Probability That the Family Has Two Children and Zero Boys

Paradox in probability theory

The Boy or Girl paradox surrounds a prepare of questions in probability theory, which are also known as The Two Child Problem,^[1] Mr. Smith's Children ^[2] and the Mrs. Smith Trouble. 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 information technology The 2 Children Problem, and phrased the paradox as follows:

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

Gardner initially gave the answers 1 / 2 and 1 / 3 , respectively, but later best-selling that the second question was ambiguous.^[1] Its answer could be 1 / ii , depending on the process by which the information "at least ane of them is a boy" was obtained. The ambivalence, depending on the exact diction and possible assumptions, was confirmed by Maya Bar-Hillel and Ruma Falk,^[iii] and Raymond S. Nickerson.^[4]

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

The paradox has stimulated a great bargain of controversy.^[four] The paradox stems from whether the trouble setup is like for the 2 questions.^{[2] [7]} The intuitive answer is ane / 2 .^[2] This answer is intuitive if the question leads the reader to believe that at that place are two every bit likely possibilities for the sex of the second child (i.eastward., boy and daughter),^{[2] [8]} and that the probability of these outcomes is absolute, not conditional.^[9]

Common assumptions [edit]

The two possible answers share a number of assumptions. Commencement, it is assumed that the infinite of all possible events can be hands enumerated, providing an extensional definition of outcomes: {BB, BG, GB, GG}.^[10] This note indicates that at that place are 4 possible combinations of children, labeling boys B and girls G, and using the first letter to represent the older child. Second, information technology is assumed that these outcomes are as 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 same chance of being male person equally of existence female.
3. The sexual practice of each child is contained of the sex of the other.

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

First question [edit]

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

Nether the aforementioned assumptions, in this problem, a random family is selected. In this sample space, there are iv as probable events:

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

Only two of these possible events meet the criteria specified in the question (i.e., GG, GB). Since both of the 2 possibilities in the new sample space {GG, GB} are equally likely, and just ane of the two, GG, includes 2 girls, the probability that the younger child is as well a girl is 1 / 2 .

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 least one of them is a boy. 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 ane is a boy" could pb to the exact same wording of the problem. But they pb to different correct answers:

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

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 process, that yields 1/iii as the answer, is reasonable for the problem as stated above. The formulation of the question they were considering specifically is the post-obit:

• Consider a family with two children. Given that 1 of the children is a boy, what is the probability that both children are boys?

In this formulation the ambiguity is nigh evidently present, considering it is non clear whether we are allowed to assume that a specific child is a boy, leaving the other child uncertain, or whether information technology should be interpreted in the aforementioned way as 'at least one boy'. This ambiguity leaves multiple possibilities that are not equivalent and leaves the necessity to make assumptions about 'how the information was obtained', equally Bar-Hillel and Falk fence, where different assumptions can pb to dissimilar outcomes (because the problem statement was non well enough defined to allow a unmarried straightforward interpretation and answer).

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

Then, while it is certainly true that every possible Mr. Smith has at least one boy (i.e., the condition is necessary), information technology cannot be assumed that every Mr. Smith with at least one boy is intended. That is, the problem statement does not say that having a boy is a sufficient condition for Mr. Smith to be identified every bit having a boy this way.

Commenting on Gardner's version of the problem, Bar-Hillel and Falk^[3] note that "Mr. Smith, unlike the reader, is presumably enlightened of the sex of both of his children when making this statement", i.e. that 'I have two children and at to the lowest degree 1 of them is a boy.' It must exist farther assumed that Mr. Smith would always written report this fact if it were truthful, and either remain silent or say he has at to the lowest degree one daughter, for the correct answer to exist 1 / 3 as Gardner apparently originally intended. But nether that assumption, if he remains silent or says he has a daughter, there is a 100% probability he has 2 daughters.

