Randomly Selecting a Three Child Family With Two Girls

Paradox in probability theory

The Boy or Daughter paradox surrounds a set of questions in probability theory, which are also known as The Two Child Problem,^[1] Mr. Smith'south 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 column" in Scientific American. He titled it The Two Children Problem, and phrased the paradox every bit 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 one of them is a boy. What is the probability that both children are boys?

Gardner initially gave the answers 1 / ii and i / 3 , respectively, simply later acknowledged that the second question was ambiguous.^[1] Its answer could exist i / 2 , depending on the procedure by which the information "at to the lowest degree 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 Due south. Nickerson.^[4]

Other variants of this question, with varying degrees of ambiguity, have been popularized past Inquire Marilyn in Parade Magazine,^[5] John Tierney of The New York Times,^[6] and Leonard Mlodinow in The Drunkard'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 one / 2 changed from 85% to 39%.^[2]

The paradox has stimulated a bang-up deal of controversy.^[4] The paradox stems from whether the problem setup is similar for the ii questions.^{[2] [7]} The intuitive answer is 1 / two .^[two] This reply is intuitive if the question leads the reader to believe that there are two equally likely possibilities for the sexual practice of the second child (i.e., male child and girl),^{[two] [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. Get-go, information technology is assumed that the space of all possible events can exist hands enumerated, providing an extensional definition of outcomes: {BB, BG, GB, GG}.^[10] This notation indicates that there are four possible combinations of children, labeling boys B and girls G, and using the get-go letter to correspond the older child. 2d, it is assumed that these outcomes are equally probable.^[10] This implies the post-obit model, a Bernoulli procedure with p = 1 / ii :

1. Each child is either male or female.
2. Each child has the same gamble of being male as of being female.
3. The sexual practice 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.

Start question [edit]

• Mr. Jones has ii children. The older child is a girl. 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 four as probable events:

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

Only ii of these possible events meet the criteria specified in the question (i.eastward., GG, GB). Since both of the ii possibilities in the new sample space {GG, GB} are as probable, and just one of the two, GG, includes ii girls, the probability that the younger kid is besides a girl is 1 / 2 .

Second question [edit]

• Mr. Smith has two children. At least one of them is a boy. 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 respond is possible without data that was not provided. Specifically, that two different procedures for determining that "at least 1 is a boy" could lead to the exact same wording of the problem. But they pb to unlike correct answers:

• From all families with two 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 i / iii .
• From all families with 2 children, 1 kid is selected at random, and the sex of that child is specified to be a boy. This would yield an answer of 1 / two .^{[iii] [4]}

Grinstead and Snell contend that the question is ambiguous in much the aforementioned fashion Gardner did.^[11] They leave it to the reader to decide whether the procedure, that yields i/3 equally the respond, is reasonable for the problem as stated above. 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 virtually obviously present, considering it is not clear whether we are immune to presume that a specific kid is a male child, leaving the other kid uncertain, or whether it should be interpreted in the aforementioned manner 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', every bit Bar-Hillel and Falk argue, where different assumptions can pb to dissimilar outcomes (because the trouble statement was non well enough defined to allow a unmarried straightforward interpretation and answer).

For case, say an observer sees Mr. Smith on a walk with merely one of his children. If he has two boys so that kid must be a boy. But if he has a boy and a girl, that child could have been a daughter. Then seeing him with a boy eliminates not only the combinations where he has two girls, merely 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 i boy (i.e., the status is necessary), it cannot be causeless that every Mr. Smith with at least 1 boy is intended. That is, the trouble argument does not say that having a boy is a sufficient condition for Mr. Smith to exist identified as having a boy this way.

