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I'm an idiot, but I'm going to throw my hat in the ring here: The video is wrong. The problem reads: Jones has 2 kids. What is P(he has a girl) given that he

I'm an idiot, but I'm going to throw my hat in the ring here: The video is wrong. The problem reads: Jones has 2 kids. What is P(he has a girl) given that he has a boy born on a Tuesday. Consider, for a moment, what information we're getting from "boy born on a Tuesday." This is no different than "boy with red hair," or "boy with 5 freckles." The fact that the BOY was born on a tuesday does not change P(day of the week girl was born). Imagine the "boy with 5 freckles" case - let 5 freckles be denoted by F5, six freckles by F6 and so on... would the appropriate calculation include enumerating P(boy F5, boy Fn) for all n? No. The "born on Tuesday" is irrelevant. Thus you have the following scenarios: - one kid is TuesdayBoy and the other is also a boy, born at any time - one kid is TuesdayBoy and the other is a girl, born at any time Out of these options P(Jones has a girl) is a flat out 50%. There is no need to bring in concepts of "which was born first" or enumerate all possible days of the week each child could have been born. Ok... now all the real smartypants here can correct me :)

Chinjut 9y ago
There are 2 * 7 * 2 * 7 ways to assign gender and birth-day-of-week to two children. By convention, all are considered equiprobable (this is the same as assuming kids' genders and birth day-of-weeks are independent of each other and of all facts about other kids, and that both genders are equally likely and all 7 days are equally likely for any given kid.) Of these possibilities, 27 are situations where one kid is a Tuesday boy. [Do you dispute this count?] Of those, 14 are situations where one kid is a girl. [Do you dispute this count?] The answer to "What proportion of cases where there is at least one Tuesday boy also have a girl?" is thus 14/27. You have stated by fiat that certain things are irrelevant to certain other things, that certain things have probability 50%, etc, but in doing so, you have not considered the count correctly. You are likely misled by phrasing such as "the boy", when there are families with two boys in which there is no proper referent of "the boy" and no particular answer to question like "Which day was 'the boy' born?".
- astrocat 9y ago
  ok ok... let me try to get this straight. Just as kind of a mental process for trying to understand whether or not something passes the smell test, I typically try to take the basic premise and turn it up to 11 and see if that still makes sense. In this problem, as you've described it, we're enumerating "ways to assign gender and birth-day-of-week." We can do this because there are a countable number of "days of the week" (so we can map to the integers: 1-6) AND there is also a surjective function of [child] -> [day of the week they were born]. Am I right so far? Now let's replace the set [1-6] with another countable set that also maintains the surjective function. We could say "day in the lunar cycle" (so ~27 options), or better "day of the year" (366 options), for example. Do we now need to consider the 23662366 ways to assign gender and birth-day-of-the-year? Take it further with whatever you want: "birth weight in milligrams" or "number of freckles" (as I previously suggested). All countable things that meet the surjective requirement. This is starting to smell funny, right? So let's take a look at the math. You say there are 2727 ways to configure day+gender, assuming independence for kid 1 (k1) and kid 2 (k2). This represents: (k1 gender options * k1 day of week options) * (k2 gender options * k2 day of week options). Right? I'm with you so far. Then you say "Of these possibilities, 27 are situations