A family has two children. One is a boy. Now, do the math!

Problem

A family has two children. One is a boy. How likely is it that the other child is a boy?

Popular yet wrong solution

"There are four possible cases: two boys, a boy and a girl, a girl and a boy, and two girls. But because one child is a boy, it can't be the last case (two girls), so there are only three cases. Therefore the probability is one-third."

This solution is popular. Among others, Nassim Nicholas Taleb (on a since deleted tweet), vlogbrother Hank Green in an old SciShow episode (IIRC), probability instructors trying to show how interesting their class is to bored undergraduates, and people interviewing job candidates have used this solution.

This solution is fun because it's counter-intuitive; because of that it also looks like a smart solution.

This solution is wrong.

It's wrong because after we use "one is a boy" to eliminate the possibility of a family with two girls, we can no longer divide the probability equally among the remaining three possibilities. Equal division of probability can be used in a case of no information, but not in a case when information has already been used to change the set of possibilities.

The more attentive reader will notice that this is the same error most people make in the Monty Hall three-door problem. As a general rule, it's a bad idea to try to solve math problems by hand-waving.

If it's a math problem, do the math.*

Frequentist approach

Let's say we have a large number of cases, 4000 families for example. That's 1000 each for each combination of children: $(B,B), (B,G), (G,B)$, and $(G,G)$. Now we look at all the possibilities where we observe one of the children at random:

1000 $(B,B)$ families yield a total of 1000 boys;
1000 $(B,G)$ families yield a total of 500 boys;
1000 $(G,B)$ families yield a total of 500 boys;
1000 $(G,G)$ families yield a total of 0 boys.

We have a total of 2000 observed boys, and 1000 of these boys come from the case when the family has two boys, $(B,B)$. Half the time we observe a boy the underlying family has two boys; therefore the probability of a second boy is 1/2.

If instead of 4000 we had generic $N$ families, and called them "cases," this argument would be the frequentist derivation of the result. In frequentist parlance, the 2000 total boys are called the "possibles" and the 1000 boys from $(B,B)$ are called the "favorables." The probability is calculated as the ratio of favorables to possibles.

(The frequentist approach is how most people learn about probability and combinatorics.)

Bayesian approach

Frequentist arguments become unwieldy with more elaborate problems, so we can use this puzzle to illustrate a more elegant approach, Bayesian inference.†

First let's call things by their name: $(B,B), (B,G), (G,B)$, and $(G,G)$ are the unobserved states of the world. "One is a boy," which we'll represent by $B$, is an observed event.

Some events are uninformative, for example "one is blond," in that they don't help answer the question. Others like "one is a boy," $B$, are informative, because they help answer the question. But how can we tell?

Event $B$ is informative because it happens with different probabilities in different states of the world; therefore observing $B$ gives information about what states we're more likely to be in:

$\Pr(B|(B,B)) = 1$;
$\Pr(B|(B,G)) = 1/2$;
$\Pr(B|(G,B)) = 1/2$;
$\Pr(B|(G,G)) = 0$.

We don't know the unobserved state of the world (that is, in which of those four states the family in question falls), so in this situation we can assign equal probabilities to all four (we could look up demographics tables and confirm the numbers, but let's keep this simple):

$\Pr((B,B)) = \Pr((B,G)) = \Pr((G,B)) = \Pr((G,G)) = 1/4$.

What we want is the probability of the state $(B,B)$ having observed the event $B$; this is the conditional probability $\Pr((B,B)|B)$, which can be computed using the Bayes formula,

\[
\Pr((B,B)|B) = \frac{\Pr(B|(B,B)) \Pr((B,B))}{\Pr(B)}.
\]
Because the $\Pr(B)$ trips a lot of people, let's be clear about what it is: it's the probability that you will observe a boy in general, not in this particular case; sometimes called the a-priori probability or the unconditional probability. This is the probability that if we picked a two-child family at random and then picked one of the children at random, that child would be a boy. It's not "one, because we observe a boy," a common error.

