Boy-or-girl problem

The boy-or-girl problem , also known as the two-child problem or the sibling problem , is a probability- related problem . The task deals with the possibility of being able to make a conditional probability statement about the gender of the other child from knowledge of the gender of one of the two children in two-child families . The original formulation of the problem was published in Scientific American by Martin Gardner in 1959 and consists of two questions:

• Mr. Müller has two children. The older child is a girl. What is the probability that both children will be girls?
• Mr. Schmidt has two children. At least one of them is a boy. What is the probability that both children will be boys?

Gardner originally gave answers 1/2 and 1/3, but later had to admit that the answer to the second question could also be 1/2, depending on how the information about the gender of one of the children was obtained.

Reformulation of the problem

Based on the fact that information can be viewed as a clear answer to a clear question, the information from the two statements about Mr. Müller and Mr. Schmidt can also be formulated in the form of answers to questions. In order to standardize the presentation, these questions are asked below for the mother of two children; The idealized assumption is assumed that in two-child families all possible gender pairs - that is, boy / boy, boy / girl, girl / boy and girl / girl - occur with exactly the same frequency. The probability for all four possible constellations is initially the same; in particular, the unconditional ( a priori ) probability that both children are girls is 1/4.

A question to the mother should now be used to determine the probability that both children are girls. Different questions give you different information, which leads to different probabilities. (The direct question "Do you have two daughters?" Including the answer "Yes" or "No" is of course trivial and does not need to be investigated further.)

First question

"What gender is your older child?" Answer: "It is a girl."

What is the probability that both children will be girls?

The answer is 1/2.

The solution can be found with the following table:

	child		Younger child is ...	probability
	1.	2.		A priori	Conditional
1	Boy	Boy	(not possible)	1/4	0
2	Boy	girl	(not possible)	1/4	0
3	girl	Boy	Boy	1/4	1/2
4th	girl	girl	girl	1/4	1/2

The first two columns show which variants are basically possible with two children. Column 3 shows the options that remain when you know that the older child is a girl - cases 1 and 2 are then not possible. This leaves two equally likely options for the second child, one of which is another girl.

In this simple case, another consideration leads to the same result: if one of the two children is a girl, then only the gender of the other child matters whether both children are girls. Since the older child is a girl here, the probability that both children are girls is the same as the probability that the younger child is also a girl. Since the sex of each child is independent of the sex of the other child, the information about the sex of the older child does not provide any information about the sex of the younger child, which could give rise to the a priori equal probability for girl or boy as sex of modify younger child. The probability that the younger child is also a girl and that both children are girls is therefore 1/2.

Second question

"Do you have two sons?" Answer: "No."

What is the probability that both children will be girls?

The answer is 1/3.

The solution can be found with the following table:

	1 child	2nd child	Answer from mother	Another child is ...
1	Boy	Boy	Yes	irrelevant
2	Boy	girl	No	Boy
3	girl	Boy	No	Boy
4th	girl	girl	No	girl

Since at least one of the children has to be a girl, case 1 is not possible. So there are only three constellations left, the last column shows the possibilities. Just counting shows that in one in three equally likely cases, both children are girls.

This is identical to the question “Do you have at least one daughter?” And the answer “Yes”.

Third question

“ What gender is one of your children? "Answer:" One of my children is a girl. "

What is the probability that both children will be girls?

The answer is 1/2.

Solution using a table

The solution can be found with the following table. Because the mother can name each of her two children, there are a total of eight options:

	1 child	2nd child	Choice of mother	Answer from mother	Another child is ...
1	Boy	Boy	1 child	"Boy"	Boy
2	Boy	Boy	2nd child	"Boy"	Boy
3	Boy	girl	1 child	"Boy"	girl
4th	Boy	girl	2nd child	"Girl"	Boy
5	girl	Boy	1 child	"Girl"	Boy
6th	girl	Boy	2nd child	"Boy"	girl
7th	girl	girl	1 child	"Girl"	girl
8th	girl	girl	2nd child	"Girl"	girl

Since there is no reason to assume that the mother has a preference for naming the sex of one of her children, it makes sense to assume that the mother chose the named child by chance ( discrete equal distribution ). Therefore, all eight possibilities are equally likely. Since we know that the mother answered that one of her children is a girl, the possible constellations are limited to the numbers 4, 5, 7 and 8. They form the condition. In half of these cases, namely in cases 7 and 8, the other child is also a girl. So the conditional probability is 1/2.

