely as $H_3.$ An odds function captures our *relative* probabilities between the hypotheses in $\boldsymbol H;$ for example, (6 : 2 : 1) odds are the same as (18 : 6 : 3) odds. We don't need to know the absolute probabilities of the $H_i$ in order to know the relative odds. All we require is that the relative odds are proportional to the absolute probabilities:
$$\mathbb O(\boldsymbol H) \propto \mathbb P(\boldsymbol H).$$

In the example with the death of Mr. Boddy, suppose $H_1$ denotes the proposition "Reverend Green murdered Mr. Boddy", $H_2$ denotes "Mrs. White did it", and $H_3$ denotes "Colonel Mustard did it". Let $\boldsymbol H$ be the vector $(H_1, H_2, H_3).$ If these propositions respectively have [prior](https://arbital.com/p/1rm) probabilities of 80%, 8%, and 4% (the remaining 8% being reserved for other hypotheses), then $\mathbb O(\boldsymbol H) = (80 : 8 : 4) = (20 : 2 : 1)$ represents our *relative* credences about the murder suspects — that Reverend Green is 10 times as likely to be the murderer as Miss White, who is twice as likely to be the murderer as Colonel Mustard.

# Likelihood functions

Suppose we discover that the victim was murdered by wrench.  Suppose we think that Reverend Green, Mrs. White, and Colonel Mustard, *if* they murdered someone, would respectively be 60%, 90%, and 30% likely to use a wrench.  Letting $e_w$ denote the observation "The victim was murdered by wrench," we would have $\mathbb P(e_w\mid \boldsymbol H) = (0.6, 0.9, 0.3).$ This gives us a [https://arbital.com/p/-56s](https://arbital.com/p/-56s) defined as $\mathcal L_{e_w}(\boldsymbol H) = P(e_w \mid \boldsymbol H).$

# Bayes' rule, odds form

Let $\mathbb O(\boldsymbol H\mid e)$ denote the [posterior](https://arbital.com/p/1rp) odds of the hypotheses $\boldsymbol H$ after observing evidence $e.$  [Bayes' rule](https://arbital.com/p/1xr) then states:

$$\mathbb O(\boldsymbol H) \times \mathcal L_{e}(\boldsymbol H) = \mathbb O(\boldsymbol H\mid e)$$

This says that we can multiply the relative prior credence $\mathbb O(\boldsymbol H)$ by the likelihood $\mathcal L_{e}(\boldsymbol H)$ to arrive at the relative posterior credence $\mathbb O(\boldsymbol H\mid e).$ Because odds are invariant under multiplication by a positive constant, it wouldn't make any difference if the _likelihood_ function was scaled up or down by a constant, because that would only have the effect of multiplying the final odds by a constant, which does not affect them. Thus, only the [relative likelihoods](https://arbital.com/p/-1rq) are necessary to perform the calculation; the absolute likelihoods are unnecessary. Therefore, when performing the calculation, we can simplify $\mathcal L_e(\boldsymbol H) = (0.6, 0.9, 0.3)$ to the relative likelihoods $(2 : 3 : 1).$

In our example, this makes the calculation quite easy. The prior odds for Green vs White vs Mustard were $(20 : 2 : 1).$ The relative likelihoods were $(0.6 : 0.9 : 0.3)$ = $(2 : 3 : 1).$ Thus, the relative posterior odds after observing $e_w$ = Mr. Boddy was killed by wrench are $(20 : 2 : 1) \times (2 : 3 : 1) = (40 : 6 : 1).$ Given the evidence, Reverend Green is 40 times as likely as Colonel Mustard to be the killer, and 20/3 times as likely as Mrs. White.

Bayes' rule states that this *relative* proportioning of odds among these three suspects will be correct, regardless of how our remaining 8% probability mass is assigned to all other suspects and possibilities, or indeed, how much probability mass we assigned to other suspects to begin with. For a proof, see [https://arbital.com/p/1xr](https://arbital.com/p/1xr).