To compute $\Pr(B)$ we must consider all four states of the world and add up ("integrate over the space of states" in expensive wording) the probability of observing a boy in each of these states weighed by the probability of the state itself:

$\begin{array}{rl}\Pr(B) =& \Pr(B|(B,B)) \Pr((B,B)) + \\
 & \Pr(B|(B,G)) \Pr((B,G)) +  \\
& \Pr(B|(G,B)) \Pr((G,B)) + \\
&\Pr(B|(G,G)) \Pr((G,G)) \\
=& 1/2
\end{array}$

(Unsurprisingly, it's 1/2, since half of the children are boys.)

Now we can compute our quantity of interest $\Pr((B,B)|B)$ by replacing the numbers in the Bayes formula. In fact, we can do that for all the states,

$\Pr((B,B)|B) = 1/2$;
$\Pr((B,G)|B) = 1/4$;
$\Pr((G,B)|B) = 1/4$;
$\Pr((G,G)|B) = 0$.

(As they used to say in the Soviet Union, trust but verify: check those numbers to be sure.)

If it's a math problem, do the math.

-- -- -- --
* "Do the math" means apply the rules of math, not just the notation and numbers.

† There's a bit of a schism in statistical modeling between frequentists and Bayesians. I'll let you figure out which side I'm on.

From value to share (in math)

There are a few places online (and in some books) where a kind of magic happens. At some point we go from something like this
\[
v(x) = w_1 x_1 + \ldots + w_K x_K
\]
to, in a single magic jump,
\[
\Pr(\text{Buy } x) = \frac{\exp(v(x))}{1 + \exp(v(x))}.
\]
The $v(x) = w_1 x_1 + \ldots + w_K x_K$ is easy enough to understand: the value of some object $x$ is a weighted sum of the dimensions of $x$; there are $K$ dimensions $x_1, \ldots, x_K$ and associated weights, $w_1, \ldots, w_K$.

But then there's something along the lines of "using appropriate assumptions, the probability of a customer purchasing $x$ is given by" that fraction.

This magic jump appears in so many places discussing analytics, data science, big data, machine learning, artificial intelligence, and other fashionable ways to refer to statistics (because that's what's there) that we could be forgiven for suspecting that the people making the jump don't know where it comes from.

It's fine not to know where something comes from, as long as one is aware of that. But it's instructive to work through the simple cases, like this one. After all, a personal trainer needn't be a super-athlete, but should at least be able to exercise.

So, how do we go from the weighted sum to those exponentials (called the logit formula)?

First, that formula is only correct if, among other things: (a) the choice is between buying $x$ or nothing at all; and (b) the $v(\cdot)$ for buying nothing is set to be zero, denote that as $v(0) =0$. Okay, the (a) makes this a simple choice and (b) is just a threshold; to move it around we can always have a constant $x_0$ added to the $v(x)$.

Now, why isn't the probability either one or zero, then? After all, either $v(x) > 0 = v(0)$ and the customer should always buy or $v(x) < 0 = v(0)$ and the customer should never buy.

That's the second step: there are things that can change the $v(x)$ that we don't observe. Maybe the customer's mood changes and the $w_i$ change with it, for example. We don't know, so we say that the customer optimizes "utility" $u(x)$ instead of value $v(x)$, and the difference is a stochastic disturbance (expensive wording for random stuff) $\epsilon$:
\[
u(x) = v(x) - \epsilon.
\]
(Let's assume that $u(0) = v(0) = 0$. It's not necessary but makes things simpler.) This second step changes the nature of the problem: instead of a decision (buy or not), we can only predict a probability:
\[
\Pr(\text{Buy } x) = \Pr(u(x)>0) = \Pr(v(x)> \epsilon).
\]
Now we can compute that probability: it's just the cumulative distribution for the $\epsilon$ variable evaluated at the point $v(x)$, denoted $F_{\epsilon}(v(x))$.

Unfortunately, by their own nature the $\epsilon$ aren't known, so we need to assume a probability distribution for $\epsilon$, one that's acceptable and is computationally convenient.

We could try the Normal distribution, which most people accept without questioning, so it gets us the first desirable characteristic. Unfortunately the $\exp(-z^2/2)$ makes integrating it difficult. What we need is a distribution that looks more or less like a Normal but can be integrated into a manageable form.

Enter the logistic distribution.

