Boy Girl One
Boy Girl One
The Boy or Girl paradox surrounds a well-known set of questions in probability theory which are also known as The Two Child Problem , Mr. Smith's Children and the Mrs. Smith Problem. The initial formulation of the question dates back to at least 1959, when Martin Gardner published one of the earliest variants of the paradox in Scientific American. Titled the The Two Children Problem , he 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 least one 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 later acknowledged that the second question was ambiguous. Its answer could be 1/2, depending on how you found out that one child was a boy. The ambiguity, depending on the exact wording and possible assumptions, was confirmed by Bar-Hillel and Falk, and Nickerson.
Other variants of this question, with varying degrees of ambiguity, have been recently popularized by Ask Marilyn in Parade Magazine, John Tierney of The New York Times, Leonard Mlodinow in Drunkard's Walk ., as well as numerous online publications. 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%.
The paradox has frequently stimulated a great deal of controversy. Many people, including professors of mathematics, argued strongly for both sides with a great deal of confidence, sometimes showing disdain for those who took the opposing view. The paradox stems from whether the problem setup is similar for the two questions. The intuitive answer is 1/2. This answer is intuitive if the question leads the reader to believe that there are two equally likely possibilities for the gender of the second child (i.e., boy and girl), and that the probability of these outcomes is absolute, not conditional.
Common assumptions
The two possible answers share a number of assumptions. First, it is assumed that the space of all possible events can be easily enumerated, providing an extensional definition of outcomes: {BB, BG, GB, GG}. This notation indicates that there are four possible combinations of children, labeling boys B and girls G, and using the first letter to represent the older child. Second, it is assumed that these outcomes are equally probable. This implies the following:
- That each child is either male or female.
- That the sex of each child is independent of the sex of the other.
- That each child has the same chance of being male as of being female.
These assumptions have been shown empirically to be false. It is worth noting that these conditions form an incomplete model. By following these rules, we ignore the possibilities that a child is intersex, the ratio of boys to girls is not exactly 50:50, and (amongst other factors) the possibility of identical twins means that sex determination is not entirely independent. However, one can see intuitively that the occurrence of each of these exceptions is sufficiently rare to have little effect on our simple analysis of the general population.
First question
- Mr. Jones has two children. The older child is a girl. What is the probability that both children are girls?
In this problem, a random family is selected. In this sample space, there are four equally probable events:
Only two of these possible events meets the criteria specified in the question (e.g., GB, GG). Since both of the two possibilities in the new sample space {GB, GG} are equally likely, and only one of the two, GG, includes two girls, the probability that the younger child is also a girl is 1/2.
Second question
- 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. If it is assumed that this information was obtained by considering both children, then there are four equally probable events for a two-child family as seen in the sample space above. Three of these families meet the necessary and sufficient condition of having at least one boy. The set of possibilities (possible combinations of children that meet the given criteria) is:
Thus, if it is assumed that both children were considered, the answer to question 2 is 1/3. In this case the critical assumption is how Mr. Smith's family was selected and how the statement was formed. One possibility is that families with two girls were excluded in which case the answer is 1/3. The other possibility is that the family was selected randomly and THEN a true statement was made about the family and IF there HAD BEEN two girls in the Smith family, the statement would have been made that "at least one is a girl". If the Smith family were selected as in the latter case, the answer to question 2 is 1/2.
However, if it is assumed that the information was obtained by considering only one child, then the problem is an isomorphism of question one, and the answer is 1/2.
Third question
- A (random) family has two children, and one of the two children is a boy named Jacob. What is the probability that the other child is a girl?
Or, the set {GJ, JG, JB, BJ}, in which two out of the four possibilities includes a girl.
Therefore we might think that the probability returns to 1/2. But this is wrong if we again assume that the information was obtained by looking at both children , because it doesn't take into account different frequencies of each of these answers. The likelihood of a boy being named Jacob and a boy not being named Jacob are not equal. Thus, we must replace our classical interpretation of probability with either a Frequentist or Bayesian interpretation. (Note that in real life child names are not independent of each other. In particular, people usually do not give the same name to two children. Thus, this discussion is purely theoretical).
Frequentist approach
Consider 10,000 families that have two children. Assume that the gender and name of each child is independent, within family and between families. Assume that the probability of each individual child being a girl is .5; otherwise the child is a boy. Assume that the probability of a child having the name Jacob is .01, and that all children with the name Jacob are also boys.
In the table above, we have a list of all possible unique outcomes. But these outcomes do not have the same frequency. If we start with the assumption that the family has two children, we get the following frequency table:
With the additional bit of information that the family has a boy named Jacob, we can break every instance of "Boy" into two: "Jacob" and "Boy not Jacob". For every 50 Boys, 1 will fall into the "Jacob" bin and 49 into the "Boy not Jacob" bin. Thus, we have the following table:
If we eliminate all instances that do not meet our given criteria ({Girl, Girl} {Girl, Boy not Jacob} {Boy not Jacob, Girl} {Boy not Jacob, Boy not Jacob}), then we eliminate 9801 of our events, leaving 199 possible events. Of those, the successful events are {Girl, Jacob} and {Jacob, Girl}, or 100 cases.
So if the probability of a boy being named Jacob is 1 in 50, then the probability that the family has a girl is 100/199, or roughly 50%. But this value will change depending on the popularity of the name. At the extreme, if all boys were given the same name, then being named Jacob would provide no more information than being a boy, and thus the probability would still be 2/3 that the family has a girl. As the likelihood of the name decreases, the likelihood of the two-Jacob case also decreases, and the probability of the family having a girl approaches the limit of 50%.
If we further assume that parents never name two children with the same name, we can eliminate {Jacob, Jacob}, leaving 198 possible events; thus it would appear that the probability of the family having a girl is 100/198, or 50/99. However, there are now 50 occurrences each of {Jacob, Boy not Jacob} and {Boy not Jacob, Jacob} making the probability of a girl 100/200, or exactly 1/2.
Bayesian approach
The 4 cases with one boy named Jacob are: Jacob and Boy not Jacob, Boy not Jacob and Jacob, Jacob and Girl, Girl and Jacob, with probability
,
,
,
, respectively, and
p
is the probability that a boy is called Jacob. Using Bayes' theorem, we know the other child is a girl with probability:
When p is quite small as in general cases, we get the result close to 1/2. However, if we change the condition of having the name Jacob to something less informative, such as day of birth is an even number , now p is very close to 1/2, and the probability that the othe
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