# Visualization

[Frequency diagrams](https://arbital.com/p/560), [waterfall diagrams](https://arbital.com/p/1wy), and [spotlight diagrams](https://arbital.com/p/1zm) may be helpful for explaining or visualizing the odds form of Bayes' rule.

Summary:
A form of [Bayes' rule](https://arbital.com/p/1lz) that uses relative [odds](https://arbital.com/p/1rb).

Suppose we're trying to solve a mysterious murder, and we [start out](https://arbital.com/p/1rm) thinking the odds of Professor Plum vs. Miss Scarlet committing the murder are 1 : 2, that is, Scarlet is twice as likely as Plum to have committed the murder.  We then observe that the victim was bludgeoned with a lead pipe.  If we think that Plum, *if* he commits a murder, is around 60% likely to use a lead pipe, and that Scarlet, *if* she commits a murder, would be around 6% likely to us a lead pipe, this implies [relative likelihoods](https://arbital.com/p/1rq) of 10 : 1 for Plum vs. Scarlet using the pipe.

The [posterior](https://arbital.com/p/1rp) odds for Plum vs. Scarlet, after observing the victim to have been murdered by a pipe, are $(1 : 2) \times (10 : 1) = (10 : 2) = (5 : 1)$.  We now think Plum is around five times as likely as Scarlet to have committed the murder.


Aaron Gertler

summary: Nothing here yet.

Automatically generated page for "Aaron Gertler" user.
If you are the owner, click [here to edit](https://arbital.com/edit/2d8).


Introduction to Effective Altruism

Effective altruism (EA) means using evidence and reason to take actions that help others as much as possible.

The world has a lot of terrible problems, but we can't work on them all at once. And we don't agree on what the worst problems are.

Despite this, most people want to "make the world a better place". What should those people actually be doing?

Some of the questions EA sets out to answer:

 - Is it better to help people in your own country, or abroad?
 - Is it better to take a high-paying job and give money to charity, or to work for a charity directly?
 - Is it better to support an intervention with a lot of strong evidence, or to support research on a promising new intervention?

The answers to these kinds of questions will be different for each person. What you should do depends on who you are: What are your skills? What are your values? How much time do you have?

But no matter who you are, effective altruism can help you figure out how to improve the world. To learn more, follow the links below.

##More on EA##

Are you a high school student? Are you a college student?

Do you work for a nonprofit or a business with a social mission?

Are you religious?

Which of the following issues most concerns you? (Politics, poverty, health, education, animals, environment, far future)

Are you interested in donating? In direct work?

Are you financially comfortable?

##Common Questions##

 - What beliefs do all EAs have in common?
 - What are the different causes EAs support?
 - What are some actions EAs take to improve the world?
 - What makes EA different from other ways of helping people?
 - What are some common criticisms of EA, and how do EAs respond?
 - What is the philosophical justification for EA?

Here are some really good articles from outside Arbital:

 - [What is Effective Altruism?](https://whatiseffectivealtruism.com) (The best general introduction to the topic.)

**This page is a work in progress. If you have questions, or you'd like to see a topic added, please contact the author: aaronlgertler (at) gmail.com.**


Strictly factual question

A *strict question of fact* has an answer determined solely by the state of the material universe, and sufficiently straightforward math, in a clearly understandable way - none of our uncertainty about this question is about definitions, values, or viewpoints; we are just wondering which quarks go where.

Questions of strict fact:

- What happens if I press the giant red button?(\*)
- Did Sally claim more charitable donations as an income tax deduction than Bob?
- Are there unicorns anywhere on Earth?
- Will Peano arithmetic prove a contradiction if I search all the proofs less than a billion steps long?

Questions not yet of strict fact:

- Should I press the giant red button?
- Is Sally a more charitable person than Bob?
- Do unicorns exist?
- Will Zermelo-Frankel set theory prove a contradiction if I search for one forever?

Why the second group aren't yet questions of strict fact:

- Because the notion of *should* has not yet been fully specified or determined.
- Because we haven't said exactly what it means to be charitable.
- Because "exist" is a much more ambiguous notion than "exist inside our galaxy".  For example, there could be unicorns $10^{1,000,000}$ lightyears away, or inside some other mathematical universe that has as much "existence" as this one does.  The notion of "X exists" is less firmly nailed down than "X is inside our closet".
- Because the notion of 'forever' is more mathematically fraught than 'for ten to the billionth power years'.  For example, nailing down exactly which infinity we're talking about can't be done in a system of first-order logic.

