A Family Has Four Children. Let X Represent

Paradox in probability theory

The Male child or Girl paradox surrounds a prepare of questions in probability theory, which are also known equally The Two Child Problem,^[1] Mr. Smith'south Children ^[ii] and the Mrs. Smith Problem. The initial formulation of the question dates back to at least 1959, when Martin Gardner featured it in his October 1959 "Mathematical Games cavalcade" in Scientific American. He titled it The Two Children Trouble, and phrased the paradox equally follows:

• Mr. Jones has 2 children. The older child is a girl. What is the probability that both children are girls?
• Mr. Smith has two children. At least one of them is a male child. What is the probability that both children are boys?

Gardner initially gave the answers one / 2 and 1 / 3 , respectively, but later acknowledged that the second question was cryptic.^[1] Its respond could exist 1 / 2 , depending on the process by which the information "at to the lowest degree one of them is a boy" was obtained. The ambivalence, depending on the verbal wording and possible assumptions, was confirmed past Maya Bar-Hillel and Ruma Falk,^[3] and Raymond S. Nickerson.^[four]

Other variants of this question, with varying degrees of ambivalence, have been popularized by Inquire Marilyn in Parade Magazine,^[5] John Tierney of The New York Times,^{[half dozen]} and Leonard Mlodinow in The Drunk's Walk.^[vii] Ane scientific study showed that when identical information was conveyed, merely with different partially ambiguous wordings that emphasized different points, that the percent of MBA students who answered 1 / two changed from 85% to 39%.^[ii]

The paradox has stimulated a nifty deal of controversy.^[4] The paradox stems from whether the problem setup is similar for the two questions.^{[2] [7]} The intuitive reply is 1 / 2 .^[ii] This answer is intuitive if the question leads the reader to believe that there are ii as likely possibilities for the sex of the 2d child (i.e., boy and daughter),^{[2] [viii]} and that the probability of these outcomes is absolute, not provisional.^[ix]

Common assumptions [edit]

The two possible answers share a number of assumptions. Start, information technology is causeless that the space of all possible events can exist easily enumerated, providing an extensional definition of outcomes: {BB, BG, GB, GG}.^[10] This notation indicates that in that location are iv possible combinations of children, labeling boys B and girls Yard, and using the showtime letter to stand for the older kid. Second, information technology is assumed that these outcomes are equally probable.^[10] This implies the post-obit model, a Bernoulli process with p = one / 2 :

1. Each kid is either male or female.
2. Each child has the aforementioned run a risk of existence male as of being female.
3. The sex activity of each child is contained of the sex of the other.

The mathematical outcome would be the same if it were phrased in terms of a coin toss.

Commencement question [edit]

• Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?

Nether the aforementioned assumptions, in this trouble, a random family unit is selected. In this sample space, in that location are 4 equally probable events:

Older child	Younger child
Girl	Daughter
Girl	Male child
Boy	Girl
Boy	Boy

Only 2 of these possible events meet the criteria specified in the question (i.east., GG, GB). Since both of the 2 possibilities in the new sample space {GG, GB} are every bit likely, and only one of the ii, GG, includes two girls, the probability that the younger child is too a girl is ane / 2 .

Second question [edit]

• Mr. Smith has two children. At least i of them is a boy. What is the probability that both children are boys?

This question is identical to question ane, except that instead of specifying that the older kid is a boy, it is specified that at least i of them is a boy. In response to reader criticism of the question posed in 1959, Gardner said that no answer is possible without information that was not provided. Specifically, that ii different procedures for determining that "at least i is a male child" could pb to the exact aforementioned wording of the problem. Only they lead to different correct answers:

• From all families with 2 children, at to the lowest degree i of whom is a boy, a family unit is chosen at random. This would yield the answer of 1 / 3 .
• From all families with ii children, one child is selected at random, and the sex of that child is specified to be a male child. This would yield an answer of i / 2 .^{[3] [4]}

Grinstead and Snell debate that the question is ambiguous in much the same way Gardner did.^[11] They get out it to the reader to decide whether the process, that yields 1/three every bit the reply, is reasonable for the problem every bit stated higher up. The formulation of the question they were considering specifically is the following:

• Consider a family with ii children. Given that one of the children is a boy, what is the probability that both children are boys?