It looks enough like a Normal (it has fat tails but it's hard to tell by looking). And its probability density function
\[
f_{\epsilon}(z) = \frac{\exp(- z)}{(1+\exp(-z))^2}
\]
integrates easily to
\[
F_{\epsilon}(v) = \int_{-\infty}^{v} f_{\epsilon}(z) \, dz = \frac{1}{1 + \exp(-v)} = \frac{\exp(v)}{1+\exp(v)},
\]
which, when replaced in the formula for probability of buy gives the logit formula:
\[
\Pr(\text{Buy } x) = \Pr(v(x)> \epsilon) = F_{\epsilon}(v(x)) = \frac{\exp(v(x))}{1+\exp(v(x))}.
\]
When you know the math, you understand the magic.

Hidden factor correlation

Correlation is not causation; everyone learns to say that. But if there's a correlation, there's probably some sort of causal relationship hiding somewhere, unless it's a spurious correlation.

If two variables, $A$ and $B$ are correlated, the three simplest causal relationships are: $A$ causes $B$; $B$ causes $A$; or $A$ and $B$ are caused by an unseen factor $C$. There are many more complicated causation relationships, but these are the three basic ones.

The third case, where an unseen variable $C$ is the real source of the correlation, is what we're interested in this post. To illustrate the case let's say $C$ is a standard normal random variable, and $A$ and $B$ are noisy measures of $C$,

$ \qquad A = C + \epsilon_A$ and $ B = C + \epsilon_B$,

where the $\epsilon_i$ are drawn from a normal distribution with $\sigma_{\epsilon} = 0.05$.

To illustrate we generate 10,000 draws of $C$ and create the 10,000 $A$ and $B$ using R:

hidden_factor = rnorm(10000)
var_A_visible = hidden_factor + 0.05 * rnorm(10000)
var_B_visible = hidden_factor + 0.05 * rnorm(10000)

Now we can plot $A$ and $B$, and the correlation is obvious

And we can regress $A$ on $B$ to get the correlation and test statistics for the estimates using a linear model,

model_no_control = lm(var_A_visible~var_B_visible)
summary(model_no_control)

With the result:

Call:
lm(formula = var_A_visible ~ var_B_visible)

Residuals:
      Min        1Q    Median        3Q       Max 
-0.271466 -0.047214 -0.000861  0.047400  0.302517 

Coefficients:
               Estimate Std. Error  t value Pr(>|t|)    
(Intercept)   0.0005294  0.0007025    0.754    0.451    
var_B_visible 0.9975142  0.0006913 1442.852  < 2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.07025 on 9998 degrees of freedom
Multiple R-squared:  0.9952, Adjusted R-squared:  0.9952 
F-statistic: 2.082e+06 on 1 and 9998 DF,  p-value: < 2.2e-16

So, both the model and the graph confirm a strong correlation ($p < 0.0001$) between $A$ and $B$. And in many real-life cases, this is used to support the idea that either $A$ causes $B$ or $B$ causes $A$.

Now we proceed to show how the hidden factor is relevant. First, let us plot the residuals, $A-C$ against $B-C$:

The apparent correlation has now disappeared. And a linear model including the hidden factor confirms this:

model_with_control = lm(var_A_visible~var_B_visible+hidden_factor)
summary(model_with_control)

With the result

Call:
lm(formula = var_A_visible ~ var_B_visible + hidden_factor)

Residuals:
     Min       1Q   Median       3Q      Max 
-0.18347 -0.03382 -0.00021  0.03410  0.17780 

Coefficients:
               Estimate Std. Error t value Pr(>|t|)    
(Intercept)   0.0007378  0.0004986   1.480    0.139    
var_B_visible 0.0004082  0.0100560   0.041    0.968    
hidden_factor 0.9997573  0.0100707  99.274  < 2e-16 ***
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.04985 on 9997 degrees of freedom
Multiple R-squared:  0.9976, Adjusted R-squared:  0.9976 
F-statistic: 2.072e+06 on 2 and 9997 DF,  p-value: < 2.2e-16

Hidden factors are easy to test for, as seen here, but they are not always apparent. For example, in nutrition papers there's often an hidden factor relating to how health-conscious an individual is that is more often than not causing both observables (say exercising regularly and eating salads; high correlation, but exercising doesn't cause eating salads and eating salads doesn't cause exercise).