(\*)  At least, this is a straightforward question so long as we don't poke too hard at the nature of [what-if counterfactuals](https://arbital.com/p/).  In most real-life situations, the question "What happens if I turn on this blender with a fork inside?" is something that has a sufficiently straightforward material answer for us to say that it's just a question of material facts.


Proving too much

An argument "proves too much" when you can just as easily or naturally extend it to prove false things as true things.  For example, the argument "Nobody should engage in genetic engineering because Nazis" can just as easily be extended to say "Nobody should eat vegetables because Hitler ate vegetables", so the general rule *proves too much.*

See also:

- http://en.wikipedia.org/wiki/Proving_too_much
- http://slatestarcodex.com/2013/04/13/proving-too-much/


Bulverism

[Bulverism](https://en.wikipedia.org/wiki/Bulverism) is when you [explain what goes so horribly wrong in people's minds](https://arbital.com/p/43h) when they believe X, before you've actually explained why X is wrong.  It can be seen as a subspecies of the fallacy of "[poisoning the well](https://en.wikipedia.org/wiki/Poisoning_the_well)".  **Bulverism is forbidden on Arbital.**  Every time you talk about what goes wrong in people's minds that makes them believe X, make sure you first discuss why X is, in fact, wrong, or route your readers to a page that does discuss it.  Also see the entry on [https://arbital.com/p/43h](https://arbital.com/p/43h).


Emphemeral premises

A purported [psychological](https://arbital.com/p/43h) fallacy wherein people confronted with an incongruent proposition X will say "Oh, not-X because Y", and perhaps a week later, say, "Oh, not-X because Z."  [Ephemeral Premises](https://www.facebook.com/yudkowsky/posts/10154268184344228) is when you make up assumptions to deflect X, don't write them down or list them anywhere, and since there's no central list, you might be making up different assumptions when talking to someone else.  It's a disposable premise being used as a parry in short-term combat, not part of a long-term edifice being made with consistent, well-defended materials.  Needs to be paired with understanding of the [Multiple-Stage Fallacy](https://arbital.com/p/multiple_stage_fallacy) so that you don't fall for the reverse fallacy of making people list out lots of Ys and Zs to make arbitrary propositions sound very improbable.


Multiple stage fallacy

In August 2015, renowned statistician and predictor Nate Silver wrote "[Trump's Six Stages of Doom](http://fivethirtyeight.com/features/donald-trumps-six-stages-of-doom/)" in which he gave Donald Trump a 2% chance of getting the Republican *nomination* (not the presidency).  Silver reasoned that Trump would need to pass through six stages to get the nomination, "Free-for-all", "Heightened scrutiny", "Iowa and New Hampshire", "Winnowing", "Delegate accumulation", and "Endgame."  Nate Silver argued that Trump had at best a 50% chance of passing each stage, implying a final nomination probability of at most 2%.

In late March, Trump had passed the first four stages, while prediction markets gave him a 75% chance of clinching the Republican nomination.  By Nate Silver's logic, Trump's probability of passing the remaining stages should have been $0.50 \cdot 0.50 = 0.25$ *conditional* on Trump passing the first four stages.

Nate Silver might not have been wrong to assign Trump a low advance probability of being nominated.  Many people were surprised by Trump's nomination.  But it seems more likely that Silver committed an error when he said, specifically, that if we'd observed Trump to pass the first four stages, this would probably be taking place in a world where Trump had a 50% probability of passing the two remaining stages.

The purported "Multiple-Stage Fallacy" is when you list multiple 'stages' that need to happen on the way to some final outcome, assign probabilities to each 'stage', multiply the probabilities together, and end up with a small final answer.  The alleged problem is that you can do this to almost any kind of proposition by staring at it hard enough, including things that actually happen.

