The First Child Is a Boy or That the Last Two Children of the Family Are Girls

Paradox in probability theory

The Boy or Daughter paradox surrounds a prepare of questions in probability theory, which are besides known every bit The Two Child Problem,^[1] Mr. Smith'southward Children ^[2] and the Mrs. Smith Problem. The initial formulation of the question dates back to at least 1959, when Martin Gardner featured information technology in his Oct 1959 "Mathematical Games column" in Scientific American. He titled it The Two Children Problem, and phrased the paradox equally 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 least 1 of them is a boy. What is the probability that both children are boys?

Gardner initially gave the answers 1 / 2 and 1 / 3 , respectively, but subsequently acknowledged that the second question was ambiguous.^[1] Its reply could be i / 2 , depending on the procedure past which the information "at to the lowest degree 1 of them is a male child" was obtained. The ambivalence, depending on the exact wording and possible assumptions, was confirmed by Maya Bar-Hillel and Ruma Falk,^[3] and Raymond South. Nickerson.^[4]

Other variants of this question, with varying degrees of ambiguity, have been popularized by Ask Marilyn in Parade Magazine,^[5] John Tierney of The New York Times,^[6] and Leonard Mlodinow in The Drunk's Walk.^[vii] Ane 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 ane / two inverse from 85% to 39%.^[two]

The paradox has stimulated a great deal of controversy.^[iv] The paradox stems from whether the problem setup is like for the two questions.^{[two] [7]} The intuitive answer is 1 / 2 .^[2] This answer is intuitive if the question leads the reader to believe that in that location are ii equally probable possibilities for the sex of the second child (i.e., boy and girl),^{[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. First, it 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 note indicates that there are four possible combinations of children, labeling boys B and girls M, and using the first letter to represent the older child. Second, it is assumed that these outcomes are as probable.^[x] This implies the post-obit model, a Bernoulli process with p = 1 / 2 :

1. Each child is either male or female.
2. Each child has the aforementioned chance of being male equally of being female.
3. The sex of each kid is contained of the sex of the other.

The mathematical consequence would be 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 girl. What is the probability that both children are girls?

Nether the aforementioned assumptions, in this trouble, a random family is selected. In this sample space, there are four equally probable events:

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

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

Second question [edit]

• Mr. Smith has 2 children. At least one 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 child is a male child, it is specified that at least one 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 information that was not provided. Specifically, that two dissimilar procedures for determining that "at least 1 is a boy" could atomic number 82 to the exact same wording of the problem. But they pb to different correct answers:

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

Grinstead and Snell contend that the question is ambiguous in much the aforementioned manner Gardner did.^[xi] They leave it to the reader to decide whether the process, that yields i/3 as 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 two 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 most evidently present, because it is non clear whether we are allowed to presume that a specific kid is a boy, leaving the other child uncertain, or whether it should be interpreted in the same way as 'at least i male child'. This ambivalence leaves multiple possibilities that are not equivalent and leaves the necessity to make assumptions virtually 'how the information was obtained', as Bar-Hillel and Falk argue, where different assumptions can atomic number 82 to unlike outcomes (because the problem statement was non well enough defined to allow a single straightforward interpretation and respond).

For instance, say an observer sees Mr. Smith on a walk with but i of his children. If he has 2 boys and then that kid must exist a boy. But if he has a boy and a girl, that child could have been a girl. And so seeing him with a boy eliminates not simply the combinations where he has two girls, but too 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.eastward., the condition is necessary), it cannot be assumed that every Mr. Smith with at to the lowest degree i 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 equally having a boy this way.

Commenting on Gardner'due south version of the problem, Bar-Hillel and Falk^[three] annotation that "Mr. Smith, different the reader, is presumably aware of the sex of both of his children when making this statement", i.eastward. that 'I have two children and at least one of them is a boy.' Information technology must exist further assumed that Mr. Smith would always report this fact if it were true, and either remain silent or say he has at least ane girl, 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 two daughters.