In this formulation the ambivalence is about obviously present, considering it is not articulate whether we are allowed to presume that a specific child is a boy, leaving the other child uncertain, or whether it should exist interpreted in the aforementioned way as 'at least one male child'. This ambivalence leaves multiple possibilities that are not equivalent and leaves the necessity to brand assumptions nigh 'how the information was obtained', as Bar-Hillel and Falk fence, where different assumptions tin can atomic number 82 to different outcomes (considering the problem statement was not well enough defined to allow a single straightforward interpretation and answer).

For instance, say an observer sees Mr. Smith on a walk with just one of his children. If he has two boys then that kid must be a boy. But if he has a male child and a daughter, that child could take been a girl. So seeing him with a boy eliminates not only the combinations where he has two girls, but likewise the combinations where he has a son and a daughter and chooses the daughter to walk with.

So, while it is certainly truthful that every possible Mr. Smith has at least 1 boy (i.e., the condition is necessary), it cannot be causeless that every Mr. Smith with at least one male child is intended. That is, the problem argument does not say that having a male child is a sufficient condition for Mr. Smith to exist identified equally having a male child this way.

Commenting on Gardner'south version of the problem, Bar-Hillel and Falk^[3] annotation that "Mr. Smith, unlike the reader, is presumably aware of the sex of both of his children when making this statement", i.e. that 'I accept ii children and at least one of them is a boy.' Information technology must be farther assumed that Mr. Smith would e'er report this fact if information technology were true, and either remain silent or say he has at to the lowest degree ane daughter, for the correct answer to be 1 / 3 as Gardner apparently originally intended. Merely under that assumption, if he remains silent or says he has a daughter, in that location is a 100% probability he has two daughters.

Assay of the ambiguity [edit]

If it is assumed that this data was obtained past looking at both children to see if there is at to the lowest degree one boy, the condition is both necessary and sufficient. Three of the iv equally probable events for a two-kid family in the sample infinite above run across the status, as in this table:

Older child	Younger child
Girl	Girl
Daughter	Male child
Boy	Girl
Boy	Boy

Thus, if it is assumed that both children were considered while looking for a boy, the reply to question 2 is one / three . Nonetheless, if the family unit was starting time selected and then a random, true statement was made near the sexual activity of one kid in that family, whether or non both were considered, the correct way to calculate the conditional probability is not to count all of the cases that include a child with that sex. Instead, one must consider only the probabilities where the statement will exist made in each example.^[eleven] So, if ALOB represents the upshot where the statement is "at to the lowest degree i boy", and ALOG represents the consequence where the statement is "at least 1 daughter", and then this table describes the sample infinite:

Older child	Younger child	P(this family)	P(ALOB given this family)	P(ALOG given this family)	P(ALOB and this family)	P(ALOG and this family)
Girl	Girl	i / 4	0	i	0	1 / 4
Girl	Male child	1 / iv	1 / 2	1 / 2	1 / eight	1 / 8
Boy	Girl	1 / iv	1 / 2	one / 2	1 / eight	one / 8
Male child	Boy	1 / 4	1	0	i / 4	0

So, if at least 1 is a boy when the fact is called randomly, the probability that both are boys is

${\displaystyle \mathrm {\frac {P(ALOB\;and\;BB)}{P(ALOB)}} ={\frac {\frac {1}{4}}{0+{\frac {1}{8}}+{\frac {1}{8}}+{\frac {1}{iv}}}}={\frac {1}{2}}\,.}$

The paradox occurs when it is non known how the statement "at least one is a male child" was generated. Either answer could be correct, based on what is assumed.^[12]

Even so, the " 1 / three " answer is obtained only by bold P(ALOB|BG) = P(ALOB|GB) =one, which implies P(ALOG|BG) = P(ALOG|GB) = 0, that is, the other kid's sex is never mentioned although information technology is present. As Marks and Smith say, "This extreme assumption is never included in the presentation of the two-child trouble, however, and is surely not what people have in mind when they present information technology."^[12]

Modelling the generative process [edit]

Another way to analyse the ambiguity (for question two) is past making explicit the generative process (all draws are independent).