On a probability-theoretic level, the three problems at work in the usual Multiple-Stage Fallacy are as follows:

- You need to multiply *conditional* probabilities rather than the absolute probabilities. When you're considering a later stage, you need to assume that the world was such that every prior stage went through. Nate Silver was [probably](https://arbital.com/p/43h) trying to simulate his prior model of Trump accumulating enough delegates in March through June, not imagining his *updated* beliefs about Trump and the world after seeing Trump be victorious up to March.
  - Even if you're aware in principle that you need to use conditional probabilities - Nate Silver certainly knew about them - it's hard to update far *enough* when you imagine the pure hypothetical possibility that Trump wins stages 1-4 for some reason - compared to how much you actually update when you actually see Trump winning! (Some sort of reverse hindsight bias or something? We don't realize how much we'd need to update our current model if we were already that surprised?)
- Often, people neglect to consider disjunctive alternatives - there may be more than one way to reach a stage, so that not *all* the listed things need to happen. Trump accumulated enough delegates in Nate's fifth stage that there was no "Endgame" convention fight in the supposed sixth stage.
- People have tendencies to assign middle-tending probabilities. So if you list enough stages, you can drive the apparent probability of anything down to zero, *even if you seem to be soliciting probabilities from the reader.*
 - If you're a [motivated skeptic](https://wiki.lesswrong.com/wiki/Motivated_skepticism), you will be tempted to list more 'stages'.

The Multiple-Stage Fallacy is particularly dangerous for people who've read studies on the dangers of probabilistic overconfidence.  In late March, the 75% prediction-market probabilities must have corresponded to, e.g., something like an 80% chance of getting enough delegates and a 90% chance of passing the convention conditional on getting enough delegates.  Imagine how overconfident this might have sounded without the prediction market to establish a definite probability - "Oh, don't you know that what people assign 90% confidence doesn't usually happen 90% of the time?"

Instances of the Multiple-Stage Fallacy may also sound more persuasive to readers who've read about the [Conjunction Fallacy](http://lesswrong.com/lw/ji/conjunction_fallacy/).

Yudkowsky [argues](https://www.facebook.com/yudkowsky/posts/10154036150109228):

> If you're not willing to make "overconfident" probability assignments like those, then you can drive the apparent probability of anything down to zero by breaking it down into enough 'stages'. In fact, even if someone hasn't heard about overconfidence, people's probability assignments often trend toward the middle, so you can drive down their "personally assigned" probability of anything just by breaking it down into more stages...

> From beginning to end, I've never used this style of reasoning and I don't recommend that you do so either. \[https://arbital.com/p/Since\](https://arbital.com/p/Since\) even Nate Silver couldn't get away with it, I think we just shouldn't try. It's a doomed methodology.

Summary:
The purported "Multiple-Stage Fallacy" is when you list multiple 'stages' that need to happen on the way to some final outcome, assign probabilities to each 'stage', multiply the probabilities together, and end up with a small final answer.  The alleged problem is that you can do this to almost *any* proposition by staring at it hard enough, including propositions that turn out to be true.

An alleged example would be that in August 2015, renowned forecaster Nate Silver wrote "[Trump's Six Stages of Doom](http://fivethirtyeight.com/features/donald-trumps-six-stages-of-doom/)" in which he gave Donald Trump a 2% chance of getting the Republican *nomination* (not the presidency), based on six stages each with at most a 50% probability of being passed.  In late March, Trump had passed the first four stages Nate listed, and prediction markets gave Trump a 75% chance of clinching the Republican nomination.  By Nate Silver's logic, Trump's probability of passing the remaining stages should have been $0.50 \cdot 0.50 = 0.25.$

The three main purported problems with the Multiple-Stage Fallacy are that:

- To get the final estimate, you logically need to multiply [conditional probabilities](https://arbital.com/p/1rj), not unconditional probabilities.  Getting your mind to give good conditional probabilities isn't easy; Nate Silver might have failed to imagine how much he'd *actually* update after *actually* seeing Trump pass the first four stages.
- Often, people fail to consider disjunctive alternatives, or ways that not all the stages may need passing.
- People tend to assign middle-tending probabilities.   So if you list enough stages, you can drive the apparent probability of anything down to zero, *even if you seem to be soliciting probabilities from the reader.*


Mind projection fallacy

The "mind projection fallacy" occurs when somebody expects an overly direct resemblance between the intuitive language of the mind, and the language of physical reality.