Analysis of the ambiguity [edit]

If it is assumed that this information was obtained by looking at both children to encounter if in that location is at least one male child, the status is both necessary and sufficient. Three of the iv equally probable events for a two-child family in the sample infinite to a higher place run across the status, as in this tabular array:

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

Thus, if information technology is causeless that both children were considered while looking for a boy, the reply to question 2 is i / 3 . Still, if the family was kickoff selected and then a random, true statement was made near the sex of ane child in that family, whether or not both were considered, the right mode to calculate the conditional probability is not to count all of the cases that include a kid with that sexual practice. Instead, i must consider only the probabilities where the statement will be made in each example.^[11] And so, if ALOB represents the event where the statement is "at to the lowest degree one boy", and ALOG represents the upshot where the statement is "at to the lowest degree 1 girl", and then this table describes the sample space:

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

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

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

The paradox occurs when it is not known how the argument "at least 1 is a boy" was generated. Either respond could be correct, based on what is causeless.^[12]

However, the " one / 3 " reply is obtained only by bold P(ALOB|BG) = P(ALOB|GB) =i, which implies P(ALOG|BG) = P(ALOG|GB) = 0, that is, the other child's sex is never mentioned although it is present. As Marks and Smith say, "This extreme assumption is never included in the presentation of the two-kid problem, yet, and is surely not what people have in mind when they present it."^[12]

Modelling the generative process [edit]

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

Bayesian assay [edit]

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

Now nosotros add the boosted supposition that "at least ane is a boy" = B. Using Bayes' Theorem, we notice

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

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

Annotation that, although the natural assumption seems to be a probability of 1 / 2 , then the derived value of 1 / 3 seems low, the bodily "normal" value for P(BB) is ane / four , so the 1 / 3 is actually a bit higher.

The paradox arises because the second supposition is somewhat artificial, and when describing the problem in an actual setting things get a scrap pasty. But how do nosotros know that "at least" one is a boy? I clarification of the trouble states that we look into a window, see but one child and it is a male child. This sounds like the same assumption. However, this ane is equivalent to "sampling" the distribution (i.eastward. removing one child from the urn, ascertaining that information technology is a boy, then replacing). Allow's phone call the argument "the sample is a male child" proposition "b". Now we accept:

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

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

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

Martingale assay [edit]

Suppose i had wagered that Mr. Smith had two boys, and received fair odds. I pays $1 and they will receive $4 if he has 2 boys. Their wager will increase in value as practiced news arrives. What testify would make them happier almost their investment? Learning that at to the lowest degree one kid out of two is a male child, or learning that at least one child out of one is a boy?

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

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 $4, then the odds are ane in 2.

On the other hand if one were acquire that at least i of two children is a boy, the investment increases equally if they had wagered on this question. Our $one is now worth $ane+ ane / 3 . To get to $4 we however have to increase our wealth threefold. So the respond is one 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 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 as his son. What is the probability that Mr. Smith'south other child is likewise a boy?

Bar-Hillel & Falk use this variant to highlight the importance of considering the underlying assumptions. The intuitive respond is 1 / 2 and, when making the nigh natural assumptions, this is correct. Even so, someone may argue that "…before Mr. Smith identifies the boy every bit his son, we know only that he is either the father of 2 boys, BB, or of 2 girls, GG, or of one of each in either birth lodge, i.e., BG or GB. Bold once more independence and equiprobability, we brainstorm with a probability of 1 / 4 that Smith is the father of two boys. Discovering that he has at least one boy rules out the event GG. Since the remaining three events were equiprobable, we obtain a probability of ane / 3 for BB."^[3]

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 boy walking companion (and combination GG has zero probability, ruling it out), the wedlock of events BG and GB becomes equiprobable with event BB, and so the hazard that the other child is also a boy is i / 2 . Bar-Hillel & Falk, notwithstanding, suggest an alternative 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 causeless equally likely to have resulted in the boy walking companion, and thus the probability that the other child is also a boy is 1 / 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 over again in a different form. The 1991 and 1996 questions, respectively were phrased:

• A shopkeeper says she has two new infant beagles to show you, but she doesn't know whether they're male, female, or a pair. You tell her that y'all want simply a male, and she telephones the boyfriend who's giving them a bathroom. "Is at to the lowest degree one a male?" 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 have two children. We know that at least one of the adult female's children is a male child and that the man'southward oldest child is a male child. Tin can you explain why the chances that the woman has two boys do not equal the chances that the man has ii boys?

