Class 10: Minding Your Ps & Qs & Monty Hall: Uncertainty & Probability Theory: The Logic of Science

# Video

Links:

**HOMEWORK:** You meet a guy who says, “I have two children, one of whom is a boy who was born on a Tuesday.” What’s the probability, given this and only this Q, that both this guy’s children are boys?

# Lecture

I don’t think we’re there yet, so let’s review.

All—as in *all*—probability and logical equations follow this schema:

Pr(P|Q)

where P is the proposition of interest, and Q the evidence, assumptions, arguments, observations, whatever, all *assumed to be true*.

This is also the form of all—as in *all*—scientific models. Or models of any kind.

This is why I stress it so strongly. We must understand what this all means and how it works or we will never get the strengths and weaknesses of models later.

Anytime you hear any claim, especially a probability claim, like “Did you know the probability of a Russian attack is 32%?” *you must ask for the Q.*

We deduce all probabilities assuming Q is true. And we can always—as in *always*—find a Q that says P has any probability we want, from certainly false to certainly true to anything in between.

From that we deduce we must always ask about the Q.

Now we often can’t make heads or tails (good statistics joke!) of the Q. Too complex, obscure, and even proprietary. But part of Q, in all science models, is subset of propositions that say “Put the world in this setting”. This we will always know. Then we can judge how well the model works.

Simple example:

Pr(Light will come on| Plug is in the outlet & Electricity theory) = high

We don’t have to know how electricity works to test the model that plugging in the light should, with high probability, light it up. Not when the world has been put in the setting as the model states.

That is our goal; this is where we are heading.

I gave us as homework find the Q for P = “Biden Wins”, and find Pr(P|Q).

If Q = “Biden can win or not”, the best we can do is deduce this is a contingent event, i.e. one that is not necessarily true or false. Logically, this Q is equivalent in truth to “You will win or not”. Or “A walrus will win or not.” It gives us nothing, with the exception we induce the tacit premise of contingency. Thus:

Pr(Biden| Bidens wins or not) = (0,1),

i.e. the unit interval not touching 0 (false) or 1 (true).

This is quite different than this Q:

Q = “Either Trump or Biden must win”. Then

Pr(Biden|Either Trump or Biden must win) = 1/2,

as long as this is the ** ONLY** information that is considered. I cannot stress that

**enough. If I could come out from your screen and beat you over the head with it, I would.**

*ONLY*For it will turn out that all disagreements over probability turn out to be people forgetting this. We shall see.

Now a fews week’s back I had a model that said, Q = “When the incumbent runs, he wins; when he doesn’t, the party out of power wins.”

Then

Pr(Biden|Incumbent/Opposite) = high (or 1).

It’s only high because, as that post showed, there is some ambiguity in the terms. But you get the idea.

There is an INFINITE choice in Q. One is a betting market. This site, which updates frequently, and I’m using the numbers as I write this (14 June in the early AM), says “The probability Biden wins is 36.7%.”

This is some aggregate over several betting markets, which each of them is an aggregate over their users. People bet and try to make money. So these are like horse-race odds. Which means we have a conflation of betting—which is always a decision—and probability. These are not the same.

This means we can’t sort out the Q. It’s too big to write down, and can’t be, since the users at those sites don’t put in all their reasons for their bets.

But we can still check the model. And it is a model. The stats guy who runs the site does this, and its fun to look at. We’ll discuss how all this works a lot later.

Next, Monty Hall.

*This is an excerpt from Chapter 4 of *Uncertainty.*All the references have been removed.*

## Assigning Probability: Seeming Paradoxes & Doomsday Arguments

Assigning, or rather deducing, probabilities isn’t always easy; indeed it could be formidably difficult, or even impossible. It’s sometimes unclear whether, in a problem, the focus is on the P or the Q in Pr(P|Q). Because of that, disputes about the nature of probability arise. This section shows how tricky assignment can be.

The Monty Hall problem is infamous. Here are the premises: A contestant on a game show is offered to choose one of three doors, A, B, or C. Behind one and only one is a prize, and behind the others there is nothing. The contestant chooses, say, B. Monty then opens, say, C, behind which is nothing. The contestant is then offered to stay with his original choice or to switch to A. What decision maximizes the chance he wins something?

This problem became infamous when Marilyn vos Savant wrote a column in which she revealed the contestant ought to switch, which gives a 2/3 chance of winning the prize. Many—all too many—irate readers wrote in, claiming various superior credentials, including being professors of mathematics. These folks said vas Savant was wrong and that the probability of winning switching or staying was “obviously” 1/2. These professors were, it might surprise you to learn, right. But then so was vos Savant. Because all probability is conditional, people were giving the right answer to the wrong question: they did not see that their premises were different than vos Savant’s.

Given *only* the premises, Q = “Before us are two doors, A and B, only one of which conceals a prize”, then the probability P = “A has the prize” is 1/2; and for door B, too. Switch or stay; it’s the same. But those aren’t the premises vos Savant used. She began with the original premises but added to them knowledge that if the contestant initially chose the right door, Monty could open either remaining (thus the contestant should stay); but if the contestant had chosen incorrectly, then Monty could only open the one remaining empty door (thus the contestant should switch). Since picking the right door on the initial premises has a 1/3 probability, vos Savant was right: switching makes more sense. This incident, which attracted nation-wide attention, proves that it’s often tough to remember what the premises are and when.

There are many other probability “paradoxes” and problems. The Sleepy Beauty, three envelopes, one child born on a Tuesday, and many, many more, all of which share the same nature as the Monty Hall. Which is to say, they all have hidden, or not readily visible, premises, or they have deductions and implications which are missed, or information that is hideously complex to grasp. But all of them prove that probability is conditional; indeed, it is their conditionality which makes them interesting and (for some of us) fun.