Analysis of the ambiguity [edit]

If it is assumed that this information was obtained by looking at both children to come across if in that location is at least i boy, the condition is both necessary and sufficient. Three of the iv equally probable events for a two-kid family in the sample space to a higher place meet the status, equally in this table:

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

Thus, if it is assumed that both children were considered while looking for a boy, the answer to question 2 is 1 / 3 . However, if the family was first selected and then a random, true statement was made about the sexual practice of one child in that family, whether or not both were considered, the correct manner to calculate the provisional probability is not 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 made in each case.^[eleven] So, if ALOB represents the result where the argument is "at least 1 boy", and ALOG represents the event where the argument is "at least one girl", so this table describes the sample space:

Older kid	Younger child	P(this family unit)	P(ALOB given this family)	P(ALOG given this family)	P(ALOB and this family)	P(ALOG and this family unit)
Daughter	Girl	1 / 4	0	one	0	1 / 4
Girl	Boy	ane / 4	one / 2	1 / 2	1 / 8	1 / viii
Boy	Girl	1 / 4	1 / two	i / 2	1 / viii	1 / 8
Boy	Boy	1 / 4	1	0	i / four	0

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

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

The paradox occurs when it is not known how the statement "at least one is a boy" was generated. Either answer could exist correct, based on what is causeless.^[12]

However, 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'due south sex is never mentioned although it is present. Every bit Marks and Smith say, "This extreme supposition is never included in the presentation of the 2-kid problem, withal, and is surely not what people have in mind when they nowadays it."^[12]

Modelling the generative procedure [edit]

Some other way to analyse the ambiguity (for question ii) is by making explicit the generative process (all draws are contained).

Bayesian analysis [edit]

Post-obit classical probability arguments, nosotros consider a large urn containing two children. We assume equal probability that either is a boy or a daughter. The three discernible cases are thus: 1. both are girls (GG) — with probability P(GG) = 1 / 4 , 2. both are boys (BB) — with probability of P(BB) = 1 / 4 , and three. 1 of each (G·B) — with probability of P(Thousand·B) = 1 / 2 . These are the prior probabilities.

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

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

where P(A|B) means "probability of A given B". P(B|BB) = probability of at least one boy given both are boys = i. P(BB) = probability of both boys = 1 / 4 from the prior distribution. P(B) = probability of at least one beingness a boy, which includes cases BB and One thousand·B = 1 / 4 + i / ii = 3 / 4 .

Notation that, although the natural assumption seems to be a probability of 1 / ii , so the derived value of ane / 3 seems depression, the actual "normal" value for P(BB) is one / iv , so the i / 3 is actually a bit higher.

The paradox arises because the second assumption is somewhat artificial, and when describing the problem in an actual setting things get a scrap gummy. Just how do nosotros know that "at least" one is a boy? One description of the problem states that we look into a window, see but one kid and it is a boy. This sounds like the same assumption. However, this ane is equivalent to "sampling" the distribution (i.e. removing 1 kid from the urn, ascertaining that it is a male child, then replacing). Let'southward call the statement "the sample is a boy" suggestion "b". Now we accept:

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

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

The Bayesian assay generalizes hands to the case in which we relax the l:50 population assumption. If we take no information almost the populations so we assume a "apartment prior", i.e. P(GG) = P(BB) = P(G·B) = i / three . In this example the "at to the lowest degree" supposition produces the result P(BB|B) = 1 / 2 , and the sampling assumption produces P(BB|b) = 2 / 3 , a consequence too derivable from the Dominion of Succession.

Martingale analysis [edit]

Suppose one had wagered that Mr. Smith had 2 boys, and received fair odds. One pays $1 and they volition receive $iv if he has two boys. Their wager will increase in value as good news arrives. What evidence would make them happier near their investment? Learning that at to the lowest degree one child out of 2 is a boy, or learning that at to the lowest degree one child out of i is a boy?

The latter is a priori less likely, and therefore ameliorate news. That is why the 2 answers cannot exist the aforementioned.

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

On the other hand if one were acquire that at least one of two children is a male child, the investment increases as if they had wagered on this question. Our $i is now worth $1+ i / 3 . To get to $4 we still have to increase our wealth threefold. So the answer is 1 in 3.