With regard to the second formulation Vos Savant gave the classic answer that the chances that the woman has 2 boys are virtually ane / three whereas the chances that the man has two boys are about 1 / 2 . In response to reader response that questioned her analysis vos Savant conducted a survey of readers with exactly two children, at least one of which is a boy. Of 17,946 responses, 35.9% reported two boys.^[10]

Vos Savant'due south articles were discussed by Carlton and Stansfield^[10] in a 2005 article in The American Statistician. The authors exercise not hash out 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 beingness a boy or girl is equal, and that the sex of the second child is contained of the first. With regard to her survey they say information technology "at to the lowest degree validates vos Savant's correct assertion that the "chances" posed in the original question, though similar-sounding, are unlike, and that the commencement probability is certainly nearer to ane in iii than to i in 2."

Carlton and Stansfield continue to talk over the common assumptions in the Boy or Girl paradox. They demonstrate that in reality male children are really more probable than female person children, and that the sex of the second child is non independent of the sexual activity of the kickoff. The authors conclude that, although the assumptions of the question run counter to observations, the paradox notwithstanding has pedagogical value, since it "illustrates one of the more intriguing applications of conditional probability."^[10] Of course, the actual probability values do non matter; the purpose of the paradox is to demonstrate seemingly contradictory logic, not actual nativity rates.

Information about the child [edit]

Suppose nosotros were told not only that Mr. Smith has two children, and one of them is a male child, 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 option procedure produced this knowledge.

Following the tradition of the problem, suppose that in the population of 2-child families, the sex activity of the ii children is independent of one another, equally probable boy or girl, and that the nascency date of each child is contained of the other child. The run a risk of being born on any given solar day of the calendar week is 1 / 7 .

From Bayes' Theorem that the probability of two boys, given that one 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 existence built-in on a Tuesday is ε  = 1 / 7 which will exist set up after arriving at the general solution. The 2nd factor in the numerator is just 1 / iv , the probability of having two boys. The first term in the numerator is the probability of at least one boy born on Tuesday, given that the family has two boys, or 1 − (i − ε)² (one minus the probability that neither boy is born 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 1 / 4 . The commencement term is already known by the previous remark, the concluding term is 0 (there are no boys). ${\displaystyle P(B_{T}\mid BG)}$ and ${\displaystyle P(B_{T}\mid GB)}$ is ε, there is 1 and only one boy, thus he has ε take chances of existence built-in on Tuesday. Therefore, the full equation is:

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

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 nigh 0.48. In fact, every bit ε approaches 0, the full probability goes to 1 / ii , which is the respond expected when one child is sampled (e.1000. the oldest kid is a boy) and is thus removed from the pool of possible children. In other words, as more than and more than details almost the boy kid are given (for example: built-in on January 1), the run a risk that the other child is a girl approaches one half.

It seems that quite irrelevant data was introduced, withal the probability of the sex of the other kid has changed dramatically from what it was before (the chance the other child was a daughter 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'due south poll of readers had asked which solar day of the week boys in the family were born. If Marilyn and so divided the whole data set into seven groups - ane for each twenty-four hours of the week a son was born - half-dozen out of seven families with two boys would be counted in two groups (the group for the 24-hour interval of the week of birth boy 1, and the group of the day of the week of birth for boy 2), doubling, in every grouping, the probability of a boy-male child combination.

However, is information technology really plausible that the family with at least 1 boy born on a Tuesday was produced by choosing just i 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 boy comes and answers the door. We enquire the male child on what twenty-four hour period of the calendar week he was born.