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Probability of 2 boys = 1/3 by either Bayes or by the possibilities of BB, GG, BG, GB. Trans issues not included.

Here is another probability question: What are the odds a U.S. governmental agency will cook the books to support the Climate Change Catastrophe narrative?

Dr. Roger Pielke, Jr. has looked at NOAA’s claims of the rise of billion dollar disasters, and concluded that they are playing games with the numbers.

https://rogerpielkejr.substack.com/p/billion-dollar-disasters-on-steroids

Homework answer: Given the Q and ONLY that Q, the probability cannot be calculated. Why? Because in order to calculate 1/2 or 1/3 or whatever, you ALSO must include something like Q = “the average observed ratio of boys to girls is 1/2” and Q = “there are 7 days in the calendar week”.

Chris brings to mind a flaw where I added girl to Q. I suppose the options can only be boy or not boy. I don’t see how the day figures in, so I stick with the same answer.

Enumerate to find what we have. Remember BTu is fixed!

Start first is Btu second Girl any day we have

First Second

Btu Gm

Gtu

Gw

GTh

GF

GSa

GSu

First is Btu second is B any day

First Second

Btu Bm

Btu

BW

BTh

BF

BSa

BSu

First G any day second Btu

First Second

Gm BTu

Gtu

Gw

GTh

GF

GSa

GSu

First is B any day Second is Btu

First Second

BM BTu

Btu

BW

BTh

BF

BSa

BSu

So we have two sets of Btu first and Btu second. One is thrown out because it would be double counted. Each scenario has seven possibilities except in one case we through out the double leaving 7+7+7+6 possible outcomes=27 and 13 with Boy Boy.

Sorry about that! Formatting these comments is impossible.

Dang this took awhile, but I think I got it.

Q includes definitions, such as definition of “child”: a child is human offspring, either a boy or a girl. Without more information, this is 1 in 2 (not 0..1). This is different from the Biden example, because “Biden can win the election or not” does not imply the number of losers of the election, etc.

I’m going to butcher the notation.

If we start with our Q1 simply as “I have a child”, find Pr(B|Q1):

Assuming Q1 is True, 2 possibilities:

1. Child is Boy

2. Child is Girl

Pr(B|Q1) = 1/2.

Changing our Q, we have Q2: “I have two children”, find Pr(BB|Q2)

Assuming Q2 is True, 4 possibilities:

1. Child 1 is Boy, Child 2 is Girl

2. Child 1 is Girl, Child 2 is Girl

3. Child 1 is Boy, Child 2 is Boy

4. Child 1 is Girl, Child 2 is Boy

Pr(BB|Q2) = 1/4

More information! Q3: “I have two children, at least one of whom is a boy”

Assuming Q3 is True, 3 possibilities:

1. Child 1 is Girl, Child 2 is Boy

2. Child 1 is Boy, Child 2 is Boy

3. Child 1 is Boy, Child 2 is Girl

Pr(BB|Q3) = 1/3

Even more information! Q4: “I have two children, at least one of whom is a boy born on a Tuesday”

Tuesday is a day of the week, of which there are seven, so we can break down the possibilities:

Assuming Q3 is True, 27?! possibilities:

01. Child 1 is Girl (Born Sunday), Child 2 is Boy (Born Tuesday)

…

07. Child 1 is Girl (Born Saturday), Child 2 is Boy (Born Tuesday)

08. Child 1 is Boy (Born Tuesday), Child 2 is Girl (Born Sunday)

…

14. Child 1 is Boy (Born Tuesday), Child 2 is Girl (Born Saturday)

…

17. Child 1 is Boy (Born Tuesday), Child 2 is Boy (Born Tuesday)

…

21. Child 1 is Boy (Born Saturday), Child 2 is Boy (Born Tuesday)

…

24. Child 1 is Boy (Born Tuesday), Child 2 is Boy (Born Tuesday)

…

28. Child 1 is Boy (Born Tuesday), Child 2 is Boy (Born Saturday)

Possibility 17 is the same as possibility 24, so we have 27 distinct possibilities, of which 13 yield two boys:

Pr(BB|Q4) = 13/27

For bonus points?

Q5: “I have two children, one of whom is a boy born on Tuesday, January 2, 2024 at 4:43:00 AM.”

1. Child 1 is Girl, Child 2 is Boy born at that instant.

2. Child 1 is Boy born at that instant, Child 2 is Girl.

3. Child 1 is Boy born at that instant, Child 2 is Boy.

4. Child 1 is Boy, Child 2 is Boy born at that instant.

Pr(BB|Q5) = 2/4 = 1/2

I suppose the guy could have knocked up two ladies a few days apart and they gave birth at exactly the same time. If it was a gal giving Q5, we could say exactly 1/2, since one lady can’t have 2 babies exactly at the same time (cue “how about a C-section with 2 doctors removing both children at exactly the same moment”). Gah there’s so much hiding in Q that shows up based on the language alone!

If we put the world in the setting, we assume that there are two genders, the gender of one kid has no effect on the gender of subsequent births, and the probability of having a kid with a given gender is 0.5.

In terms of kid 2’s gender, the gender of kid 1 is as irrelevant as the day of the week on which he was born. But it does tell us that there’s already a boy, which is relevant to the question of whether both kids are boys.

We can then reframe the question as “What is the probability that kid 2 is a boy?” (and in consequence, both kids are boys).

So the answer would be 0.5.