By far the most unintuitive paradox for me personally is the one presented here: https://youtu.be/go3xtDdsNQM?t=3m27s "Mr. Jones has 2 children. What is the probability he has a girl if he has a boy born on Tuesday?" Somehow knowing the day of the week the boy was born changes the result. It's completely bizarre.
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 we…
Paradoxes of Probability and Other Statistical Strangeness
81–90 of 93 posts
Re: Paradoxes of Probability and Other Statistical Strangeness
#82By far the most unintuitive paradox for me personally is the one presented here: https://youtu.be/go3xtDdsNQM?t=3m27s "Mr. Jones has 2 children. What is the probability he has a girl if he has a boy born on Tuesday?" Somehow knowing the day of the week the boy was born changes the result. It's completely bizarre.
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 we…
Re: Paradoxes of Probability and Other Statistical Strangeness
#83Earlier quoted context omitted.
I think this only sounds like a paradox if it is phrased poorly. The accurate way to state it is "The probability of randomly picking a specific number is 0" and that sounds reasonable. The probability of successfully picking any number is 1.
The paradox is that, after picking a random number, you have just done a thing which has probability zero. Doing a thing that has zero probability shouldn't be possible. Ever.
Re: Paradoxes of Probability and Other Statistical Strangeness
#84Earlier quoted context omitted.
Non-countable sets defy intuition on several ways. The silver lining is that we don't have any evidence a non-countable thing exist on the real world. I don't think anybody even has a procedure for gathering that kind of evidence.
No. Rational number are countable, but have the same property of zero probability for an item in a uniform distribution.
Re: Paradoxes of Probability and Other Statistical Strangeness
#85Earlier quoted context omitted.
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 we…
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…
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.
Re: Paradoxes of Probability and Other Statistical Strangeness
#86Earlier quoted context omitted.
Non-countable sets defy intuition on several ways. The silver lining is that we don't have any evidence a non-countable thing exist on the real world. I don't think anybody even has a procedure for gathering that kind of evidence.
No. Rational number are countable, but have the same property of zero probability for an item in a uniform distribution.
Infinities are problematic too.
Re: Paradoxes of Probability and Other Statistical Strangeness
#87Earlier quoted context omitted.
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 we…
I adamantly agreed with you. Then I made a simple spreadsheet that proves us wrong: http://cl.ly/kuQE
Re: Paradoxes of Probability and Other Statistical Strangeness
#88Earlier quoted context omitted.
I adamantly agreed with you. Then I made a simple spreadsheet that proves us wrong: http://cl.ly/kuQE
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).
Re: Paradoxes of Probability and Other Statistical Strangeness
#89Earlier quoted context omitted.
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…
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…
If K1 is a girl and K2 is a boy born on Tuesday, this still counts as the family (Mr. Jones, if you like) having a boy born on Tuesday. There are 27 cases that count as the family having a boy born on Tuesday, all equiprobable. And out of those, 14 also count as the family having a girl.
As for your noting that we can split these cases even more finely, so that there's no distinguished end-all, be-all partitioning of cases, sure, you can do that. What I'm really saying is this:
1/2 of two-child families have their elder child being a boy. 1/2 of two-child families have their elder child being a girl. [On conventional idealizations for these problems. You surely do not dispute this, yes? You may not care about this number, but you don't dispute it, right?]
In each of those subgroups, 1/7 of families have their elder child born on Sunday, 1/7 have their elder child born on Monday, etc. [Do you dispute this?]
In each of THOSE subgroups, 1/2 of families have their younger child a boy, and 1/2 have their younger child a girl. [Any dispute?]
And in each of THOSE subgroups, 1/7 of families have their younger child born on Sunday, 1/7 of families have their younger child born on Monday, etc. [Any dispute?]
And some amount of those have low birth weight, some have high birthweight, some have 5 freckles, etc., but we needn't figure out those numbers.
So now I've carved the world up into 2 * 7 * 2 * 7 groups, based on gender and birth-date-of-week for older and younger child. We can carve the world up into groups in different ways also, more finely or more coarsely or just differently. But making the four conventional assumptions we just made, the 2 * 7 * 2 * 7 grouping based on gender and birth-date-of-week for older and younger child is such that each particular such group takes up 1/2 * 1/7 * 1/2 * 1/7 of all families; these are all equifrequent groups.
And that having been done, we find that in 27 of these groups, there is at least one boy born on a Tuesday. In 13, the elder child is a boy born on Tuesday but not the younger child; in 13, the younger child is a boy born on Tuesday but not the elder child; in 1, both children are boys born on Tuesday.
But the question was not intended to be about a specific boy. The question was intended to be "Out of families that have a boy born on Tuesday (meaning at least one boy born on Tuesday), what proportion have a girl?". Any family with at least one boy born on Tuesday counts as having "a boy born on Tuesday", and even families with two boys born on Tuesday count, with no particular of their two boys given any distinguished status.
Perhaps you read the question differently; that, then, is a problem with the phrasing of the question for communicating to you its intent. But when it was asked "What is the probability Mr. Jones has a girl, given that he has a boy born on Tuesday?", what the author indeed intended this to mean, and would be generally taken in the conventional language of probability to mean, was "Out of families that have at least one boy born on Tuesday, what proportion have a girl?".
And we find that, out of the 27 equally sized groups of families that have at least one boy born on Tuesday, 14 of them have a girl, so that the answer to this question becomes 14/27.
Re: Paradoxes of Probability and Other Statistical Strangeness
#90Earlier quoted context omitted.
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…
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…
problem 1) You go up to a person and ask them if they have exactly 2 children, at least one of which is a boy born on Tuesday. They say yes. What is the probability that they have a girl?
problem 2) You go up to a person and ask them if they have exactly 2 children, at least one of which is a boy. They say yes. You then ask them which day of the week a boy they have was born on. They say Tuesday. What is the probability that they have a girl?
The original problem that was posed is equivalent to problem 1, but not equivalent to problem 2. This could be what is confusing you, because in problem 2 the extra information plays no role in the selection process, while it does play a role in problem 1. In problem 2, the answer is the standard 2/3. Why are the probabilities different between problem 1 and 2? Here's why:
Think about the set of people who could answer yes to the question in problem 2. The ratio of these groups is important. A parent with BB (two boys) is equally likely to answer yes to problem 2 (100% likely to be exact) as a parent with BG and GB (also 100% likely to answer yes), which leads to the correct solution of 2/3. However, in problem 1 a parent with BB is NOT EQUALLY LIKELY to answer yes as a parent with BG. This is because we added an extra qualifier (must be born on Tuesday). The parent with BB has two chances to meet this qualifier because they have two boys, so the parent with BB is actually more likely to answer yes to the question than the parent with BG. As the qualifier becomes more and more rare (day of lunar cycle), the probability of the BB parent answer yes P(yes|BB) approaches twice the value of P(yes|BG). So now you're left with some subset of parents with BB, BG, and GB, but in this scenario you've sampled from BB approximately twice as much as you've sampled from each of the BG and GB groups, leaving you with approximately the same number of people from group BB as the combined amount from groups BG and GB. This is why the probability approaches 50%
I spend a while writing this, so hopefully it helps!