The simplicity of your explanations just drive home the stupidity involved in the massive wastes of time and money involved in the entirety of the "Climate" kerfuffle. I understand that a some of the people driving it are doing so because of the tsunami of government money being poured into it, or because of the power it gives them, but I absolutely cannot understand how any rational person can buy into the hysteria surrounding anthropogenic climate change given the 100% failure rate of the predictions to this point, and given the obviousness of the vast numbers of vague assumptions and bad data poured into all the models used as justification.
I just don't understand how we have gotten to the point where the government can use "climate change" as an excuse to dictate to me what kind of light bulb I'm allowed to have, what kind of dishwasher I'm allowed to buy, and to make the building codes restrictive to the point of making the building of a home ruinously expensive, and so many people simply nod their heads without the slightest moment of doubt.
There are so many things wrong with all of it, and it's all built on such purely speculative assumptions. I just don't get it.
Woohoo, I got it. Will have to come back to the card question. Thinking that both value and suite are relevant but will need to work that out on paper.
> Now suppose instead of assuming the first child is a boy, you assume the first or the second can be a boy. We must now consider only those instances in which the first or second child was a boy. This is three possibilities, only one of which resulted in two boys. That gives a probability of 1/3.
Not to be too "conformist" here, but the "divide probability equally among all cases" approach (aka maximum entropy rule) only applies before any informative events. "One is a boy" is an informative event (it happens with different probability for different states of the world); doing the math step-by-step gets the correct, "boring," 1/2 answer.
Some post-2000 textbooks use this precise problem as an example of how people get informativeness wrong (by getting the 1/3 answer).
In your post, your Pr(B) HAS conditions on it - it's the probability of a single boy when picking one child, GIVEN the condition that you're picking from families with two kids..
Instead, you can say, "we want to observe two boys in the set of all two child families which have at least one boy in them."
Just write down your 'states of the world' given ALL of the conditions (two kids, at minimum one is a boy)
(B,B) (B,G) (G,B)
Only in one of the three states are there two boys. 1 of 3.
Here's an example:
Imagine that you're working at a "Dads of Two Kids Convention."
10,000 dads are here, they're all fathers of 2 kids.
Each dad has a card hanging around their neck.
On the card, each of the dads that have at least one boy in their family have written "I have at least one boy", otherwise they have written nothing.
You walk into the room and are given a task:
1. Each dad with writing on the front of their card will walk up to your table.
2. You must predict if he has two boys, and he will tell you truthfully if he does or does not.
3. For every successful prediction that you make, you will be given a dollar. For every failure, you will lose a dollar. You will get paid (or owe) at the end of the guessing.
If you always predict that a given dad does NOT have two boys, you will earn ~$2,500. Because the probability of guessing "NOT two boys" correctly, *given* the information you have, is 2/3.
Simulate this in a computer. 10,000 dads, 2 kids, 50% chance first is boy, 50% chance second is boy. About 7500 of them will have 'at least one boy' and ~2500 will have two boys.
Compute your winnings for each iteration, *skipping* the dads that didn't come to your table, because they had no boys!
The probability of 1 in 2, which you calculated, is the answer to a different problem, such as this:
Instead of writing "I have at least one boy" on the card, each dad writes "My first child is a boy" on their card, or they write nothing.
This time, you have no good strategy, because ~5000 dads's first child will be boy, and of those, ~2500 will have their second child as a boy.
> In your post, your Pr(B) HAS conditions on it - it's the probability of a single boy when picking one child, GIVEN the condition that you're picking from families with two kids..
Kudos for trying to do the correct thing when presented with math, which is to see if there are unsupported steps in the derivation.
Alas, the formula for P(B) is not _my_ formula for this problem, it's the general formula for computing the unconditional probability of an event given the conditional probabilities and the probabilities of the states of the world. This is why we set the problem as a math formula and do the math with no regard for what the symbols mean outside of the math, which is how math works.
(It's easy to see that it could never be 1/3 since the problem is exactly the same as asking "we flip a fair coin twice, once it came heads, how likely is it that if came heads the other time too?" But we should always avoid words-thinking when given a math problem.)
If anyone wants an illustration of how nonsensical it is to divide probably equally among states after an informative event, here's a picture I made for a post that will probably never happen:
So the example is wrong? How? 10k men. 7.5k of them have at least one boy. Picking from the 7.5k, how many have 2 boys? 1 of 3. We don't *want* to pick from the 10k. Change boys to coins.
10,000 people in a room. They all flip two coins and everyone writes down YES if they got at least one head, and NO if not. You randomly select a man that wrote down YES. You guess if he flipped two heads. Your odds of being correct are 1 in 3, not 1 in 2.
This example shows how it is *not* like the "odds of getting a second heads on a coin flip when first flip is heads" because the probability being computed is *not* that of a single flip as relates to the other flip, but that of the outcome of *two flips* given we know that *either or both* came up heads, not that "coin 1 came up heads" or "coin 2 came up heads"
Question about the homework: Does the psychic know what cards have already been turned over? If so then I assert a bald 1/42 chance.