Consider the [map and territory](https://arbital.com/p/map_territory) metaphor, in which the world is a like a territory and your mental model of the world is like a map of that territory. In this metaphor, the  mind projection fallacy is analogous to thinking that the territory can be folded up and put into your pocket.

As an archetypal example:  Suppose you flip a coin, slap it against your wrist, and don't yet look at it.  Does it make sense to say that the probability of the coin being heads is 50%?  How can this be true, when the coin itself is already either definitely heads or definitely tails?

One who says "the coin is fundamentally uncertain; it is a feature of the coin that it is always 50% likely to be heads" commits the mind projection fallacy. Uncertainty is in the mind, not in reality.  If you're ignorant about a coin, that's not a fact about the coin, it's a fact about you. It makes sense that your brain, the map, has an internal measure of how it's more or less sure of something.  But that doesn't mean the coin itself has to contain a corresponding quantity of increased or decreased sureness; it is just heads or tails.

The [https://arbital.com/p/-ontology](https://arbital.com/p/-ontology) of a system is the elementary or basic components of that system. The ontology of your model of the world may include intuitive measures of uncertainty that it can use to represent the state of the coin, used as primitives like [floating-point numbers](https://arbital.com/p/float) are primitive in computers. The mind projection fallacy occurs whenever someone reasons as if the territory, the physical universe and its laws, must have the same sort of ontology as the map, our models of reality.

See also:

- [https://arbital.com/p/4vr](https://arbital.com/p/4vr)
- The LessWrong sequence on [Reductionism](https://wiki.lesswrong.com/wiki/Reductionism_%28sequence%29), especially:
 - [How an algorithm feels from the inside](http://lesswrong.com/lw/no/how_an_algorithm_feels_from_inside/)
 - [The Mind Projection Fallacy](http://lesswrong.com/lw/oi/mind_projection_fallacy/)
 - [Probability is in the mind](http://lesswrong.com/lw/oj/probability_is_in_the_mind/)

Summary:
One commits the mind projection fallacy when they postulate that features of how their model of the world works are actually features of the world.

Suppose you flip a coin, slap it against your wrist, and don't look at the result.  Does it make sense to say that the probability of the coin being heads is 50%?  How can this be true, when the coin has already landed, and is either definitely heads or definitely tails? One who says "the coin is fundamentally uncertain; it is a feature of the coin that it is always 50% likely to be heads" commits the mind projection fallacy. Uncertainty is in the mind, not in reality. It makes sense that brains have an internal measure of how uncertain they are about the world, but that uncertainty is not a fact about the coin, it's a fact about the uncertain person. The coin itself is not sure or unsure.


Invisible background fallacies

Suppose that you went outdoors and jumped up into the air, and then, as you were jumping, the law of gravity suddenly switched off--for whatever weird reason, masses everywhere in the Universe stopped attracting one another.  Would your feet hit the ground again?

As an initial thought about this problem, you might imagine that gravity no longer pulls you down toward the Earth, and that therefore your jump would keep you continuing up into the air.  If you were thinking more broadly, you might realize that the concept of 'air pressure' means that the air on Earth's surface is kept pressurized by gravity, and in the absence of gravity holding down all that air, the atmosphere might start to rush off into space as it expanded to relieve the pressure.  Although (you then reason further) the speed of sound is finite, so the atmosphere might start rushing out at the edges, and would need to disappear there before the air started to expand away from the Earth's surface.  And then that rushing air would only carry you further away from the ground.

This answer is incorrect.  Why?  Because the ground under your feet is also currently being pulled downward by gravity.