where one kid is a Tuesday boy." Hold up. We are given two pieces of information: that one of the kids is a boy, and that particular boy was born on a Tuesday. Let's say the boy is k1 (this is an assignment of enumeration, not of "who came first;" just like Sunday = 1 does not mean that any kid born on a Sunday was born before every kid born on Monday = 2). So now the k1 options are [11] (boy, tuesday), and the total number of options are: [11] * [27] = 14. Of those 14, 7 are girl options. And we're back to a straight 50%. So yes, I dispute the 27 number. It seems like it is arrived at by 2127, minus one for an apparent duplicate. But the 212*7 represents maintaining gender non-specificity for Tuesday boy, which should be incorrect, no? > You have stated by fiat that certain things are irrelevant to certain other things... Yes, but that's what "independent" means, right? You also stated that you're assuming these two things are independent, hence equiprobability. But independence is defined by P(A) = P(A|B). The probability of A is completely unaffected by B. Yet the outcome you arrive at is that P(A) IS affected by B, so the math presented is internally inconsistent. What am I missing here? I'm fascinated by the uncertainty around this little problem.
  - Chinjut 9y ago
ubernostrum 9y ago
First step back and consider the possibilities given no knowledge whatsoever: For each child the problem constrains to one of two possible sexes and one of seven possible days of birth. 2 * 7 = 14 possible sex/day combinations for a single child. (2 * 7) * (2 * 7) = 196 possible sex/day combinations for a pairing of two children. To see why, you could write a program to enumerate all of them, starting with the pairing "Boy/Monday + Boy/Monday", then "Boy/Monday + Boy/Tuesday" and so on until you exhaust all possible options at "Girl/Sunday + Girl/Sunday". You'll see there are 196 options. Now start applying the facts given to us: one of the children is born on a Tuesday (eliminate all possibilities which don't have at least one Tuesday child), and that child is a boy (eliminate all possibilities in which there is not a Tuesday child who is also a boy). This leaves exactly 27 possible cases: Boy/Sunday + Boy/Tuesday, Boy/Monday + Boy/Tuesday, Boy/Tuesday + Boy/Tuesday, Boy/Wednesday + Boy/Tuesday, Boy/Thursday + Boy/Tuesday, Boy/Friday + Boy/Tuesday, Boy/Saturday + Boy/Tuesday, Girl/Sunday + Boy/Tuesday, Girl/Monday + Boy/Tuesday, Girl/Tuesday + Boy/Tuesday, Girl/Wednesday + Boy/Tuesday, Girl/Thursday + Boy/Tuesday, Girl/Friday + Boy/Tuesday, Girl/Saturday + Boy/Tuesday, Boy/Tuesday + Boy/Sunday, Boy/Tuesday + Boy/Monday, Boy/Tuesday + Boy/Wednesday, Boy/Tuesday + Boy/Thursday, Boy/Tuesday + Boy/Friday, Boy/Tuesday + Boy/Saturday, Boy/Tuesday + Girl/Sunday, Boy/Tuesday + Girl/Monday, Boy/Tuesday + Girl/Tuesday, Boy/Tuesday + Girl/Wednesday, Boy/Tuesday + Girl/Thursday, Boy/Tuesday + Girl/Friday, Boy/Tuesday + Girl/Saturday If you count, you'll see that of those 27, there are 13 with two boys and 14 with a boy and a girl. The probability of two boys, given that one child is a boy born on Tuesday, is thus 13/27.
deleted 9y ago
[deleted]
sxv 9y ago
I adamantly agreed with you. Then I made a simple spreadsheet that proves us wrong: http://cl.ly/kuQE http://cl.ly/kuQE
- astrocat 9y ago
  Ah, but see... you're counting (B2,B2) as one item because "order doesn't matter", but then counting (G2,B2) and (B2,G2) independently. If (G2,B2) is different than (B2,G2), then (B2,B'2) is distinct from (B'2,B2).
  - sxv 9y ago
    Think of the x-axis as the first child and the y-axis as the second child. One in fourteen chance of choosing a column and one in fourteen chance to choose a row. I fail to see how there could be any additional outcomes or that any square has a greater chance of occurring than another..