Variants of the question [edit]

Following the popularization of the paradox past Gardner it has been presented and discussed in diverse forms. The first variant presented by Bar-Hillel & Falk^[3] is worded as follows:

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

Bar-Hillel & Falk use this variant to highlight the importance of considering the underlying assumptions. The intuitive answer is 1 / 2 and, when making the near natural assumptions, this is correct. However, someone may debate that "…before Mr. Smith identifies the boy as his son, we know only that he is either the begetter of two boys, BB, or of two girls, GG, or of i of each in either nascency order, i.due east., BG or GB. Assuming again independence and equiprobability, we begin with a probability of 1 / iv that Smith is the male parent of two boys. Discovering that he has at least ane boy rules out the event GG. Since the remaining three events were equiprobable, we obtain a probability of ane / iii for BB."^[3]

The natural assumption is that Mr. Smith selected the child companion at random. If and then, every bit combination BB has twice the probability of either BG or GB of having resulted in the boy walking companion (and combination GG has goose egg probability, ruling it out), the union of events BG and GB becomes equiprobable with event BB, and then the take chances that the other kid is also a male child is 1 / 2 . Bar-Hillel & Falk, withal, suggest an alternative scenario. They imagine a culture in which boys are invariably called over girls every bit walking companions. In this case, the combinations of BB, BG and GB are causeless equally probable to have resulted in the boy walking companion, and thus the probability that the other kid is too a boy is 1 / 3 .

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

• A shopkeeper says she has ii new baby beagles to show you, but she doesn't know whether they're male person, female, or a pair. You tell her that you want only a male, and she telephones the fellow who'southward giving them a bathroom. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?
• Say that a adult female and a man (who are unrelated) each have ii children. Nosotros know that at to the lowest degree one of the adult female'southward children is a boy and that the man'southward oldest kid is a male child. Tin can you explicate why the chances that the adult female has two boys do non equal the chances that the human being has 2 boys?

With regard to the second formulation Vos Savant gave the classic reply that the chances that the woman has 2 boys are about 1 / 3 whereas the chances that the man has 2 boys are about 1 / 2 . In response to reader response that questioned her assay vos Savant conducted a survey of readers with exactly 2 children, at to the lowest degree one of which is a boy. Of 17,946 responses, 35.9% reported ii boys.^[ten]

Vos Savant'southward manufactures were discussed by Carlton and Stansfield^[ten] 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 right from a mathematical perspective, given the assumptions that the likelihood of a child being a boy or daughter is equal, and that the sexual practice of the 2d child is contained of the first. With regard to her survey they say it "at least validates vos Savant's right exclamation that the "chances" posed in the original question, though like-sounding, are different, and that the starting time probability is certainly nearer to 1 in 3 than to 1 in ii."

Carlton and Stansfield continue to discuss the common assumptions in the Boy or Girl paradox. They demonstrate that in reality male children are actually more likely than female person children, and that the sex of the second child is non independent of the sex of the first. 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 than intriguing applications of conditional probability."^[10] Of form, the actual probability values practice not matter; the purpose of the paradox is to demonstrate seemingly contradictory logic, not actual birth rates.

Information well-nigh the child [edit]

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

Following the tradition of the problem, suppose that in the population of two-kid families, the sex of the two children is independent of 1 another, equally likely male child or girl, and that the birth engagement of each child is independent of the other kid. The adventure of being built-in on whatsoever given day of the week is 1 / 7 .

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

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

Presume that the probability of being born on a Tuesday is ε  = 1 / 7 which will be prepare later arriving at the general solution. The 2d factor in the numerator is simply 1 / 4 , the probability of having two boys. The kickoff term in the numerator is the probability of at least one boy born on Tuesday, given that the family has two boys, or i − (ane − ε)² (one minus the probability that neither boy is built-in on Tuesday). For the denominator, permit 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 1 / four . The first term is already known by the previous remark, the terminal 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 merely one boy, thus he has ε chance of existence born on Tuesday. Therefore, the full equation is:

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

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

If ε is at present set to ane / 7 , the probability becomes xiii / 27 , or about 0.48. In fact, as ε approaches 0, the total probability goes to 1 / 2 , 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 and more details about the boy child are given (for instance: born on January 1), the chance that the other child is a daughter approaches 1 half.

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

To understand why this is, imagine Marilyn vos Savant's poll of readers had asked which day of the week boys in the family were born. If Marilyn then divided the whole data set into vii groups - 1 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 day of the week of birth boy ane, and the grouping of the mean solar day of the week of nascency for male child 2), doubling, in every grouping, the probability of a boy-boy combination.