Solution using an urn model

There are four urns, one with two black balls , one with two white balls and two mixed urns with one black and one white ball: and . (This corresponds to the assumption of the equal distribution of the sexes in two children.) A ball is drawn from a randomly selected urn (a randomly selected mother of two children is interviewed) (the mother names the sex of one of her two children at random), the ball is white ( ). What is the probability that the other ball in this urn is also white? ${\displaystyle U_{ss}}$ ${\displaystyle U_{ww}}$ ${\displaystyle U_{sw}}$ ${\displaystyle U_{ws}}$ ${\displaystyle K_{1}=W}$

We are looking for the conditional probability

${\displaystyle P(K_{2}=W\mid K_{1}=W)}$.

The second ball can only be white if the first ball has been drawn from the urn with the two white balls. Since all four urns are a priori equally probable, is . Since all urns contain the same number of balls, the probability of being drawn is the same for all balls; the total probability of pulling a white ball is therefore equal to the proportion of white balls in all balls and is . According to Bayes' theorem, this results in the probability being sought ${\displaystyle P(U_{ww})=1/4}$ ${\displaystyle P(K_{1}=W)}$ ${\displaystyle 4/8=1/2}$

${\displaystyle P(K_{2}=W\mid K_{1}=W)=P(U_{ww}\mid K_{1}=W)={\frac {P(K_{1}=W\mid U_{ww})\cdot P(U_{ww})}{P(K_{1}=W)}}={{1\cdot {1 \over 4}} \over {1 \over 2}}={1 \over 2}.}$

Cross check: The other ball in the urn is black if and only if the white ball comes from one of the two mixed urns. Since the condition can be met by independent events, the probabilities of these events must be added. In this case, where two equally probable urns meet the condition, the probability is consequently twice the probability for a single one of the two mixed urns. The probability that the other ball in the urn is black is therefore

${\displaystyle P(K_{2}=S\mid K_{1}=W)=P(U_{sw}\mid K_{1}=W)+P(U_{ws}\mid K_{1}=W)=2\;P(U_{sw}\mid K_{1}=W)}$

${\displaystyle =2\;{\frac {P(K_{1}=W\mid U_{sw})\cdot P(U_{sw})}{P(K_{1}=W)}}=2\cdot {{{1 \over 2}\cdot {1 \over 4}} \over {1 \over 2}}={1 \over 2}.}$

Modeling the question with an urn model therefore leads to the same result as the probability table.

Similar problems

• Goat problem
• Prisoner's paradox

literature

• Norbert Henze : Stochastics for Beginners: An Introduction to the Fascinating World of Chance . 10th edition. Springer Spectrum, Wiesbaden 2013, ISBN 978-3-658-03076-6 , p. 114 f . ( limited preview in Google Book search).

Web links

• At least one girl on MathPages
• Avoiding thought traps - using the example of the sibling problem . From Stochastics in School 31 (2011), pp. 30–32

Individual evidence

1. ^ Martin Gardner: The Second Scientific American Book of Mathematical Puzzles and Diversions . Simon & Schuster, 1954, ISBN 978-0226282534 ..
2. ^ Maya Bar-Hillel and Ruma Falk: Some teasers concerning conditional probabilities . In: Cognition . 11, No. 2, 1982, pp. 109-122. doi : 10.1016 / 0010-0277 (82) 90021-X . PMID 7198956 .
3. ^ Raymond S. Nickerson: Cognition and Chance: The Psychology of Probabilistic Reasoning . Psychology Press, May 2004, ISBN 0-8058-4899-1 .