Commenting on Gardner's version of the problem, Bar-Hillel and Falk^[iii] 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 take two children and at least one of them is a boy.' It must be further assumed that Mr. Smith would always report this fact if it were true, and either remain silent or say he has at to the lowest degree one daughter, for the right respond to be 1 / three as Gardner apparently originally intended. Only under 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 data was obtained past looking at both children to see if at that place is at least 1 boy, the condition is both necessary and sufficient. Iii of the four equally probable events for a two-kid family in the sample infinite in a higher place run across the condition, as in this table:

Older child	Younger child
Girl	Girl
Girl	Boy
Boy	Girl
Male child	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 . Nevertheless, if the family was get-go selected and then a random, true statement was made about the sex of one child in that family, whether or not 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 be made in each example.^[11] And then, if ALOB represents the event where the statement is "at least one boy", and ALOG represents the issue where the statement is "at least 1 daughter", and then this table describes the sample space:

Older kid	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)
Girl	Daughter	1 / four	0	1	0	1 / 4
Girl	Male child	i / 4	1 / 2	1 / 2	1 / viii	1 / viii
Male child	Girl	one / 4	1 / 2	1 / two	1 / viii	one / 8
Boy	Boy	one / 4	1	0	1 / 4	0

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

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

The paradox occurs when it is non known how the statement "at to the lowest degree i is a boy" was generated. Either answer could exist right, based on what is assumed.^[12]

However, the " 1 / 3 " answer is obtained just by assuming P(ALOB|BG) = P(ALOB|GB) =1, which implies P(ALOG|BG) = P(ALOG|GB) = 0, that is, the other kid's sex activity is never mentioned although it is nowadays. As Marks and Smith say, "This extreme assumption is never included in the presentation of the two-child problem, however, and is surely not what people have in mind when they present information technology."^[12]

Modelling the generative process [edit]

Another mode 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. We presume equal probability that either is a boy or a girl. The iii 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) = one / 4 , and iii. one of each (Grand·B) — with probability of P(Thou·B) = ane / two . These are the prior probabilities.

At present we add the additional assumption that "at least one is a boy" = B. Using Bayes' Theorem, we observe

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

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

Note that, although the natural supposition seems to be a probability of 1 / 2 , and so the derived value of 1 / 3 seems low, the bodily "normal" value for P(BB) is i / 4 , so the 1 / 3 is actually a bit higher.

The paradox arises because the second assumption is somewhat bogus, and when describing the trouble in an actual setting things get a scrap sticky. Merely how do we know that "at least" one is a boy? I description of the problem states that we expect into a window, see only one child and it is a boy. This sounds like the same assumption. However, this i is equivalent to "sampling" the distribution (i.east. removing one child from the urn, ascertaining that it is a boy, then replacing). Let's call the argument "the sample is a male child" proffer "b". Now we accept:

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

The difference hither 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 ane / ii .

The Bayesian analysis generalizes easily to the instance in which we relax the l:50 population assumption. If we accept no information almost the populations and then we assume a "flat prior", i.e. P(GG) = P(BB) = P(G·B) = i / 3 . In this instance the "at to the lowest degree" assumption produces the result P(BB|B) = 1 / 2 , and the sampling supposition produces P(BB|b) = 2 / 3 , a result also derivable from the Dominion 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 volition receive $4 if he has two boys. Their wager will increment in value as practiced news arrives. What evidence would make them happier about their investment? Learning that at least one kid out of two is a boy, or learning that at least one kid out of ane is a boy?

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

At present for the numbers. If we bet on one kid and win, the value of their investment has doubled. It must double once more to get to $4, so the odds are 1 in two.

On the other mitt if ane were larn that at least one of two children is a boy, the investment increases equally if they had wagered on this question. Our $1 is now worth $ane+ 1 / 3 . To get to $four we yet accept to increment 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 various forms. The first variant presented by Bar-Hillel & Falk^[three] is worded every bit follows:

• Mr. Smith is the father of two. Nosotros meet him walking forth the street with a young boy whom he proudly introduces as his son. What is the probability that Mr. Smith's other child is also a boy?

Bar-Hillel & Falk use this variant to highlight the importance of because the underlying assumptions. The intuitive respond is 1 / 2 and, when making the most natural assumptions, this is correct. However, someone may argue that "…before Mr. Smith identifies the boy equally his son, we know only that he is either the begetter of two boys, BB, or of 2 girls, GG, or of one of each in either birth club, i.east., BG or GB. Bold once again independence and equiprobability, we begin with a probability of 1 / 4 that Smith is the begetter of two boys. Discovering that he has at to the lowest degree one boy rules out the effect GG. Since the remaining three events were equiprobable, we obtain a probability of 1 / 3 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 zip probability, ruling information technology out), the wedlock of events BG and GB becomes equiprobable with upshot BB, and so the chance that the other child is also a male child is 1 / 2 . Bar-Hillel & Falk, however, suggest an alternative scenario. They imagine a civilization 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 have resulted in the boy walking companion, and thus the probability that the other kid is as well a male child is 1 / 3 .