Correlation is not causation, but generally one can find a causal relationship behind a correlation, possibly one that involves hidden factors or more complex relationships.

Calculating God?

I don't believe that the God of any earthly religion is the creator of the Universe.

But I really dislike a lazy and innumerate argument commonly used to "prove" the non-existence of God, which can be summarized in the following false dichotomy:

Either there is no God and the universe just 'poofed' into existence, or there's an infinite number of Gods, because the plane of existence for each God has to be created by a higher-level God.

This is a false dichotomy: it could well be that our universe was created by a powerful being from a higher-order universe, but that universe poofed into existence without a creator. Or maybe it did have a creator, whose universe poofed into existence; or that third universe may have had a creator...

Hey, this looks like dynamic programming. I know dynamic programming.

Let's say that universes are recursively nested until one of them just poofs into existence. Of course we can't see outside our universe, but we can build simple models.

So, our universe either poofed into existence (say with probability $p$) or it was created by some higher being (with probability $1-p$). Now we iterate the process: 'level 2' universe either poofed into existence (with some probability $q$) or was created by a 'level 3' universe being (with probability $1-q$); and so on.

Time for a simplifying assumption, or as non-mathematicians call it, making things up. Let's assume that all these universes share the poofed/created probabilities, so that for any 'level $k$' universe, it poofed into existence with probability $p$ and was created by a being from a 'level $k+1$' universe with probability $1-p$.

Note that it's still possible to have an infinite number of universes, but with this formulation, the probability of a 'level $k$' universe (with us being 'level 1') being the last level is

$p (1-p)^{k-1}$.

This probability gets small pretty quickly, which suggests the 'infinite regress of universes' argument gets thin very fast.

Now we can compute the expected number of universes as a function of $p$:

$\mathbb{E}(n) = p + 2(1-p)p + 3 (1-p)^2 p + \ldots N (1-p)^{N-1} p + \ldots$

or

$\mathbb{E}(n) = p/(1-p) \times ( \text{ sum of series } N (1-p)^N )$

The sum of series $N (1-p)^N$ is $(1-p)/p^2$, so

$E(n) = 1/p$

Therefore, if we believe that the probability of a universe poofing into existence is 0.1, there are an expected ten universes; for 0.2, five universes; for 0.5, two universes.

Very far from 'turtles all the way down.'

Of course, these calculations were unnecessary, because as we know from the revelations of the prophet Terry Pratchett, it's four elephants on the back of the Great A'Tuin swimming in the Sea of Stars.

A statistical analysis of reviews of L.A. Finest: audience vs. critics

"If numbers are available, let's use the numbers. If all we have are opinions, let's go with mine." -- variously attributed to a number of bosses.

There's a new police procedural this season, L.A. Finest, and Rotten Tomatoes has done it again: critics and audience appear to be at loggerheads. Like with The Orville, Star Trek Discovery, and the last season of Doctor Who.

But "appear to be" is a dequantified statement. And Rotten Tomatoes has numbers; so, what can these numbers tell us?

Before they can tell us anything, we need to write our question: first in words, then as a math problem. Then we can solve the math problem and that solution gets translated into a "words" answer, but now a quantified "words" answer.

The question, which is suggested by the above numbers is:

Do the critics and the audience use similar or opposite criteria to rate this show?

One way to answer this question, which would have been feasible in the past when Rotten Tomatoes had user reviews, would be to do text analytics on the reviews themselves. But now the user reviews are gone so that's no longer possible.

Another way, a simpler and cleaner way, is to use the data above.

To simplify we'll assume that all ratings are either positive or negative, 0 or 1; there are some unobservable random factors that make some people like a show more or less, so these ratings are random variables. For a given person $i$, the probability that that person likes L.A. Finest is captured in some parameter $\theta_i$ (we don't observe that, of course), which is the probability of that person giving a positive rating.

So, our question above is whether the $\theta_i$ of the critics and the $\theta_i$ of the audience are the same or "opposed." And what is "opposed"? If $i$ and $j$ use opposite criteria, the probability that $i$ gives a 1 is the probability that $j$ gives a 0, so $\theta_i = 1-\theta_j$.

We don't have the individual parameters $\theta_i$ but we can simplify again by assuming that all variation within each group (critics or audience) is random, so we really only need two $\theta$.

