Mar 18, 2015 - sample space, and also to provoke discussion about probabilistic assumptions ... cell represents Thursday Boy followed by Wednesday. Girl.
The Two Boys problem revisited Rina Zazkis and Chanakya Wijeratne explain
T
he Two Boys problem
The Tuesday Boy mystery
Consider a well-know probability problem, often credited to Martin Gardner.
Consider the following variation on part (b):
(a) Mr. Smith has two children. The older child is a boy. What is the probability that the other child is also a boy? (b) Mr. Smith has two children. One of them is a boy. What is the probability that the other child is also a boy? The first intuitive reaction is to disregard the minor difference in phrasing of the two scenarios and to claim that the requested probability is ½, as the gender of one child does not, at least in probability problems, influence the gender of the other. However, the particular question what is the probability that the other child is also a boy? could be misleading. This is because the reference to ‘other child’ implies equal outcome for a boy or a girl. However, while ½ is a correct answer to scenario in (a), a variation in (b) requires further consideration. One possible way to interpret and reformulate the problem in (b) is as follows: Among families with 2 children, with at least one boy, what is the probability that some families have 2 boys? Rephrasing the question this way, and considering Mr. Smith as a randomly chosen representative from families with 2 children it is easy to see that the correct answer is 1/3, as having 2 boys (BB) is one option out of 3 equally probable options, boy-boy, boy-girl and girl-boy, denoted BB, BG and GB, respectively. Now, returning to case (a), ½ is indeed the correct answer, but not because the chances for a boy or a girl are equal, but because when the older child is a boy, having two boys is one of two equally probable events, BG and BB. We often use this problem in teaching probability in order to highlight the importance of considering an appropriate sample space, and also to provoke discussion about probabilistic assumptions that guided different solutions. However, recently we came across a fascinating variation, Tuesday Boy (Rehmeyer, 2010), which we explore here. We proceed with the underlying conventional and convenient assumptions, not corresponding to reality, but reasonable approximations, that sons and daughters are equally probable and all days of the week are equally probably birthdays.
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(c) Mr. Smith has two children. One of them is a boy born on Tuesday. What is the probability that Mr. Smith has two boys? “What has Tuesday got to do with it?”, is the first intuitive reaction of many people, regardless of their mathematics background. The information about Tuesday appears at first irrelevant, even for mathematicians (Rehmeyer, 2010). However, acknowledging the weekday of birth for either a boy or a girl, that is, children born on different weekdays are considered as different events, we have 14 options for the first child and 14 options for the second child. This information is presented in Figure 1, see page 19, which provides a clear alternative to listing all the possible options of pairs of events, gender, day; gender, day, where gender = Boy, Girl and Day = Sun, Mon, Tue, Wed, Thu, Fri, Sat, Sun. Each cell in Figure 1 represents one event out of 196 in the sample space. For example, the shaded cell represents Thursday Boy followed by Wednesday Girl. The decision to first attend to the row and then to the column is arbitrary and can be reversed. The highlighted cells on Figure 1 represent all the possibilities, total of 27, in which one of the children is Tuesday Boy. The 13 highlighted cells marked with X are those in which we find two boys, where at least one of the boys is a Tuesday Boy. As such, the solution to the Tuesday Boy problem is 13/27. The task can also be considered using conditional probability and Bayesian formula: P(A∩B)
P(A|B) =
P(B)
From here, acknowledging the symmetry in P(A∩B) P(A∩B) = P(A|B) P(B) = P(B|A) x P(A) As such, P(A|B) =
P(B|A) x P(A) P(B)
(*)
Let us denote X an event of Tuesday Boy in a family of two children. We are interested in calculating P(BB|X), that is, the probability of two boys in a family when it is known that one of the boys was born on Tuesday. Substituting this particular notation in (*) we get: P(BB|X) =
P(X|BB) x P(BB) P(X)
(**)
Further, applying a conventional assumption that the events B (boy) or G (girl) are independent and equiprobable, and P(B) = P(G) = ½, we get:
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The Two Boys problem revisited