Indeed, the ground under your feet is currently in a state of [equilibrium](https://arbital.com/p/) despite being pulled downward by gravity at what would otherwise be an acceleration of $9.8m/s^2.$  It follows that other pressures on the ground immediately below your feet--from the air above it, and from the rock below--must net out to an upward force that would, if not for gravity, push the ground upward at $9.8m/s^2.$  This logic applies to the dirt below your feet, the air around you, and the rocks immediately underneath the dirt under your feet.  Everything around you that seemed motionless was previously in equilibrium with gravity, so as soon as gravity vanishes, it all accelerates upward in near unison.

So your feet hit the ground again, as the Earth expands beneath you.  And then the Earth continues to accelerate upward at $9.8m/s^2$, pressuring your feet and accelerating you as well.  So in fact, if gravity had switched off everywhere in the universe 10 seconds earlier as you read this, you probably would not have noticed anything different, yet.  (Though you *would* notice shortly afterward as things began to expand far enough to change the balance of forces, and as the Sun finished exploding.)

In many situations and scenarios, there are rules or laws or considerations that properly apply to *everything*--including background objects and processes and concepts we don't usually think about--and then considering the *complete* effect gives us a very different picture from thinking only about the effect on *visible* things, with the latter being a common species of fallacy.

An essay now famous in the field of [economics](https://arbital.com/p/) is Frederic Bastiat's "[That Which Is Seen, And That Which Is Not Seen](http://bastiat.org/en/twisatwins.html)" which illustrated this idea and introduced the now-famous 'broken window fallacy'.  Suppose you heave a rock through somebody's window; have you done a good deed or a bad deed?  You can imagine somebody arguing that they've done a good deed, despite the initial inconvenience, because the homeowner needs to pay a glazier to fix the window, and then the glazier spends money at the baker, and so the whole economy is helped.  This is a fallacy because, if the window hadn't been broken, the homeowner would have spent the money somewhere else.  But the broken window and the payment to the glazier are highly visible; while the disappearing of whatever economic activity would otherwise have occurred, is not seen.  If we consider the effect on the whole picture, including things that are ordinarily hard to see or fading into the background or invisible for some other reason, we arrive at a very different picture of the net effect of breaking somebody's window.

Summary:
Considering *all* the implications often gives us a very different picture from considering *some* of the implications, and the latter is a common species of fallacy.

Suppose that you went outdoors and jumped up into the air, and then, as you were jumping, gravity suddenly switched off.  Would your feet hit the ground again?

You might initially imagine that gravity no longer pulls you down toward the Earth, and that therefore your jump would keep you continuing up into the air.  If you were thinking more broadly, you might imagine that the air on Earth's surface, previously kept pressurized by gravity, might start to rush off into space.

Actually, the ground under your feet is also currently being pulled downward by gravity.  It follows that other pressures on the ground immediately below your feet--from the air above it, and from the rock below--must net out to an upward force that would, if *not* for gravity, push the ground upward at $9.8m/s^2.$  So if you jumped up into the air, and then gravity switched off, the ground and everything previously motionless around you would accelerate upward at $9.8m/s^2,$ causing your feet to hit the ground at the expected time.  That is, if gravity switched off everywhere in the universe, you wouldn't notice anything for a couple of minutes.


Ideological Turing test

The [Ideological Turing Test](https://en.wikipedia.org/wiki/Ideological_Turing_Test) is when, to try to make sure you really understand an opposing position, you explain it such that *an advocate of the position cannot tell whether the explanation was written by you or a real advocate.*  This is often run online as a formal test where an audience tries to distinguish which pro-X and anti-X essays have been written by real pro-X and anti-X advocates, and which essays were actually written by advocates for the opposing side.  If you're pro-X, and you write an intelligent essay advocating anti-X, and other people can't tell that you were really pro-X, you win the Ideological Turing Test by having demonstrated that you really and honestly understand the arguments for anti-X.


Gotcha button

A "gotcha button" is a conversational point which, when pressed, causes some substantial fraction of listeners to leap up and shout "Gotcha!  That's obviously wrong and shows you don't know what you're talking about!", after which they have [license to dismiss](https://arbital.com/p/motivated_skepticism) the conversation.