As I read this, it seems to me like so much of what we currently call The Science, is dragging in non-probative evidence to try and juice the number of potential positive cases and move the probabilities. Obviously this only changes the calculation and doesn't actually change anything real. Reminds me of my favorite rule of thumb for the corporate world, Goodhart's Law 'When a measure becomes a target it ceases to be a useful measure.'
1/[52 - (c + n)] where c is number of correct guesses that she eliminates from her guess pool and n is the number of previous times that she has made that same guess.
So, she needs to make use of the fact that she knows that previous cards were not a particular card and repeatedly guess the same card. Her odds of being right go up until she calls it correctly. And then she is at 1/(52-c) but c has increased by 1. This might make her claim to psychic powers harder to support though. But c increasing depends on her remembering the previously guessed cards, which is uncertain.
Regarding the weekend/weekday example. My first instinct was to use knowledge of days such that ratio of weekend to weekday births would be 2:5. Same approach but enumerating possibilities at the DAY level then counting according to weekend/weekday categories. Would this be a reasonable approach? Would not the knowledge of what a weekend is imply knowledge of the days that make it up? But I suppose this way assumes births are equally likely on any given day, which the problem didn't state. Thoughts?
I may misunderstand you, but I don't think so. The example for Weekday and Weekend also work if you use, say, MTWT and FSS. So the ratio of days is no longer the same.
Or for months JFMAMJ vs JASOND. There the ratio is 1:1.
Oops I didn't count Friday as Weekend in first post, no idea why! So let's say we use MTWT and FSS for "weekday" and "weekend", and we enumerate at the day level such that MTWT more likely than FSS since one extra day / chance. Would you consider this a valid approach to the problem even though the answers would be different (vs your original way / equal weighting of the possibilities "weekday" and "weekend")?
Sorry, all. I goofed and left the HOMEWORK out of the original. It's now added. But those who got emails won't see it.
The simplicity of your explanations just drive home the stupidity involved in the massive wastes of time and money involved in the entirety of the "Climate" kerfuffle. I understand that a some of the people driving it are doing so because of the tsunami of government money being poured into it, or because of the power it gives them, but I absolutely cannot understand how any rational person can buy into the hysteria surrounding anthropogenic climate change given the 100% failure rate of the predictions to this point, and given the obviousness of the vast numbers of vague assumptions and bad data poured into all the models used as justification.
I just don't understand how we have gotten to the point where the government can use "climate change" as an excuse to dictate to me what kind of light bulb I'm allowed to have, what kind of dishwasher I'm allowed to buy, and to make the building codes restrictive to the point of making the building of a home ruinously expensive, and so many people simply nod their heads without the slightest moment of doubt.
There are so many things wrong with all of it, and it's all built on such purely speculative assumptions. I just don't get it.
Murphy's Law explained....
(To an engineer the answer is simple; Overbuild as much as the accountants will permit. :-)
Optimist: It's half full.
Pessimist: It's half empty.
Engineer: It's twice as big as it needs to be ... (but I had the funds).
Woohoo, I got it. Will have to come back to the card question. Thinking that both value and suite are relevant but will need to work that out on paper.
I've missed three of these. Will catch up!
> Now suppose instead of assuming the first child is a boy, you assume the first or the second can be a boy. We must now consider only those instances in which the first or second child was a boy. This is three possibilities, only one of which resulted in two boys. That gives a probability of 1/3.
Not to be too "conformist" here, but the "divide probability equally among all cases" approach (aka maximum entropy rule) only applies before any informative events. "One is a boy" is an informative event (it happens with different probability for different states of the world); doing the math step-by-step gets the correct, "boring," 1/2 answer.
Some post-2000 textbooks use this precise problem as an example of how people get informativeness wrong (by getting the 1/3 answer).
(Yes, my book also uses this example, for the same purpose as the post-2000 textbooks. But here's a free version: http://sitacuisses.blogspot.com/2019/07/a-family-has-two-children-one-is-boy.html)
In your post, your Pr(B) HAS conditions on it - it's the probability of a single boy when picking one child, GIVEN the condition that you're picking from families with two kids..
Instead, you can say, "we want to observe two boys in the set of all two child families which have at least one boy in them."
Just write down your 'states of the world' given ALL of the conditions (two kids, at minimum one is a boy)
(B,B) (B,G) (G,B)
Only in one of the three states are there two boys. 1 of 3.
Here's an example:
Imagine that you're working at a "Dads of Two Kids Convention."
10,000 dads are here, they're all fathers of 2 kids.
Each dad has a card hanging around their neck.
On the card, each of the dads that have at least one boy in their family have written "I have at least one boy", otherwise they have written nothing.
You walk into the room and are given a task:
1. Each dad with writing on the front of their card will walk up to your table.
2. You must predict if he has two boys, and he will tell you truthfully if he does or does not.
3. For every successful prediction that you make, you will be given a dollar. For every failure, you will lose a dollar. You will get paid (or owe) at the end of the guessing.