P(BB) = P(BG)= P (GB) = P(GG) = ¼ ……………..(1) P(X|BB) = 1/7 + 1/7 – 1/49 = 13/49 ………………… (2) The probability P(X|BB) – that is, the probability of Tuesday Boy in a family of 2 boys – is calculated by noting that either the first boy born on Tuesday ( 1/7) OR the second boy born on Tuesday (1/7), avoiding double counting of the case that both are born Tuesday (1/49). We stated that X is an event of Tuesday Boy in a family of two children. Note that this is different from the question, what is the probability that a child is a boy born on a Tuesday, in considering one child only. In the case of one child the probability is: 1/2 x 1/7 = 1/14 But, bearing in mind a family of two children, the event Tuesday Boy has to be considered in 4 different equiprobable sequences of two children: Since P(BB) = P(BG) = P(GB) = P(GG) = 1/4 , we get P(X) = P(X|BB)P(BB) + P(X|BG)P(BG) + P(X|GB)P(GB) + P(X|GG)P(GG)
= [P(X|BB) + P(X|BG) + P(X|GB) + P(X|GG)] x 1/4
= (13/49 + 1/7 + 1/7 + 0) x 1/4 = 27/196…………….........(3)
Substituting (1), (2) and (3) into (**) we get P(BB/X) =
13/49 x 1/4 13/196 = 13/27 = 27/196 27/196
Mon Boy MonBoy TueBoy
MonBoy TueBoy
Thu Boy
Fri Boy
Sat Boy
Sun Boy
x
x
x
x
x
x x
ThuBoy
x
FriBoy
x
SatBoy
x
SunBoy
x
Figure 2: Sample space for 2 boys and weekdays However, having confirmed the result on the chart with more complicated considerations of conditional probability does not solve the presumed paradox. The students struggle with this rather counterintuitive result: why does this knowledge of the day of birth change the probability? There is a strong desire to find an explanation that will help the intuition. In what follows we consider variations on the Tuesday boy problem first by changing what is known about one of the children and then by changing the context. Spring boy and other variations Before suggesting what can be helpful in adjusting the intuition, we invite the reader to consider several additional scenarios. • Mr. Smith has two children. One of them is a boy born in Spring. What is the probability that the other child is also a boy?
Note that P(X) = 27/196 corresponds to the 27 highlighted cells on a grid of 196 in Figure 1. Further, the number P(X|BB) = 13/49 is shown in 13 highlighted cells in the “boys only” quadrant (See Figure 2).
Tue Boy
Wed Boy
x x
WedBoy
[An additional approach to this calculation is shown at the end.]
Mon Boy
Tue Boy
• Mr. Smith has two children. One of them is a boy born in January. What is the probability that the other child is also a boy? • Mr. Smith has two children. One of them is a boy born on Valentine’s Day. What is the probability that the other child is also a boy?
Wed Boy
Thu Boy
Fri Boy
Sat Boy
Sun Boy
X
X
X
X
X
Mon Girl
Tue Girl
Wed Girl
Thu Girl
Fri Girl
Sat Girl
Sun Girl
X X
X
WedBoy
X
ThuBoy
X
FriBoy
X
SatBoy
X
SunBoy
X
MonGirl
TueGirl
WedGirl
ThuGirl
FriGirl
SatGirl
SunGirl
Figure 1: Sample space for 2 children and weekdays
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The Two Boys problem revisited
WinterBoy WinterBoy
SpringBoy
SummerBoy
FallBoy
WinterGirl
SpringGirl
SummerGirl
FallGirl
x
x
x
SpringBoy
x
x
SummerBoy
x
FallBoy
x
WinterGirl
SpringGirl
SummerGirl
FallGirl
Figure 3: Sample space for 2 children and seasons Starting with the Spring Boy case, similar to the previous case, each cell in Figure 3 represents one of the 64 events that show the gender and the season of birth for the two children. The 15 highlighted cells show all the possibilities in which there is a Spring Boy. In 7 of these possibilities – marked with X – there are two boys. As such, 7/15 is the solution to the Spring Boy variation. With a similar line of reasoning, the probability of having two boys in a family when one is a January Boy is 23/47. Furthermore, acknowledging a particular day of birth, such as Valentine’s Day, for one boy, the probability of having two boys in a family is 729/1459 in a non-leap year. We note that when additional information about Mr. Smith’s boy includes more possibilities the result gets close to 1/2. In cases we considered so far
males in adult populations. To avoid biological and cultural connotations we invite the reader to consider the following experiment with a coin and a 7-region spinner. A 7-region spinner is rolled and a coin is tossed. The spinning results in a number from 1 to 7, the coin falls on heads (H) or tails (T). Then again 7-region spinner is rolled and a coin is tossed. The results of both experiments are recorded as sequences of the form [(coin, number), (coin, number)], for example 4H-5T, 1T-6T. You pick a sequence at random and note that it includes 2H. What is the probability that this sequence includes another H?