This behavior does not *necessarily* mean the listener is wrong.  If you're describing your brilliant invention to a physics-literate person, and you mention that it can accelerate forward without pushing anything else in the opposite direction, the listener will shout "Gotcha!  You can't violate Conservation of Momentum!"  They'll be right, and their implied inference about your level of physics literacy will probably also be right.

Nonetheless, there are a *lot* of ideas that function as Gotcha Buttons for some significant fraction of various audiences, which do *not* correspond to a claimed known falsity.

- Some true/plausible claims will be seen as 'obviously false' by many listeners.
- Some phrases or ideas will be predictably misinterpreted to a claim that actually is false.

Example of the first type:  If you were explaining evolutionary biology to a novice audience, you might be wise to start with some example other than how hominids in particular evolved from primate ancestors.  This isn't wrong, but a fraction of the audience thinks it's obviously wrong and diagnostic of you as someone they shouldn't listen to.

Example of the second type:  If you were discussing [Artificial General Intelligence](https://arbital.com/p/42g), or AI with [par-human performance](https://arbital.com/p/7mt) in some domain, you might be wise to avoid the term "human-level AI"; because that particular phrase will cause some fraction of the audience to jump up and shout "Gotcha!  AIs won't have a humanlike balance of abilities!  There's no such thing as the human level!"  (They can then dismiss the rest of your argument, without trying to carefully consider whether your further arguments are actually refuted by accepting "AIs won't have a humanlike balance of abilities", because you've made such an obvious and severe mistake that you just lose.)


Harmless supernova fallacy

Harmless supernova fallacies are a class of arguments, usually a subspecies of false dichotomy or continuum fallacy, which [can equally be used to argue](https://arbital.com/p/3tc) that almost any physically real phenomenon--including a supernova--is harmless / manageable / safe / unimportant.

- **Bounded, therefore harmless:**  "A supernova isn't infinitely energetic--that would violate the laws of physics! Just wear a flame-retardant jumpsuit and you'll be fine."  (All physical phenomena are finitely energetic; some are nonetheless energetic enough that a flame-retardant jumpsuit won't stop them.  "Infinite or harmless" is a false dichotomy.)
- **Continuous, therefore harmless:**  "Temperature is continuous; there's no qualitative threshold where something becomes 'super' hot. We just need better versions of our existing heat-resistant materials, which is a well-understood engineering problem."  (Direct instance of the standard continuum fallacy: the existence of a continuum of states between two points does not mean they are not distinct.  Some temperatures, though not qualitatively distinct from lower temperatures, exceed what we can handle using methods that work for lower temperatures.  "Quantity has a quality all of its own.")
- **Varying, therefore harmless:**  "A supernova wouldn't heat up all areas of the solar system to exactly the same temperature!  It'll be hotter closer to the center, and cooler toward the outside. We just need to stay on a part of Earth that's further way from the Sun."  (The temperatures near a supernova vary, but they are all quantitatively high enough to be far above the greatest manageable level; there is no nearby temperature low enough to form a survivable valley.)
- **Mundane, therefore harmless** or **straw superpower**: "Contrary to what many non-astronomers seem to believe, a supernova can't burn hot enough to sear through time itself, so we'll be fine."  (False dichotomy: the divine ability is not required for the phenomenon to be dangerous / non-survivable.)
- **Precedented, therefore harmless:**  "Really, we've already had supernovas around for a while: there are already devices that produce 'super' amounts of heat by fusing elements low in the periodic table, and they're called thermonuclear weapons.  Society has proven well able to regulate existing thermonuclear weapons and prevent them from being acquired by terrorists; there's no reason the same shouldn't be true of supernovas."  (Reference class tennis / noncentral fallacy / continuum fallacy: putting supernovas on a continuum with hydrogen bombs doesn't make them able to be handled by similar strategies, nor does finding a class such that it contains both supernovas and hydrogen bombs.)
- **Undefinable, therefore harmless:**  "What is 'heat', really?  Somebody from Greenland might think 295 Kelvin was 'warm', somebody from the equator might consider the same weather 'cool'.  And when exactly does a 'sun shining' become a 'supernova'?  This whole idea seems ill-defined."  (Someone finding it difficult to make exacting definitions about a physical process doesn't license the conclusion that the physical process is harmless. %%note: It also happens that the distinction between the runaway process in a supernova and a sun shining is empirically sharp; but this is not why the argument is invalid--a super-hot ordinary fire can also be harmful even if we personally are having trouble defining "super" and "hot". %%)

Summary:
"Harmless supernova" fallacies are arguments that [can equally be used to conclude](https://arbital.com/p/3tc) than any real phenomenon, including a supernova, is harmless.

