If Baby Rests on Left Side Is It a Girl or Boy

Paradox in probability theory

The Boy or Girl paradox surrounds a set of questions in probability theory, which are also known as The 2 Child Problem,^[1] Mr. Smith's Children ^[2] and the Mrs. Smith Problem. The initial formulation of the question dates dorsum to at to the lowest degree 1959, when Martin Gardner featured it in his October 1959 "Mathematical Games column" in Scientific American. He titled it The Two Children Problem, and phrased the paradox as follows:

• Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?
• Mr. Smith has two children. At to the lowest degree one of them is a boy. What is the probability that both children are boys?

Gardner initially gave the answers i / 2 and 1 / 3 , respectively, but later best-selling that the 2nd question was ambiguous.^[1] Its answer could be one / 2 , depending on the procedure by which the information "at least one of them is a boy" was obtained. The ambiguity, depending on the exact wording and possible assumptions, was confirmed by Maya Bar-Hillel and Ruma Falk,^[3] and Raymond Southward. Nickerson.^[4]

Other variants of this question, with varying degrees of ambivalence, have been popularized by Ask Marilyn in Parade Magazine,^[v] John Tierney of The New York Times,^[6] and Leonard Mlodinow in The Boozer's Walk.^[7] One scientific study showed that when identical information was conveyed, but with different partially ambiguous wordings that emphasized different points, that the percentage of MBA students who answered 1 / 2 changed from 85% to 39%.^[2]

The paradox has stimulated a great bargain of controversy.^[four] The paradox stems from whether the problem setup is similar for the two questions.^{[ii] [7]} The intuitive answer is ane / two .^[2] This answer is intuitive if the question leads the reader to believe that at that place are 2 every bit likely possibilities for the sex activity of the second child (i.e., boy and daughter),^{[2] [viii]} and that the probability of these outcomes is absolute, not conditional.^[9]

Common assumptions [edit]

The two possible answers share a number of assumptions. First, it is causeless that the space of all possible events can be easily enumerated, providing an extensional definition of outcomes: {BB, BG, GB, GG}.^[10] This annotation indicates that at that place are four possible combinations of children, labeling boys B and girls G, and using the first letter to represent the older kid. 2nd, it is causeless that these outcomes are equally probable.^[10] This implies the following model, a Bernoulli process with p = i / 2 :

1. Each kid is either male or female.
2. Each child has the same chance of being male as of being female person.
3. The sexual practice of each child is independent of the sex activity of the other.

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

Outset question [edit]

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

Under the aforementioned assumptions, in this trouble, a random family is selected. In this sample infinite, there are 4 equally likely events:

Older child	Younger kid
Girl	Girl
Girl	Boy
Boy	Girl
Boy	Boy

Merely 2 of these possible events encounter the criteria specified in the question (i.eastward., GG, GB). Since both of the two possibilities in the new sample space {GG, GB} are equally likely, and just one of the two, GG, includes two girls, the probability that the younger child is also a daughter is 1 / ii .

Second question [edit]

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

This question is identical to question one, except that instead of specifying that the older child is a boy, it is specified that at to the lowest degree i of them is a male child. In response to reader criticism of the question posed in 1959, Gardner said that no answer is possible without data that was not provided. Specifically, that two different procedures for determining that "at least one is a boy" could atomic number 82 to the exact same wording of the problem. Just they lead to unlike correct answers:

• From all families with two children, at least 1 of whom is a boy, a family is chosen at random. This would yield the reply of 1 / three .
• From all families with two children, ane child is selected at random, and the sex of that child is specified to exist a boy. This would yield an answer of 1 / two .^{[three] [four]}

Grinstead and Snell argue that the question is ambiguous in much the same way Gardner did.^[11] They leave it to the reader to decide whether the procedure, that yields 1/iii as the answer, is reasonable for the problem every bit stated above. The conception of the question they were considering specifically is the post-obit:

• Consider a family unit with 2 children. Given that one of the children is a male child, what is the probability that both children are boys?

In this conception the ambiguity is most obviously nowadays, because it is not clear whether we are allowed to assume that a specific child is a boy, leaving the other child uncertain, or whether it should be interpreted in the same way as 'at to the lowest degree 1 boy'. This ambiguity leaves multiple possibilities that are not equivalent and leaves the necessity to make assumptions about 'how the information was obtained', as Bar-Hillel and Falk debate, where unlike assumptions can lead to different outcomes (because the trouble statement was non well enough defined to let a single straightforward interpretation and reply).