In 1991, Marilyn vos Savant responded to a reader who asked her to answer 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 bear witness y'all, but she doesn't know whether they're male, female, or a pair. Yous tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male person?" she asks him. "Yes!" she informs y'all with a smiling. What is the probability that the other one is a male?
• Say that a adult female and a human (who are unrelated) each have two children. We know that at to the lowest degree ane of the adult female's children is a male child and that the man'southward oldest child is a boy. Can you lot explain why the chances that the adult female has two boys exercise not equal the chances that the man has two boys?

With regard to the second formulation Vos Savant gave the archetype reply that the chances that the adult female has two boys are about 1 / 3 whereas the chances that the man has two boys are nearly 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 male child. Of 17,946 responses, 35.9% reported two boys.^[10]

Vos Savant's articles were discussed by Carlton and Stansfield^[10] in a 2005 article in The American Statistician. The authors practice not talk over 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 boy or girl is equal, and that the sex of the second child is contained of the offset. With regard to her survey they say information technology "at least validates vos Savant's correct assertion that the "chances" posed in the original question, though similar-sounding, are dissimilar, and that the first probability is certainly nearer to 1 in 3 than to ane in 2."

Carlton and Stansfield become on to discuss the common assumptions in the Boy or Girl paradox. They demonstrate that in reality male person children are really more probable than female children, and that the sexual practice of the second kid is not independent of the sexual practice of the first. The authors conclude that, although the assumptions of the question run counter to observations, the paradox nevertheless has pedagogical value, since it "illustrates one of the more intriguing applications of conditional probability."^[10] Of class, the actual probability values do not matter; the purpose of the paradox is to demonstrate seemingly contradictory logic, non actual nascence rates.

Data nearly the child [edit]

Suppose nosotros were told not only that Mr. Smith has two children, and one of them is a boy, but also that the boy was built-in on a Tuesday: does this change the previous analyses? Again, the reply depends on how this data was presented - what kind of selection process produced this knowledge.

Following the tradition of the problem, suppose that in the population of two-child families, the sex of the 2 children is contained of 1 another, equally likely boy or girl, and that the birth appointment of each child is contained of the other kid. The take chances of being built-in on whatever given 24-hour interval of the calendar week is 1 / vii .

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

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

Assume that the probability of being born on a Tuesday is ε  = 1 / 7 which will be ready after arriving at the general solution. The second factor in the numerator is simply 1 / iv , the probability of having two boys. The first term in the numerator is the probability of at least 1 male child born on Tuesday, given that the family has 2 boys, or ane − (one − ε)ⁱⁱ (one minus the probability that neither boy is born on Tuesday). For the denominator, allow 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 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 simply one boy, thus he has ε take chances of beingness born on Tuesday. Therefore, the full equation is:

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

For ${\displaystyle \varepsilon >0}$ , this reduces to ${\displaystyle \mathrm {P(BB\mid B_{T})} ={\frac {2-\varepsilon }{iv-\varepsilon }}}$

If ε is at present gear up to i / 7 , the probability becomes 13 / 27 , or about 0.48. In fact, every bit ε approaches 0, the full probability goes to 1 / 2 , which is the answer expected when one child is sampled (e.k. the oldest child is a boy) and is thus removed from the puddle of possible children. In other words, as more and more details near the male child child are given (for example: built-in on Jan ane), the hazard that the other child is a girl approaches one one-half.

It seems that quite irrelevant information was introduced, yet the probability of the sex of the other child has changed dramatically from what it was before (the gamble the other child was a daughter was 2 / 3 , when information technology was non known that the boy was born on Tuesday).