Bayesian analysis [edit]

Following classical probability arguments, we consider a big urn containing two children. We assume equal probability that either is a boy or a girl. The 3 discernible cases are thus: 1. both are girls (GG) — with probability P(GG) = one / four , ii. both are boys (BB) — with probability of P(BB) = 1 / 4 , and 3. ane of each (One thousand·B) — with probability of P(Thousand·B) = i / ii . These are the prior probabilities.

Now we add the additional assumption that "at least one is a boy" = B. Using Bayes' Theorem, nosotros detect

${\displaystyle \mathrm {P(BB\mid B)} =\mathrm {P(B\mid BB)\times {\frac {P(BB)}{P(B)}}} =1\times {\frac {\left({\frac {1}{4}}\right)}{\left({\frac {three}{4}}\right)}}={\frac {ane}{iii}}\,.}$

where P(A|B) means "probability of A given B". P(B|BB) = probability of at least one boy given both are boys = ane. P(BB) = probability of both boys = 1 / 4 from the prior distribution. P(B) = probability of at to the lowest degree one beingness a male child, which includes cases BB and G·B = one / four + 1 / 2 = 3 / 4 .

Note that, although the natural assumption seems to exist a probability of 1 / two , so the derived value of 1 / 3 seems low, the actual "normal" value for P(BB) is 1 / 4 , so the 1 / three is actually a fleck college.

The paradox arises because the 2d assumption is somewhat artificial, and when describing the trouble in an bodily setting things get a bit sticky. Just how do nosotros know that "at least" one is a boy? 1 description of the trouble states that we look into a window, see only one child and information technology is a boy. This sounds like the same assumption. Nonetheless, this ane is equivalent to "sampling" the distribution (i.e. removing ane child from the urn, ascertaining that information technology is a boy, so replacing). Permit's call the statement "the sample is a boy" suggestion "b". Now we have:

${\displaystyle \mathrm {P(BB\mid b)} =\mathrm {P(b\mid BB)\times {\frac {P(BB)}{P(b)}}} =i\times {\frac {\left({\frac {1}{4}}\right)}{\left({\frac {1}{2}}\right)}}={\frac {1}{two}}\,.}$

The difference hither is the P(b), which is simply the probability of cartoon a boy from all possible cases (i.e. without the "at least"), which is clearly i / two .

The Bayesian analysis generalizes easily to the case in which nosotros relax the 50:l population supposition. If we take no information almost the populations and then nosotros presume a "flat prior", i.e. P(GG) = P(BB) = P(G·B) = one / 3 . In this instance the "at least" assumption produces the outcome P(BB|B) = 1 / two , and the sampling assumption produces P(BB|b) = 2 / 3 , a effect also derivable from the Rule of Succession.

Martingale analysis [edit]

Suppose one had wagered that Mr. Smith had two boys, and received off-white odds. I pays $1 and they will receive $iv if he has two boys. Their wager volition increase in value as skilful news arrives. What show would brand them happier most their investment? Learning that at least one child out of two is a boy, or learning that at least one child out of 1 is a boy?

The latter is a priori less probable, and therefore better news. That is why the two answers cannot exist the same.

Now for the numbers. If we bet on one child and win, the value of their investment has doubled. It must double once again to get to $iv, so the odds are 1 in 2.

On the other hand if one were learn that at to the lowest degree 1 of 2 children is a boy, the investment increases every bit if they had wagered on this question. Our $i is now worth $1+ i / 3 . To get to $four we nonetheless take to increase our wealth threefold. So the answer is 1 in 3.

Variants of the question [edit]

Following the popularization of the paradox by Gardner it has been presented and discussed in various forms. The get-go variant presented past Bar-Hillel & Falk^[iii] is worded as follows:

• Mr. Smith is the father of ii. We encounter him walking along the street with a young boy whom he proudly introduces as his son. What is the probability that Mr. Smith's other child is as well a boy?