For example, 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 male child. But if he has a male child and a daughter, that kid could take been a girl. So seeing him with a boy eliminates not but the combinations where he has two girls, simply also 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 one male child (i.e., the condition is necessary), information technology cannot be assumed that every Mr. Smith with at to the lowest degree one boy is intended. That is, the trouble statement does not say that having a boy is a sufficient condition for Mr. Smith to be identified as having a male child this fashion.

Commenting on Gardner'due south version of the problem, Bar-Hillel and Falk^[3] notation that "Mr. Smith, different the reader, is presumably aware of the sexual practice of both of his children when making this statement", i.e. that 'I have two children and at to the lowest degree one of them is a boy.' It must exist further assumed that Mr. Smith would always report this fact if it were truthful, and either remain silent or say he has at least i daughter, for the correct respond to exist 1 / iii equally Gardner plain originally intended. Only nether that assumption, if he remains silent or says he has a daughter, there is a 100% probability he has ii daughters.

Analysis of the ambiguity [edit]

If it is assumed that this information was obtained by looking at both children to see if at that place is at least one boy, the condition is both necessary and sufficient. Three of the four equally probable events for a two-child family in the sample space above run into the condition, as in this table:

Older child	Younger kid
Girl	Girl
Girl	Boy
Boy	Girl
Boy	Boy

Thus, if information technology is assumed that both children were considered while looking for a boy, the answer to question 2 is 1 / three . Yet, if the family unit was commencement selected and then a random, true argument was fabricated about the sex of i kid in that family, whether or non both were considered, the correct way to calculate the conditional probability is non to count all of the cases that include a child with that sex. Instead, 1 must consider only the probabilities where the statement volition be fabricated in each case.^[xi] So, if ALOB represents the event where the statement is "at least i boy", and ALOG represents the consequence where the statement is "at least 1 girl", so 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	1 / iv	0	i	0	ane / iv
Girl	Boy	ane / 4	1 / 2	1 / 2	one / eight	1 / 8
Boy	Girl	one / iv	i / 2	i / 2	1 / 8	i / 8
Boy	Boy	1 / 4	1	0	1 / 4	0

So, if at least i is a male child when the fact is called randomly, the probability that both are boys is

${\displaystyle \mathrm {\frac {P(ALOB\;and\;BB)}{P(ALOB)}} ={\frac {\frac {1}{4}}{0+{\frac {1}{eight}}+{\frac {i}{eight}}+{\frac {1}{4}}}}={\frac {1}{ii}}\,.}$

The paradox occurs when it is not known how the argument "at least ane is a male child" was generated. Either reply could be right, based on what is causeless.^[12]

Yet, the " i / 3 " answer is obtained only by assuming P(ALOB|BG) = P(ALOB|GB) =1, which implies P(ALOG|BG) = P(ALOG|GB) = 0, that is, the other child's sexual practice is never mentioned although it is nowadays. As Marks and Smith say, "This extreme supposition is never included in the presentation of the two-child trouble, still, and is surely not what people have in mind when they present it."^[12]

Modelling the generative process [edit]

Another fashion to analyse the ambiguity (for question 2) is by making explicit the generative process (all draws are independent).

Bayesian analysis [edit]

Following classical probability arguments, we consider a large urn containing two children. Nosotros assume equal probability that either is a male child or a girl. The three discernible cases are thus: 1. both are girls (GG) — with probability P(GG) = ane / 4 , 2. both are boys (BB) — with probability of P(BB) = 1 / 4 , and 3. i of each (G·B) — with probability of P(Thousand·B) = 1 / ii . These are the prior probabilities.

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

${\displaystyle \mathrm {P(BB\mid B)} =\mathrm {P(B\mid BB)\times {\frac {P(BB)}{P(B)}}} =1\times {\frac {\left({\frac {1}{4}}\right)}{\left({\frac {3}{iv}}\right)}}={\frac {one}{3}}\,.}$

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

Note that, although the natural assumption seems to be a probability of 1 / 2 , and so the derived value of 1 / 3 seems depression, the actual "normal" value for P(BB) is one / iv , so the i / three is really a flake higher.