Assume that which of the ii children answers the door is determined past gamble. So the procedure was (ane) choice a two-child family unit at random from all two-kid families (2) pick one of the 2 children at random, (three) meet if it is a boy and enquire on what solar day he was born. The risk the other child is a daughter is 1 / 2 . This is a very different procedure from (one) picking a ii-child family unit at random from all families with 2 children, at least ane a male child, born on a Tuesday. The chance the family consists of a boy and a girl is fourteen / 27 , most 0.52.

This variant of the boy and girl problem is discussed on many net blogs and is the subject of a paper past Ruma Falk.^[13] The moral of the story is that these probabilities do not just depend on the known data, only on how that information was obtained.

Psychological investigation [edit]

From the position of statistical analysis the relevant question is often cryptic and as such there is no "correct" reply. Nevertheless, this does not frazzle 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 prefer an understanding of Gardner's trouble that if they were consequent would lead them to the 1 / 3 probability answer only overwhelmingly people intuitively get in at the 1 / 2 probability reply. Ambiguity notwithstanding, this makes the problem of involvement to psychological researchers who seek to empathize how humans estimate probability.

Fox & Levav (2004) used the problem (called the Mr. Smith problem, credited to Gardner, but not worded exactly the same as Gardner's version) to exam theories of how people estimate provisional probabilities.^[ii] In this report, the paradox was posed to participants in two ways:

• "Mr. Smith says: 'I have two children and at least one of them is a boy.' Given this data, what is the probability that the other child is a boy?"
• "Mr. Smith says: 'I accept ii children and information technology 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 get-go formulation gives the reader the mistaken impression that there are ii possible outcomes for the "other child",^[two] whereas the second formulation gives the reader the impression that there are four possible outcomes, of which one has been rejected (resulting in 1 / 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 institute that 85% of participants answered 1 / 2 for the first formulation, while simply 39% responded that manner to the 2nd formulation. The authors argued that the reason people respond differently to each question (along with other like problems, such equally the Monty Hall Problem and the Bertrand's box paradox) is because of the employ of naive heuristics that fail to properly define the number of possible outcomes.^[2]

Come across also [edit]

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

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 thou h} Craig R. Fob & 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 2020-04-10.
3. ^ ^{a b c d east} Bar-Hillel, Maya; Falk, Ruma (1982). "Some teasers concerning conditional probabilities". Cognition. eleven (2): 109–122. doi:ten.1016/0010-0277(82)90021-X. PMID 7198956. S2CID 44509163.
4. ^ ^{a b c} Raymond South. Nickerson (May 2004). Cognition and Chance: The Psychology of Probabilistic Reasoning. Psychology Printing. ISBN0-8058-4899-1.
5. ^ ^{a b} "Inquire Marilyn". Parade Magazine. October xiii, 1991 [January 5, 1992; May 26, 1996; December one, 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 Feb 2009.
7. ^ ^{a b} Leonard Mlodinow (2008). The Drunkard's Walk: How Randomness Rules our Lives. Pantheon. ISBN978-0-375-42404-five.
8. ^ Nikunj C. Oza (1993). "On The Confusion in Some Popular Probability Problems". CiteSeerX10.1.one.44.2448.
9. ^ P.J. Laird; et al. (1999). "Naive Probability: A Mental Model Theory of Extensional Reasoning". Psychological Review. 106 (i): 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 (two): 180–182. doi:x.1198/000313005x42813. S2CID 43825948.
11. ^ ^{a b} Charles Yard. Grinstead and J. Laurie Snell. "Grinstead and Snell's Introduction to Probability" (PDF). The Gamble Project.
12. ^ ^{a b} Stephen Marks and Gary Smith (Winter 2011). "The Ii-Child Paradox Reborn?" (PDF). Chance (Mag of the American Statistical Clan). 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 clash: Coping with a counterintuitive problem concerning the notorious two-child family unit". Thinking & Reasoning. 17 (four): 353–366. doi:ten.1080/13546783.2011.613690. S2CID 145428896.

External links [edit]

• At Least 1 Girl at MathPages
• A Problem With 2 Bear Cubs
• Lewis Carroll's Pillow Problem
• When intuition and math probably look wrong

rollinsstiverrom.blogspot.com

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

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