Bar-Hillel & Falk use this variant to highlight the importance of considering the underlying assumptions. The intuitive answer is ane / ii and, when making the about natural assumptions, this is correct. Withal, someone may argue that "…before Mr. Smith identifies the boy as his son, we know only that he is either the father of two boys, BB, or of two girls, GG, or of one of each in either nascence order, i.e., BG or GB. Assuming again independence and equiprobability, we brainstorm with a probability of ane / four that Smith is the father of two boys. Discovering that he has at to the lowest degree one boy rules out the issue GG. Since the remaining three events were equiprobable, we obtain a probability of one / 3 for BB."^[iii]

The natural assumption is that Mr. Smith selected the child companion at random. If and then, as combination BB has twice the probability of either BG or GB of having resulted in the male child walking companion (and combination GG has zilch probability, ruling it out), the union of events BG and GB becomes equiprobable with event BB, and so the chance that the other child is also a boy is i / ii . Bar-Hillel & Falk, however, suggest an culling scenario. They imagine a culture in which boys are invariably chosen over girls as walking companions. In this case, the combinations of BB, BG and GB are assumed equally likely to take resulted in the male child walking companion, and thus the probability that the other child is too a boy is 1 / 3 .

In 1991, Marilyn vos Savant responded to a reader who asked her to respond a variant of the Male child or Girl paradox that included beagles.^[5] In 1996, she published the question again in a different form. The 1991 and 1996 questions, respectively were phrased:

• A shopkeeper says she has two new baby beagles to bear witness yous, but she doesn't know whether they're male, female, or a pair. You tell her that you want only a male, and she telephones the fellow who'southward giving them a bath. "Is at least one a male person?" she asks him. "Aye!" she informs yous with a grin. What is the probability that the other one is a male?
• Say that a woman and a man (who are unrelated) each accept two children. We know that at to the lowest degree one of the adult female's children is a boy and that the man'southward oldest child is a boy. Tin you explain why the chances that the adult female has two boys practice not equal the chances that the man has two boys?

With regard to the 2d formulation Vos Savant gave the classic answer that the chances that the woman has 2 boys are about 1 / three whereas the chances that the human being has two boys are about i / 2 . In response to reader response that questioned her assay vos Savant conducted a survey of readers with exactly 2 children, at least ane of which is a boy. Of 17,946 responses, 35.9% reported two boys.^[ten]

Vos Savant's manufactures were discussed by Carlton and Stansfield^[10] in a 2005 article in The American Statistician. The authors do not discuss the possible ambiguity in the question and conclude that her answer is correct from a mathematical perspective, given the assumptions that the likelihood of a child being a male child or girl is equal, and that the sex of the second child is independent of the kickoff. With regard to her survey they say it "at least validates vos Savant's right assertion that the "chances" posed in the original question, though similar-sounding, are dissimilar, and that the first probability is certainly nearer to one in 3 than to ane in 2."

Carlton and Stansfield go along to discuss the common assumptions in the Boy or Daughter paradox. They demonstrate that in reality male person children are actually more likely than female children, and that the sexual practice of the second kid is non independent of the sexual activity of the first. The authors conclude that, although the assumptions of the question run counter to observations, the paradox all the same has pedagogical value, since it "illustrates ane of the more intriguing applications of conditional probability."^[10] Of course, the actual probability values do not affair; the purpose of the paradox is to demonstrate seemingly contradictory logic, not actual nativity rates.

Data about the child [edit]

Suppose nosotros were told not only that Mr. Smith has two children, and ane of them is a boy, just also that the male child was born on a Tuesday: does this change the previous analyses? Again, the respond depends on how this information was presented - what kind of choice process produced this knowledge.

Following the tradition of the trouble, suppose that in the population of two-child families, the sex of the two children is independent of one another, equally probable boy or girl, and that the nativity engagement of each kid is contained of the other child. The chance of being born on any given day of the week is ane / seven .

From Bayes' Theorem that the probability of two boys, given that 1 male child was born on a Tuesday is given past:

${\displaystyle \mathrm {P(BB\mid B_{T})={\frac {P(B_{T}\mid BB)\times P(BB)}{P(B_{T})}}} }$

Assume that the probability of being born on a Tuesday is ε  = 1 / 7 which will be ready after arriving at the full general solution. The second factor in the numerator is simply i / iv , the probability of having two boys. The first term in the numerator is the probability of at to the lowest degree one boy built-in on Tuesday, given that the family has ii boys, or one − (1 − ε)² (one minus the probability that neither male child is born on Tuesday). For the denominator, let united states decompose: ${\displaystyle \mathrm {P(B_{T})=P(B_{T}\mid BB)P(BB)+P(B_{T}\mid BG)P(BG)+P(B_{T}\mid GB)P(GB)+P(B_{T}\mid GG)P(GG)} }$ . Each term is weighted with probability one / iv . The kickoff term is already known by the previous remark, the concluding term is 0 (at that place are no boys). ${\displaystyle P(B_{T}\mid BG)}$ and ${\displaystyle P(B_{T}\mid GB)}$ is ε, at that place is one and simply one boy, thus he has ε chance of being built-in on Tuesday. Therefore, the full equation is:

${\displaystyle \mathrm {P(BB\mid B_{T})} ={\frac {\left(i-(1-\varepsilon )^{2}\correct)\times {\frac {1}{four}}}{0+{\frac {1}{4}}\varepsilon +{\frac {1}{four}}\varepsilon +{\frac {one}{4}}\left(\varepsilon +\varepsilon -\varepsilon ^{two}\right)}}={\frac {1-(1-\varepsilon )^{2}}{4\varepsilon -\varepsilon ^{2}}}}$

For ${\displaystyle \varepsilon >0}$ , this reduces to ${\displaystyle \mathrm {P(BB\mid B_{T})} ={\frac {two-\varepsilon }{four-\varepsilon }}}$

If ε is now fix to 1 / 7 , the probability becomes xiii / 27 , or about 0.48. In fact, as ε approaches 0, the total probability goes to 1 / two , which is the respond expected when one kid is sampled (e.g. the oldest kid is a boy) and is thus removed from the pool of possible children. In other words, equally more and more than details about the boy child are given (for instance: born on Jan 1), the chance that the other kid is a girl approaches one one-half.

Information technology seems that quite irrelevant data was introduced, notwithstanding the probability of the sex of the other child has inverse dramatically from what information technology was before (the chance the other child was a girl was 2 / 3 , when it was not known that the boy was born on Tuesday).

To empathise why this is, imagine Marilyn vos Savant'southward poll of readers had asked which solar day of the week boys in the family were born. If Marilyn so divided the whole information prepare into seven groups - one for each day of the week a son was built-in - six out of seven families with two boys would be counted in two groups (the group for the twenty-four hours of the week of birth boy one, and the group of the twenty-four hours of the week of nascence for male child 2), doubling, in every group, the probability of a boy-boy combination.

However, is it really plausible that the family with at to the lowest degree one boy born on a Tuesday was produced past choosing simply one of such families at random? It is much more like shooting fish in a barrel to imagine the following scenario.

• We know Mr. Smith has ii children. Nosotros knock at his door and a male child comes and answers the door. Nosotros ask the boy on what 24-hour interval of the week he was built-in.

Assume that which of the two children answers the door is adamant by gamble. So the procedure was (i) choice a 2-child family at random from all ii-child families (2) option one of the ii children at random, (iii) run across if it is a boy and ask on what day he was born. The chance the other child is a girl is ane / 2 . This is a very different procedure from (i) picking a ii-child family at random from all families with two children, at to the lowest degree ane a boy, born on a Tuesday. The chance the family unit consists of a boy and a girl is xiv / 27 , about 0.52.

This variant of the boy and daughter problem is discussed on many internet blogs and is the subject field of a paper past Ruma Falk.^[thirteen] The moral of the story is that these probabilities do not just depend on the known data, but on how that information was obtained.

Psychological investigation [edit]

From the position of statistical analysis the relevant question is often ambiguous and as such there is no "right" answer. Still, this does not exhaust the boy or girl paradox for information technology is not necessarily the ambiguity that explains how the intuitive probability is derived. A survey such as vos Savant's suggests that the bulk of people adopt an agreement of Gardner'south problem that if they were consistent would pb them to the ane / three probability answer but overwhelmingly people intuitively arrive at the ane / 2 probability answer. Ambiguity notwithstanding, this makes the problem of interest to psychological researchers who seek to understand how humans judge probability.

Fox & Levav (2004) used the trouble (called the Mr. Smith trouble, credited to Gardner, but not worded exactly the same as Gardner's version) to test theories of how people approximate conditional probabilities.^[2] In this study, the paradox was posed to participants in two ways:

• "Mr. Smith says: 'I have two children and at least 1 of them is a boy.' Given this information, what is the probability that the other child is a boy?"
• "Mr. Smith says: 'I have ii children and information technology is not the case that they are both girls.' Given this data, what is the probability that both children are boys?"