The paradox arises because the second supposition is somewhat artificial, and when describing the problem in an bodily setting things get a bit sticky. Just how do we know that "at least" one is a boy? I description of the trouble states that we look into a window, encounter only one child and it is a boy. This sounds like the same assumption. However, this ane is equivalent to "sampling" the distribution (i.e. removing one kid from the urn, ascertaining that it is a boy, then replacing). Permit'southward call the argument "the sample is a boy" proposition "b". Now we have:

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

The difference here is the P(b), which is just the probability of cartoon a boy from all possible cases (i.e. without the "at least"), which is clearly 1 / 2 .

The Bayesian analysis generalizes easily to the example in which nosotros relax the 50:50 population assumption. If nosotros have no information about the populations then nosotros assume a "flat prior", i.eastward. P(GG) = P(BB) = P(G·B) = 1 / 3 . In this instance the "at least" assumption produces the result P(BB|B) = 1 / 2 , and the sampling assumption produces P(BB|b) = 2 / iii , a result also derivable from the Rule of Succession.

Martingale assay [edit]

Suppose one had wagered that Mr. Smith had two boys, and received off-white odds. Ane pays $1 and they will receive $iv if he has two boys. Their wager will increase in value equally good news arrives. What evidence would brand them happier nearly their investment? Learning that at to the lowest degree one child out of two is a male child, or learning that at to the lowest degree one kid out of one is a boy?

The latter is a priori less likely, and therefore better news. That is why the ii answers cannot be the same.

Now for the numbers. If we bet on ane child and win, the value of their investment has doubled. It must double over again to get to $4, so the odds are 1 in 2.

On the other hand if 1 were larn that at to the lowest degree one of two children is a boy, the investment increases as if they had wagered on this question. Our $1 is now worth $1+ 1 / 3 . To become to $4 we all the same have to increase our wealth threefold. Then 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 diverse forms. The showtime variant presented by Bar-Hillel & Falk^[3] is worded as follows:

• Mr. Smith is the father of two. We meet him walking along the street with a young boy whom he proudly introduces every bit his son. What is the probability that Mr. Smith'southward other child is also a male child?

Bar-Hillel & Falk use this variant to highlight the importance of considering the underlying assumptions. The intuitive answer is ane / 2 and, when making the most natural assumptions, this is correct. Still, 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 2 girls, GG, or of one of each in either nascency order, i.e., BG or GB. Assuming once more independence and equiprobability, we begin with a probability of 1 / 4 that Smith is the father of ii boys. Discovering that he has at to the lowest degree one boy rules out the event GG. Since the remaining three events were equiprobable, we obtain a probability of i / three for BB."^[3]

The natural assumption is that Mr. Smith selected the child companion at random. If so, as combination BB has twice the probability of either BG or GB of having resulted in the boy walking companion (and combination GG has null probability, ruling it out), the matrimony of events BG and GB becomes equiprobable with result BB, so the chance that the other child is likewise a boy is 1 / two . Bar-Hillel & Falk, however, propose 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 as probable to have resulted in the boy walking companion, and thus the probability that the other child is also a boy is ane / 3 .

In 1991, Marilyn vos Savant responded to a reader who asked her to answer a variant of the Boy or Daughter 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 show yous, but she doesn't know whether they're male person, female, or a pair. Y'all tell her that you want merely a male person, and she telephones the swain who's giving them a bath. "Is at least ane a male person?" she asks him. "Aye!" she informs you with a grinning. What is the probability that the other one is a male?
• Say that a woman and a human (who are unrelated) each have 2 children. Nosotros know that at least one of the adult female's children is a boy and that the homo's oldest child is a male child. Can you explain why the chances that the woman has 2 boys practice not equal the chances that the man has ii boys?

With regard to the second formulation Vos Savant gave the classic reply that the chances that the adult female has ii boys are about 1 / 3 whereas the chances that the man has ii boys are about i / 2 . In response to reader response that questioned her analysis vos Savant conducted a survey of readers with exactly two children, at least i of which is a male child. Of 17,946 responses, 35.9% reported two boys.^[ten]

Vos Savant'south articles were discussed by Carlton and Stansfield^[10] in a 2005 article in The American Statistician. The authors practise not discuss the possible ambivalence in the question and conclude that her reply is correct from a mathematical perspective, given the assumptions that the likelihood of a child being a boy or girl is equal, and that the sex of the 2d child is independent of the first. 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 like-sounding, are different, and that the kickoff probability is certainly nearer to 1 in three than to 1 in ii."