To understand why this is, imagine Marilyn vos Savant'southward poll of readers had asked which twenty-four hour period of the week boys in the family were built-in. If Marilyn then divided the whole data set into seven groups - ane for each day of the week a son was born - six out of seven families with two boys would be counted in two groups (the group for the day of the week of nativity boy i, and the group of the day of the week of birth for male child 2), doubling, in every group, the probability of a boy-boy combination.

However, is it really plausible that the family unit with at least one 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. Nosotros ask the boy on what solar day of the week he was born.

Assume that which of the ii children answers the door is adamant past chance. Then the procedure was (1) selection a two-child family at random from all two-child families (two) option ane of the two children at random, (3) see if information technology is a boy and enquire on what day he was born. The chance the other kid is a girl is 1 / two . This is a very unlike process from (1) picking a 2-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 , well-nigh 0.52.

This variant of the boy and girl problem is discussed on many internet blogs and is the subject of a paper by Ruma Falk.^[xiii] The moral of the story is that these probabilities do not merely 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 oftentimes ambiguous and every bit such there is no "correct" answer. However, 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'southward 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 reply but overwhelmingly people intuitively arrive at the 1 / 2 probability answer. Ambiguity yet, this makes the problem of interest to psychological researchers who seek to understand 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 test theories of how people estimate conditional probabilities.^[ii] In this study, the paradox was posed to participants in two ways:

• "Mr. Smith says: 'I have two children and at to the lowest degree 1 of them is a boy.' Given this information, what is the probability that the other child is a boy?"
• "Mr. Smith says: 'I accept ii children and it is not the instance that they are both girls.' Given this information, what is the probability that both children are boys?"

The authors contend that the first formulation gives the reader the mistaken impression that there are 2 possible outcomes for the "other kid",^[ii] 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 one / 3 beingness the probability of both children beingness boys, as there are 3 remaining possible outcomes, only one of which is that both of the children are boys). The written report establish that 85% of participants answered i / 2 for the start formulation, while only 39% responded that way to the second 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 because of the use of naive heuristics that fail to properly define the number of possible outcomes.^[2]

See also [edit]

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

References [edit]

1. ^ ^{a b} Martin Gardner (1961). The 2d Scientific American Volume of Mathematical Puzzles and Diversions. Simon & Schuster. ISBN978-0-226-28253-4.
2. ^ ^{a b c d e f grand h} Craig R. Fox & Jonathan Levav (2004). "Partition–Edit–Count: Naive Extensional Reasoning in Judgment of Conditional Probability" (PDF). Periodical 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-ten.
3. ^ ^{a b c d e} Bar-Hillel, Maya; Falk, Ruma (1982). "Some teasers concerning provisional probabilities". Cognition. xi (two): 109–122. doi:x.1016/0010-0277(82)90021-X. PMID 7198956. S2CID 44509163.
4. ^ ^{a b c} Raymond S. Nickerson (May 2004). Noesis and Adventure: The Psychology of Probabilistic Reasoning. Psychology Printing. ISBN0-8058-4899-i.
5. ^ ^{a b} "Ask Marilyn". Parade Magazine. Oct xiii, 1991 [January five, 1992; May 26, 1996; Dec 1, 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 Drunkard'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". CiteSeerX10.i.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 (two): 180–182. doi:10.1198/000313005x42813. S2CID 43825948.
11. ^ ^{a b} Charles Thou. Grinstead and J. Laurie Snell. "Grinstead and Snell's Introduction to Probability" (PDF). The CHANCE Projection.
12. ^ ^{a b} Stephen Marks and Gary Smith (Winter 2011). "The Ii-Kid Paradox Reborn?" (PDF). Chance (Mag of the American Statistical Clan). 24: 54–nine. doi:ten.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 trouble concerning the notorious two-child family". Thinking & Reasoning. 17 (four): 353–366. doi:10.1080/13546783.2011.613690. S2CID 145428896.

External links [edit]

• At To the lowest degree I Daughter at MathPages
• A Problem With Two Bear Cubs
• Lewis Carroll's Pillow Problem
• When intuition and math probably look wrong

lehmannlaught.blogspot.com

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

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