However, is it actually plausible that the family with at least one male child born on a Tuesday was produced by choosing just one of such families at random? It is much more easy 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 hours of the calendar week he was born.

Assume that which of the two children answers the door is adamant past take chances. Then the procedure was (i) pick a two-child family at random from all two-child families (ii) pick i of the 2 children at random, (3) see if it is a male child and ask on what day he was born. The risk the other child is a girl is one / ii . This is a very different procedure from (1) picking a ii-child family at random from all families with two children, at least one a boy, born on a Tuesday. The chance the family consists of a boy and a girl is 14 / 27 , about 0.52.

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

Psychological investigation [edit]

From the position of statistical analysis the relevant question is ofttimes cryptic and every bit such there is no "right" respond. However, this does non exhaust the boy or girl paradox for it is not necessarily the ambiguity that explains how the intuitive probability is derived. A survey such as vos Savant's suggests that the majority of people adopt an understanding of Gardner's problem that if they were consistent would lead them to the 1 / 3 probability answer but overwhelmingly people intuitively make it at the 1 / 2 probability answer. Ambiguity withal, this makes the problem of involvement to psychological researchers who seek to sympathize how humans gauge probability.

Fox & Levav (2004) used the problem (called the Mr. Smith trouble, credited to Gardner, only not 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 means:

• "Mr. Smith says: 'I have 2 children and at least one of them is a boy.' Given this data, what is the probability that the other kid is a boy?"
• "Mr. Smith says: 'I take ii children and information technology is not the case that they are both girls.' Given this information, what is the probability that both children are boys?"

The authors contend that the offset formulation 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 four possible outcomes, of which one has been rejected (resulting in 1 / iii being the probability of both children being boys, as there are 3 remaining possible outcomes, just one of which is that both of the children are boys). The written report found that 85% of participants answered 1 / 2 for the first formulation, while but 39% responded that way to the 2d formulation. 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's box paradox) is considering of the utilise of naive heuristics that fail to properly define the number of possible outcomes.^[2]

See also [edit]

• Bertrand paradox (probability)
• Necktie paradox
• Sleeping Beauty problem
• St. Petersburg paradox
• Two envelopes problem

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 chiliad h} Craig R. Fox & Jonathan Levav (2004). "Partition–Edit–Count: Naive Extensional Reasoning in Judgment of Conditional Probability" (PDF). Journal of Experimental Psychology. 133 (four): 626–642. doi:10.1037/0096-3445.133.4.626. PMID 15584810. S2CID 391620. Archived from the original (PDF) on 2020-04-x.
3. ^ ^{a b c d due east} Bar-Hillel, Maya; Falk, Ruma (1982). "Some teasers apropos provisional probabilities". Cognition. 11 (two): 109–122. doi:10.1016/0010-0277(82)90021-X. PMID 7198956. S2CID 44509163.
4. ^ ^{a b c} Raymond S. Nickerson (May 2004). Knowledge and Adventure: The Psychology of Probabilistic Reasoning. Psychology Printing. ISBN0-8058-4899-1.
5. ^ ^{a b} "Ask Marilyn". Parade Magazine. October 13, 1991 [January 5, 1992; May 26, 1996; December i, 1996; March 30, 1997; July 27, 1997; Oct 19, 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 Drunk'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 Bug". CiteSeerX10.1.1.44.2448.
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 (2): 180–182. doi:x.1198/000313005x42813. S2CID 43825948.
11. ^ ^{a b} Charles M. Grinstead and J. Laurie Snell. "Grinstead and Snell's Introduction to Probability" (PDF). The Take a chance Project.
12. ^ ^{a b} Stephen Marks and Gary Smith (Winter 2011). "The Two-Kid Paradox Reborn?" (PDF). Chance (Magazine of the American Statistical Association). 24: 54–9. doi:x.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 two-kid family unit". Thinking & Reasoning. 17 (four): 353–366. doi:x.1080/13546783.2011.613690. S2CID 145428896.

External links [edit]

• At Least 1 Girl at MathPages
• A Problem With Two Comport Cubs
• Lewis Carroll's Pillow Trouble
• When intuition and math probably look wrong

onealboymor50.blogspot.com

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