Carlton and Stansfield continue to discuss the common assumptions in the Boy or Daughter paradox. They demonstrate that in reality male children are actually more probable than female children, and that the sex of the 2d child is not independent of the sex of the start. The authors conclude that, although the assumptions of the question run counter to observations, the paradox still has pedagogical value, since it "illustrates one of the more intriguing applications of conditional probability."^[10] Of class, the bodily probability values do not affair; the purpose of the paradox is to demonstrate seemingly contradictory logic, not actual birth rates.

Information about the child [edit]

Suppose nosotros were told not only that Mr. Smith has 2 children, and i of them is a boy, merely also that the boy was born on a Tuesday: does this change the previous analyses? Again, the answer depends on how this information was presented - what kind of selection process produced this knowledge.

Following the tradition of the problem, suppose that in the population of ii-child families, the sexual activity of the two children is contained of ane another, equally probable boy or girl, and that the birth date of each child is independent of the other child. The take chances of being born on whatever given day of the week is 1 / seven .

From Bayes' Theorem that the probability of two boys, given that one boy 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 built-in on a Tuesday is ε  = 1 / vii which will be set after arriving at the full general solution. The second factor in the numerator is simply one / iv , the probability of having ii boys. The offset term in the numerator is the probability of at least ane male child born on Tuesday, given that the family unit has ii boys, or ane − (1 − ε)² (one minus the probability that neither boy is built-in on Tuesday). For the denominator, let us 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 i / 4 . The first term is already known by the previous remark, the last term is 0 (there are no boys). ${\displaystyle P(B_{T}\mid BG)}$ and ${\displaystyle P(B_{T}\mid GB)}$ is ε, there is one and but one boy, thus he has ε chance of being born on Tuesday. Therefore, the total equation is:

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

For ${\displaystyle \varepsilon >0}$ , this reduces to ${\displaystyle \mathrm {P(BB\mid B_{T})} ={\frac {ii-\varepsilon }{4-\varepsilon }}}$

If ε is now set to 1 / 7 , the probability becomes 13 / 27 , or most 0.48. In fact, equally ε approaches 0, the full probability goes to 1 / ii , which is the answer expected when one child is sampled (e.g. the oldest child is a boy) and is thus removed from the pool of possible children. In other words, as more than and more than details near the boy kid are given (for instance: born on January 1), the chance that the other child is a girl approaches one half.

Information technology seems that quite irrelevant information was introduced, nonetheless the probability of the sexual practice of the other kid has changed dramatically from what it was before (the take chances the other child was a girl was 2 / 3 , when it was not known that the male child was born on Tuesday).

To understand why this is, imagine Marilyn vos Savant'south poll of readers had asked which day of the calendar week boys in the family were born. If Marilyn then divided the whole data set into seven groups - one for each day of the week a son was born - six out of vii families with two boys would exist counted in two groups (the group for the twenty-four hours of the calendar week of birth boy 1, and the group of the day of the calendar week of birth for boy 2), doubling, in every group, the probability of a boy-boy combination.

However, is information technology really plausible that the family with at least i boy born on a Tuesday was produced by choosing just one of such families at random? It is much more piece of cake to imagine the following scenario.

• We know Mr. Smith has two children. We knock at his door and a male child comes and answers the door. We ask the boy on what twenty-four hour period of the week he was built-in.

Presume that which of the ii children answers the door is determined past hazard. And so the procedure was (1) pick a two-kid family unit at random from all ii-child families (two) pick one of the 2 children at random, (three) see if it is a boy and ask on what day he was built-in. The chance the other child is a girl is 1 / 2 . This is a very unlike procedure from (1) picking a ii-child family at random from all families with two children, at least i a boy, born on a Tuesday. The take a chance the family consists of a boy and a girl is 14 / 27 , about 0.52.

This variant of the male child and girl problem is discussed on many internet blogs and is the subject field of a newspaper by Ruma Falk.^[13] The moral of the story is that these probabilities exercise not just depend on the known information, but on how that information was obtained.

Psychological investigation [edit]

From the position of statistical assay the relevant question is often ambiguous and as such there is no "right" answer. However, this does not exhaust the boy or girl paradox for it is not necessarily the ambivalence that explains how the intuitive probability is derived. A survey such every bit vos Savant'due south suggests that the majority of people adopt an understanding of Gardner'southward problem that if they were consistent would atomic number 82 them to the ane / 3 probability answer simply overwhelmingly people intuitively make it at the i / 2 probability answer. Ambiguity notwithstanding, this makes the problem of interest to psychological researchers who seek to empathise how humans estimate probability.