We are comparing two situations, call them: hypothesis zero, $H_0$, meaning the critics and the audience use the same criteria, that is they have the same $\theta$, call it $\theta_0$; and hypothesis one, $H_1$, meaning the critics use criteria opposite to those of the audience, so if the critics $\theta$ is $\theta_1$, the audience $\theta$ is $(1-\theta_1)$.

Yes, I know, we don't have $\theta_0$ or $\theta_1$. We'll get there.

Our "words" question now becomes the following math problem: how much more likely is it that the data we observe is created by $H_1$ versus created by $H_0$, or in a formula: what is the likelihood ratio

$LR = \frac{\Pr(\mathrm{Data}| H_1)}{\Pr(\mathrm{Data}| H_0)} $?

Observation: This is different from the usual statistics test: the usual test is whether the two distributions are different; we are testing for a specific type of difference, opposition. So there are in fact three states of the world: same, opposite, and different but not opposite; we want to compare the likelihood of the first two. If same is much more likely than opposite, then we conclude 'same.' If opposite is much more likely than same, we conclude 'opposite.' If same and opposite have similar likelihoods (for some notion of 'similar' we'd have to investigate), then we conclude 'different but not opposite.'

Our data is four numbers: number of critics $N_C = 10$, number of positive reviews by critics $k_C = 1$, number of audience members $N_A = 40$, number of positive reviews by audience members $k_A = 30$.

But what about the $\theta_0$ and $\theta_1$?

This is where the lofty field of mathematics gives way to the down and dirty world of estimation. We estimate $\theta$ by maximum likelihood, and the maximum likelihood estimator for the probability of a positive outcome of a binary random variable (called a Bernoulli variable) is the sample mean.

Yep, all those words to say "use the share of 1s as the $\theta$."

Not so fast. True, for $H_0$, we use the share of ones

$\theta_0 = (k_C + k_A)/(N_C + N_A) = 31/50 = 0.62$;

but for $H_1$, we need to address the audience's $1-\theta_1$ by reverse coding the zeros and ones, in other words,

$\theta_1 = (k_C + (N_A - k_A))/(N_C + N_A) = 11/50 = 0.22$.

Yes, those two fractions are "estimation." Maximum likelihood estimation, at that.

Now that we are done with the dirty statistics, we come back to the shiny world of math, by using our estimates to solve the math problem. That requires a small bit of combinatorics and probability theory, all in a single sentence:

If each individual data point is an independent and identically distributed Bernoulli variable, the sum of these data points follows the binomial distribution.

Therefore the desired probabilities, which are joint probabilities of two binomial distributions, one for the critics, one for the audience, are

$\Pr(\mathrm{Data}| H_0) = c(N_C,k_C) (\theta_0)^{k_C} (1- \theta_0)^{N_C- k_C} \times c(N_A,k_A) (\theta_0)^{k_A} (1- \theta_0)^{N_A- k_A}$

and

$\Pr(\mathrm{Data}| H_1) = c(N_C,k_C) (\theta_1)^{k_C} (1- \theta_1)^{N_C- k_C} \times c(N_A,k_A) (1 -\theta_1)^{k_A} (\theta_1)^{N_A- k_A}$.

Replacing the symbols with the estimates and the data we get

$\Pr(\mathrm{Data}| H_0) = 3.222\times 10^{-5}$;
$\Pr(\mathrm{Data}| H_1) = 3.066\times 10^{-2}$.

We can now compute the likelihood ratio,

$LR = \frac{\Pr(\mathrm{Data}| H_1)}{\Pr(\mathrm{Data}| H_0)} = 915$,

and translate that into words to make the statement

It's 915 times more likely that critics are using criteria opposite to those of the audience than the same criteria.

Isn't that a lot more satisfying than saying they "appear to be at loggerheads"?

Numbers as props vs numbers as information

Once you learn to tell the difference, you'll know whom to trust.

A very long time ago, in 2018, Elon Musk announced that Tesla would be ramping up production to 6000 vehicles per week. An anchor for a business program played the video, then addressed their co-host with:

"That's like four full parking structures a week. Wow!"
Co-host makes assenting noises.