The authors argue that the first conception gives the reader the mistaken impression that in that location are ii possible outcomes for the "other child",^[2] whereas the 2d conception gives the reader the impression that at that place are 4 possible outcomes, of which one has been rejected (resulting in ane / 3 existence the probability of both children being boys, as there are 3 remaining possible outcomes, only one of which is that both of the children are boys). The study constitute that 85% of participants answered 1 / ii for the first formulation, while only 39% responded that way to the second conception. The authors argued that the reason people respond differently to each question (along with other similar problems, such as the Monty Hall Problem and the Bertrand'southward box paradox) is because of the employ of naive heuristics that neglect to properly ascertain the number of possible outcomes.^[two]

See as well [edit]

• Bertrand paradox (probability)
• Necktie paradox
• Sleeping Dazzler trouble
• Leningrad paradox
• Two envelopes trouble

References [edit]

1. ^ ^{a b} Martin Gardner (1961). The 2nd Scientific American Book of Mathematical Puzzles and Diversions. Simon & Schuster. ISBN978-0-226-28253-4.
2. ^ ^{a b c d due east f g h} Craig R. Fox & Jonathan Levav (2004). "Partition–Edit–Count: Naive Extensional Reasoning in Judgment of Provisional Probability" (PDF). Journal of Experimental Psychology. 133 (4): 626–642. doi:10.1037/0096-3445.133.4.626. PMID 15584810. S2CID 391620. Archived from the original (PDF) on 2020-04-ten.
3. ^ ^{a b c d e} Bar-Hillel, Maya; Falk, Ruma (1982). "Some teasers concerning conditional probabilities". Knowledge. 11 (2): 109–122. doi:ten.1016/0010-0277(82)90021-X. PMID 7198956. S2CID 44509163.
4. ^ ^{a b c} Raymond S. Nickerson (May 2004). Noesis and Chance: The Psychology of Probabilistic Reasoning. Psychology Press. ISBN0-8058-4899-1.
5. ^ ^{a b} "Ask Marilyn". Parade Magazine. October thirteen, 1991 [January 5, 1992; May 26, 1996; December i, 1996; March 30, 1997; July 27, 1997; October nineteen, 1997].
6. ^ Tierney, John (2008-04-x). "The psychology of getting suckered". The New York Times . Retrieved 24 February 2009.
7. ^ ^{a b} Leonard Mlodinow (2008). The Drunkard's Walk: How Randomness Rules our Lives. Pantheon. ISBN978-0-375-42404-5.
8. ^ Nikunj C. Oza (1993). "On The Confusion in Some Pop Probability Problems". CiteSeerXx.1.1.44.2448.
9. ^ P.J. Laird; et al. (1999). "Naive Probability: A Mental Model Theory of Extensional Reasoning". Psychological Review. 106 (1): 62–88. doi:10.1037/0033-295x.106.1.62. PMID 10197363.
10. ^ ^{a b c d e} Matthew A. Carlton and William D. Stansfield (2005). "Making Babies by the Flip of a Coin?". The American Statistician. 59 (ii): 180–182. doi:10.1198/000313005x42813. S2CID 43825948.
11. ^ ^{a b} Charles G. Grinstead and J. Laurie Snell. "Grinstead and Snell's Introduction to Probability" (PDF). The CHANCE Project.
12. ^ ^{a b} Stephen Marks and Gary Smith (Winter 2011). "The Two-Child Paradox Reborn?" (PDF). Run a risk (Mag of the American Statistical Association). 24: 54–ix. doi:10.1007/s00144-011-0010-0. Archived from the original (PDF) on 2016-03-04. Retrieved 2015-01-27 .
13. ^ Falk Ruma (2011). "When truisms clash: Coping with a counterintuitive problem concerning the notorious 2-child family unit". Thinking & Reasoning. 17 (iv): 353–366. doi:ten.1080/13546783.2011.613690. S2CID 145428896.

External links [edit]

• At To the lowest degree One Girl at MathPages
• A Trouble With Ii Bear Cubs
• Lewis Carroll's Pillow Problem
• When intuition and math probably wait wrong

ashleyitth1978.blogspot.com

Source: https://en.wikipedia.org/wiki/Boy_or_Girl_paradox

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