Play a trick on & Levav (2004) used the trouble (called the Mr. Smith problem, credited to Gardner, but non worded exactly the aforementioned as Gardner'south version) to test theories of how people estimate conditional probabilities.^[2] In this study, the paradox was posed to participants in two ways:

• "Mr. Smith says: 'I take two children and at least one of them is a boy.' Given this information, what is the probability that the other child is a boy?"
• "Mr. Smith says: 'I have two children and it is non the case that they are both girls.' Given this information, what is the probability that both children are boys?"

The authors argue that the first conception gives the reader the mistaken impression that there are 2 possible outcomes for the "other child",^[2] whereas the second formulation gives the reader the impression that at that place are 4 possible outcomes, of which one has been rejected (resulting in 1 / three being the probability of both children being boys, every bit there are 3 remaining possible outcomes, merely one of which is that both of the children are boys). The study constitute that 85% of participants answered i / 2 for the offset formulation, while only 39% responded that way to the 2d formulation. The authors argued that the reason people respond differently to each question (forth with other like problems, such as the Monty Hall Problem and the Bertrand's box paradox) is because of the utilise of naive heuristics that fail to properly define the number of possible outcomes.^[2]

See also [edit]

• Bertrand paradox (probability)
• Tie paradox
• Sleeping Beauty problem
• St. Petersburg paradox
• 2 envelopes problem

References [edit]

1. ^ ^{a b} Martin Gardner (1961). The Second Scientific American Book of Mathematical Puzzles and Diversions. Simon & Schuster. ISBN978-0-226-28253-4.
2. ^ ^{a b c d e f g h} Craig R. Fox & Jonathan Levav (2004). "Partition–Edit–Count: Naive Extensional Reasoning in Judgment of Conditional Probability" (PDF). Journal of Experimental Psychology. 133 (4): 626–642. doi:ten.1037/0096-3445.133.iv.626. PMID 15584810. S2CID 391620. Archived from the original (PDF) on 2020-04-10.
3. ^ ^{a b c d e} Bar-Hillel, Maya; Falk, Ruma (1982). "Some teasers concerning provisional probabilities". Cognition. xi (2): 109–122. doi:10.1016/0010-0277(82)90021-Ten. PMID 7198956. S2CID 44509163.
4. ^ ^{a b c} Raymond South. Nickerson (May 2004). Cognition and Chance: The Psychology of Probabilistic Reasoning. Psychology Press. ISBN0-8058-4899-ane.
5. ^ ^{a b} "Ask Marilyn". Parade Magazine. October thirteen, 1991 [January five, 1992; May 26, 1996; December 1, 1996; March 30, 1997; July 27, 1997; Oct nineteen, 1997].
6. ^ Tierney, John (2008-04-10). "The psychology of getting suckered". The New York Times . Retrieved 24 February 2009.
7. ^ ^{a b} Leonard Mlodinow (2008). The Boozer's Walk: How Randomness Rules our Lives. Pantheon. ISBN978-0-375-42404-5.
8. ^ Nikunj C. Oza (1993). "On The Defoliation in Some Popular Probability Problems". CiteSeerXten.i.1.44.2448.
9. ^ P.J. Laird; et al. (1999). "Naive Probability: A Mental Model Theory of Extensional Reasoning". Psychological Review. 106 (one): 62–88. doi:ten.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 (2): 180–182. doi:10.1198/000313005x42813. S2CID 43825948.
11. ^ ^{a b} Charles M. Grinstead and J. Laurie Snell. "Grinstead and Snell'south Introduction to Probability" (PDF). The Take chances Projection.
12. ^ ^{a b} Stephen Marks and Gary Smith (Winter 2011). "The Two-Child Paradox Reborn?" (PDF). Chance (Magazine of the American Statistical Association). 24: 54–9. 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 disharmonism: Coping with a counterintuitive problem concerning the notorious two-child family unit". Thinking & Reasoning. 17 (4): 353–366. doi:10.1080/13546783.2011.613690. S2CID 145428896.

External links [edit]

• At Least One Daughter at MathPages
• A Problem With Two Conduct Cubs
• Lewis Carroll's Pillow Problem
• When intuition and math probably expect wrong

reynoldstwok1976.blogspot.com

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

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