Chinjut 9y ago

There are 2 * 7 * 2 * 7 ways to assign gender and birth-day-of-week to two children. By convention, all are considered equiprobable (this is the same as assuming kids' genders and birth day-of-weeks are independent of each other and of all facts about other kids, and that both genders are equally likely and all 7 days are equally likely for any given kid.) Of these possibilities, 27 are situations where one kid is a Tuesday boy. [Do you dispute this count?] Of those, 14 are situations where one kid is a girl. [Do you dispute this count?] The answer to "What proportion of cases where there is at least one Tuesday boy also have a girl?" is thus 14/27. You have stated by fiat that certain things are irrelevant to certain other things, that certain things have probability 50%, etc, but in doing so, you have not considered the count correctly. You are likely misled by phrasing such as "the boy", when there are families with two boys in which there is no proper referent of "the boy" and no particular answer to question like "Which day was 'the boy' born?".

astrocat 9y ago
ok ok... let me try to get this straight. Just as kind of a mental process for trying to understand whether or not something passes the smell test, I typically try to take the basic premise and turn it up to 11 and see if that still makes sense. In this problem, as you've described it, we're enumerating "ways to assign gender and birth-day-of-week." We can do this because there are a countable number of "days of the week" (so we can map to the integers: 1-6) AND there is also a surjective function of [child] -> [day of the week they were born]. Am I right so far? Now let's replace the set [1-6] with another countable set that also maintains the surjective function. We could say "day in the lunar cycle" (so ~27 options), or better "day of the year" (366 options), for example. Do we now need to consider the 23662366 ways to assign gender and birth-day-of-the-year? Take it further with whatever you want: "birth weight in milligrams" or "number of freckles" (as I previously suggested). All countable things that meet the surjective requirement. This is starting to smell funny, right? So let's take a look at the math. You say there are 2727 ways to configure day+gender, assuming independence for kid 1 (k1) and kid 2 (k2). This represents: (k1 gender options * k1 day of week options) * (k2 gender options * k2 day of week options). Right? I'm with you so far. Then you say "Of these possibilities, 27 are situations where one kid is a Tuesday boy." Hold up. We are given two pieces of information: that one of the kids is a boy, and that particular boy was born on a Tuesday. Let's say the boy is k1 (this is an assignment of enumeration, not of "who came first;" just like Sunday = 1 does not mean that any kid born on a Sunday was born before every kid born on Monday = 2). So now the k1 options are [11] (boy, tuesday), and the total number of options are: [11] * [27] = 14. Of those 14, 7 are girl options. And we're back to a straight 50%. So yes, I dispute the 27 number. It seems like it is arrived at by 2127, minus one for an apparent duplicate. But the 212*7 represents maintaining gender non-specificity for Tuesday boy, which should be incorrect, no? > You have stated by fiat that certain things are irrelevant to certain other things... Yes, but that's what "independent" means, right? You also stated that you're assuming these two things are independent, hence equiprobability. But independence is defined by P(A) = P(A|B). The probability of A is completely unaffected by B. Yet the outcome you arrive at is that P(A) IS affected by B, so the math presented is internally inconsistent. What am I missing here? I'm fascinated by the uncertainty around this little problem.
- Chinjut 9y ago

ubernostrum 9y ago

First step back and consider the possibilities given no knowledge whatsoever: For each child the problem constrains to one of two possible sexes and one of seven possible days of birth. 2 * 7 = 14 possible sex/day combinations for a single child. (2 * 7) * (2 * 7) = 196 possible sex/day combinations for a pairing of two children. To see why, you could write a program to enumerate all of them, starting with the pairing "Boy/Monday + Boy/Monday", then "Boy/Monday + Boy/Tuesday" and so on until you exhaust all possible options at "Girl/Sunday + Girl/Sunday". You'll see there are 196 options. Now start applying the facts given to us: one of the children is born on a Tuesday (eliminate all possibilities which don't have at least one Tuesday child), and that child is a boy (eliminate all possibilities in which there is not a Tuesday child who is also a boy). This leaves exactly 27 possible cases: Boy/Sunday + Boy/Tuesday, Boy/Monday + Boy/Tuesday, Boy/Tuesday + Boy/Tuesday, Boy/Wednesday + Boy/Tuesday, Boy/Thursday + Boy/Tuesday, Boy/Friday + Boy/Tuesday, Boy/Saturday + Boy/Tuesday, Girl/Sunday + Boy/Tuesday, Girl/Monday + Boy/Tuesday, Girl/Tuesday + Boy/Tuesday, Girl/Wednesday + Boy/Tuesday, Girl/Thursday + Boy/Tuesday, Girl/Friday + Boy/Tuesday, Girl/Saturday + Boy/Tuesday, Boy/Tuesday + Boy/Sunday, Boy/Tuesday + Boy/Monday, Boy/Tuesday + Boy/Wednesday, Boy/Tuesday + Boy/Thursday, Boy/Tuesday + Boy/Friday, Boy/Tuesday + Boy/Saturday, Boy/Tuesday + Girl/Sunday, Boy/Tuesday + Girl/Monday, Boy/Tuesday + Girl/Tuesday, Boy/Tuesday + Girl/Wednesday, Boy/Tuesday + Girl/Thursday, Boy/Tuesday + Girl/Friday, Boy/Tuesday + Girl/Saturday If you count, you'll see that of those 27, there are 13 with two boys and 14 with a boy and a girl. The probability of two boys, given that one child is a boy born on Tuesday, is thus 13/27.

deleted 9y ago

[deleted]

sxv 9y ago

I adamantly agreed with you. Then I made a simple spreadsheet that proves us wrong: http://cl.ly/kuQE http://cl.ly/kuQE

astrocat 9y ago
Ah, but see... you're counting (B2,B2) as one item because "order doesn't matter", but then counting (G2,B2) and (B2,G2) independently. If (G2,B2) is different than (B2,G2), then (B2,B'2) is distinct from (B'2,B2).
- sxv 9y ago
  Think of the x-axis as the first child and the y-axis as the second child. One in fourteen chance of choosing a column and one in fourteen chance to choose a row. I fail to see how there could be any additional outcomes or that any square has a greater chance of occurring than another..