- **Bounded, therefore harmless:**  "A supernova isn't *infinitely* energetic--that would violate the laws of physics! Just wear a flame-retardant jumpsuit and you'll be fine."
- **Continuous, therefore harmless:**  "Temperature is continuous; there's no qualitative threshold where something becomes 'super' hot. We just need better versions of our existing heat-resistant materials."
- **Varying, therefore harmless:**  "A supernova wouldn't heat up all areas of the solar system to exactly the same temperature!  It'll be hotter closer to the center, and cooler toward the outside. We just need to stay on a part of Earth that's further way from the Sun."
- **Mundane, therefore harmless** aka **straw superpower**:  "Contrary to what many non-astronomers seem to believe, a supernova can't burn hot enough to sear through time itself.  So we'll be fine."
- **Precedented, therefore harmless:** "Really, we've already had supernovas around for a while; there already exist devices that produce tons of energy by fusion, and they're called thermonuclear weapons.  Society has proven itself able to handle those."
- **Undefinable, therefore harmless:**  "What exactly is heat?  Different people could call different temperatures 'too warm'.  When exactly does a sun shining become so 'super' that it's a supernova?"


Epistemology

"Epistemology" is the subject matter that deals with truth on a meta level: e.g., questions such as "What is truth?" and "What methods of reasoning are most likely to lead to true beliefs?"  Epistemology can be considered as a child subject of [https://arbital.com/p/9l](https://arbital.com/p/9l).  Many [Arbital Discussion Practices](https://arbital.com/p/9n) have their roots in a thesis about epistemology - about what forms of reasoning are most likely to lead to true beliefs - and you may have ended up looking at this subject after trying to track down how a Discussion Practice is justified.

Alexei Andreev
Eliezer Yudkowsky
Eric Rogstad
Eric Bruylant


Conceivability

A hypothetical scenario is 'conceivable' or 'imaginable' when it is not *immediately* incoherent, although it may still contain some subtle flaw that makes it [logically impossible](https://arbital.com/p/).  If you haven't yet checked for factors, it's conceivable to you that 91 is either a [prime number](https://arbital.com/p/) or a [composite number](https://arbital.com/p/), even though only one of these scenarios is [logically possible](https://arbital.com/p/).

Whether 91 is prime or compositive is a logical consequence of the definitions of 'prime' and 'composite', but it's not literally and explicitly written into our brain's representations of the definitions of 'prime' and 'composite' that 91 is a prime/composite number. If you can conceive of 91 being prime, this illustrates that, in your own mind, the property of 91-ness is not the same writing on your map as the property of composite-ness, but it doesn't demonstrate that it's logically or physically possible for 91 to be composite.  If you can *conceive* of a scenario in which X is true and Y is false, this demonstrates mainly that X and Y have different [intensional definitions](https://arbital.com/p/10b) inside your own mind; you have separate concepts in your mind for X and Y.  It doesn't demonstrate that a world where 'X but not Y' is logically possible, let alone physically possible, or that X and Y refer to two different things, etcetera.

Labeling a scenario 'conceivable' or saying 'we can conceive that' introduces a scenario for discussion or refutation, *without* assuming in the introduction that the scenario is logically possible (let alone physically possible, or at all plausible in our world).  Since many true facts are also plausible, physically possible, logically possible, and conceivable, introducing a scenario as 'conceivable' also does not imply that it is false.  It means that someone found themselves able to imagine something, perhaps a true thing, perhaps without visualizing it in the complete detail that would reveal its impossibility; but if all we know at the start is that someone felt like they could imagine something, we'll call it 'conceivable' or 'imaginable'.