If you always predict that a given dad does NOT have two boys, you will earn ~$2,500. Because the probability of guessing "NOT two boys" correctly, *given* the information you have, is 2/3.
Simulate this in a computer. 10,000 dads, 2 kids, 50% chance first is boy, 50% chance second is boy. About 7500 of them will have 'at least one boy' and ~2500 will have two boys.
Compute your winnings for each iteration, *skipping* the dads that didn't come to your table, because they had no boys!
The probability of 1 in 2, which you calculated, is the answer to a different problem, such as this:
Instead of writing "I have at least one boy" on the card, each dad writes "My first child is a boy" on their card, or they write nothing.
This time, you have no good strategy, because ~5000 dads's first child will be boy, and of those, ~2500 will have their second child as a boy.
> In your post, your Pr(B) HAS conditions on it - it's the probability of a single boy when picking one child, GIVEN the condition that you're picking from families with two kids..
Kudos for trying to do the correct thing when presented with math, which is to see if there are unsupported steps in the derivation.
Alas, the formula for P(B) is not _my_ formula for this problem, it's the general formula for computing the unconditional probability of an event given the conditional probabilities and the probabilities of the states of the world. This is why we set the problem as a math formula and do the math with no regard for what the symbols mean outside of the math, which is how math works.
(It's easy to see that it could never be 1/3 since the problem is exactly the same as asking "we flip a fair coin twice, once it came heads, how likely is it that if came heads the other time too?" But we should always avoid words-thinking when given a math problem.)
If anyone wants an illustration of how nonsensical it is to divide probably equally among states after an informative event, here's a picture I made for a post that will probably never happen:
https://x.com/josecamoessilva/status/1806110268063375856
So the example is wrong? How? 10k men. 7.5k of them have at least one boy. Picking from the 7.5k, how many have 2 boys? 1 of 3. We don't *want* to pick from the 10k. Change boys to coins.
10,000 people in a room. They all flip two coins and everyone writes down YES if they got at least one head, and NO if not. You randomly select a man that wrote down YES. You guess if he flipped two heads. Your odds of being correct are 1 in 3, not 1 in 2.
This example shows how it is *not* like the "odds of getting a second heads on a coin flip when first flip is heads" because the probability being computed is *not* that of a single flip as relates to the other flip, but that of the outcome of *two flips* given we know that *either or both* came up heads, not that "coin 1 came up heads" or "coin 2 came up heads"
Breaking apart the flips is where your math breaks, because the math answers the wrong problem when you break them apart.
Question about the homework: Does the psychic know what cards have already been turned over? If so then I assert a bald 1/42 chance.
As I read this, it seems to me like so much of what we currently call The Science, is dragging in non-probative evidence to try and juice the number of potential positive cases and move the probabilities. Obviously this only changes the calculation and doesn't actually change anything real. Reminds me of my favorite rule of thumb for the corporate world, Goodhart's Law 'When a measure becomes a target it ceases to be a useful measure.'
She knows if she was right or wrong.
1/[52 - (c + n)] where c is number of correct guesses that she eliminates from her guess pool and n is the number of previous times that she has made that same guess.
So, she needs to make use of the fact that she knows that previous cards were not a particular card and repeatedly guess the same card. Her odds of being right go up until she calls it correctly. And then she is at 1/(52-c) but c has increased by 1. This might make her claim to psychic powers harder to support though. But c increasing depends on her remembering the previously guessed cards, which is uncertain.
So about this psychic problem, are the cards returned to the deck after each attempt or removed?
No, used only once.
I know this is all basic for you, Professor, but it’s “gold” to me. So enlightening. Thank you.
Many thanks.
Hmm. Solved the Tuesday problem last Monday. It was fun but all I did was enumerate and count and elminate duplicates. Presto.
So William, why can't I do "Like" on your substack? It just stopped working.
No idea.
Regarding the weekend/weekday example. My first instinct was to use knowledge of days such that ratio of weekend to weekday births would be 2:5. Same approach but enumerating possibilities at the DAY level then counting according to weekend/weekday categories. Would this be a reasonable approach? Would not the knowledge of what a weekend is imply knowledge of the days that make it up? But I suppose this way assumes births are equally likely on any given day, which the problem didn't state. Thoughts?
No. That won't work. There are two categories of "days".
I may misunderstand you, but I don't think so. The example for Weekday and Weekend also work if you use, say, MTWT and FSS. So the ratio of days is no longer the same.
Or for months JFMAMJ vs JASOND. There the ratio is 1:1.
Oops I didn't count Friday as Weekend in first post, no idea why! So let's say we use MTWT and FSS for "weekday" and "weekend", and we enumerate at the day level such that MTWT more likely than FSS since one extra day / chance. Would you consider this a valid approach to the problem even though the answers would be different (vs your original way / equal weighting of the possibilities "weekday" and "weekend")?