That statement is true for parking structures that have 1500 spots, which most in San Francisco (where the show is produced) don't. Typical numbers here are closer to 500 than 1500. But that's not the important part.

The important part is that the number was used as a prop, not information.

More precisely, the anchor first bought into the idea that 6000 is a large number for a car company's weekly production, then looked for a way to make that number look big to the show's audience; parking structures are big buildings and are related to cars, so that was a good way to create the perception of "bigness." [1]

In other words, the process of using a number as a prop is:

1. Make a decision based on something other than the number
2. Look for a number to support that decision
3. Choose context to present the number that molds perception in favor of the decision.

The alternative to using numbers as props is using them as information.

The metric '6000 per week' is just data. It becomes information when it answers a question. A few of these questions that come to mind, considering that this is a business program focussing on technology for a mostly finance and finance-adjacent audience would be:

a. How does this production level compare to that of the competitors that Musk repeatedly states he's going to put out of business?
b. How does this production level compare to that of Toyota when it was running the factory that is now Tesla's?
c. How does this production level compare to the demand for electric vehicles in general, possibly by geographical area and brand of vehicle?

Note that these questions extract information from the number 6000, by comparing it to other numbers that are of business interest. This illustrates a very important principle of data-processing for decision-making:

What is informative about data depends on what decision is to be made.

Choosing question a for illustration, and using Wikipedia data for 2016, because it's publicly available so anyone can check this computation without having to pay financial information service fees, here are the production rates for the top 15 car companies by number of vehicles produced:

Those numbers put Tesla's production in context; they suggest that Tesla, relative to the competitors that Musk repeatedly taunts as "dinosaurs" and "on their way out," is a niche player and not a serious business threat. [2]

Note the process for using numbers as information:

1. Determine what decisions are to be informed by the number
2. Find the context that is relevant for that decision
3. Compare number with the numbers from that context

Using numbers as information is important primarily for decision-makers. Realizing when others are using numbers as props, not information, is important for everyone. Especially regarding whether you can trust the numbers -- and the other person.

Just because someone uses numbers as props, that doesn't necessarily mean their intent is to deceive you. Our society, particularly our news and edutainment, are full of prop-use of numbers for non-nefarious reasons: ignorance, desire to connect abstract numbers to concrete objects, laziness.

But there are people whose intent is to deceive, and often you can tell who they are by calling them on their use of numbers as props. [3]

When faced with the above table, many Tesla fans on twitter, some of whom manage third-party money, either resorted to ad hominem ("how big is your short position?" is a common one, even used by Musk) or changing the subject ("these cars will save the planet").

This is how you identify someone who's not making a good-faith mistake of using numbers as props, but rather someone who deliberately avoids using the appropriate context for the numbers to use them as props: they never address the relevant comparison.

Because most people don't process numbers as they hear or read them, but are still influenced by the perceived authority of the number, this behavior (deliberately using numbers as props to deceive, that is) is usually effective as a persuasion tool. And people who deliberately use numbers as props know about that effectiveness and that's why they do it. Which brings us to an important insight about people we get from their use of numbers:

People who deliberately use numbers as props are not to be trusted.



-- -- -- --  FOOTNOTES -- -- -- --

[1] More likely the choice was made by a writer or a producer, not the anchor; but the anchor is the face of the show, so we'll keep referring to them.

[2] Or, if we want to apply strategic thinking, Tesla should build itself by market expansion starting from its niche, instead of a frontal assault on the much larger companies (its current strategy)

[3] For what it's worth, I don't think the anchor, or the TV channel, were trying to deceive their audience. They were just caught in Musk's Reality Distortion Field, which in 2018 was much stronger than Steve Jobs's ever was.


-- -- -- -- ADDENDUM -- -- -- --

Later that year, numbers-as-props sophistry continued unimpeded by any sense of shame on the part of Tesla fans:

Blogging again?

For the last few months I've been using Twitter more and more like a blogging-in-threads platform. That makes no sense, so I've decided that perhaps a repurposing of the old blog as something of an escape valve for the innumeracy, numberphobia, and acalculia I see might be in order.

Also, I get to use mathjax  for pretty math instead of post-it notes: \(e^{i \pi}+1=0\).

(In keeping with the new purpose, some purging of the past will occur.)