P(BB|Boy) < P(BB/Spring Boy) < P(BB|Tuesday Boy) < P(BB|Valentine’s Boy)
1/3 < 7/15 < 13/27 < 23/47 < 729/1459 < 1/2 In fact, more precise information about the boy refers to a wider variety of possibilities and attending to this variety brings the probability close to the initial intuition of 1/2. In general, if 1/p is the probability of the additional information (AI) about the boy such as 1/7 for Tuesday and 1/12 for January, then: P(BB|AI) = (2p - 1)/(4p - 1) And when p increases this probability approaches 1/2 lim
p→∞
2p – 1 4p – 1
= 1/2
The results appear surprising as more information about one of the boys brings the probability closer to the case of no additional information at all, as in scenario (a) in the beginning of this paper. Changing the setting
We acknowledge that this formulation is isomorphic to the Tuesday Boy problem. However, this formulation does not elicit the immediate response of 1/2, as does consideration of boys and girls. The setting of coins and spinners is not conflated with contextual considerations. It helps to bring students’ attention to the intended probabilistic interpretation. In particular, comparing Tuesday Boy with the result 2H in the spinner-coin experiment highlights the intended interpretation of Tuesday Boy as a compound event that combines two independent events, that of a day and that of a gender. The table in Figure 4 is the same table as in Figure 1, only with the change of column and row labels. It is evident from this table that 2H appears in 27 sequences, and in 13 of those H appears twice, so the sought probability is 13/27.
We believe that the setting of probability questions with boys, girls and birthdays affects our thinking. For example, Peled, Peled and Peled (2013) found that college students in an introductory biology class had a skewed conception of gender birth ratio, at times estimating the number of female births significantly larger than a number of male births. This kind of belief was justified by considering personal families or observation of more females than
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The Two Boys problem revisited
1H 2H 3H 4H 5H 6H 7H 1T 2T 3T 4T 5T 6T 7T 1H 2H
x x
x
3H
x
4H
x
5H
x
6H
x
7H
x
1T
2T
3T
4T
5T
6T
7T
x
x
x
x
x
Remember that we denoted X as being the event of having a Tuesday Boy in a family of two children, when this could be either the first or the second child. Denoting BT the boy born on Tuesday and acknowledging the birth order of two children, we get: ∩ ∩ ∩ X = (BTG GBT BTB BBT) That is, Tuesday Boy can be either first-born or secondborn, when the other child is either a boy or a girl. Substituting this into conditional probability Bayesian formula, we get: … ∩ ∩ ∩ ∩ P = BB|X = P((BTB BBT)|(BTG GBT BTB BBT)) =
∩ ∩ ∩ ∩ P((BTB BBT)∩(BTG GBT BTB BBT)) ∩ ∩ ∩ P(BTG GBT BTB BBT) ∩ P(BTB BBT) ∩ ∩ ∩ = P(BTG GBT BTB BBT)
Figure 4: Sample space of spinner-coin experiment Using the terminology introduced by Ejersbo and Leron (2013) we consider the spinner-coin problem as a “bridging task” to the Tuesday Boy problem.
=
P(BTB) + P(BBT) – P(BTB∩BBT) P(BTG) + P(GBT) + P(BTB) + P(BBT) – P(BTB∩BBT)
A bridging task has to be logically equivalent to, but psychologically easier than, the given task. In our case the bridging task clearly identifies the events, the probability of each event and the appropriate sample space. We believe that one of the factors in perceiving the Tuesday Boy solution as counterintuitive is that “Tuesday boy” is not seen as a combination of two independent events. Acknowledging the similarity between the spinner-coin task and the Tuesday Boy problem will help in resolving the Tuesday Boy mystery.
=
P(BTB) + P(BBT) – P(BTBT) P(BTG) + P(GBT) + P(BTB) + P(BBT) – P(BTBT)
=
=
P(BT)P(B) + P(B)P(BT) – P(BT)P(BT) P(BT)P(G) + P(G)P(BT) – P(BT)P(B) + P(B) P(BT) – P(BT) P(BT) (1/2 x 1/7 x 1/2) + (1/2 x 1/2 x 1/7) – (1/2 x 1/7 x 1/2 x 1/7) (1/2 x 1/7 x 1/2) + (1/2 x 1/2 x 1/7) + (1/2 x 1/7 x 1/2) + (1/2 x 1/2 x 1/7) – (1/2 x 1/7 x 1/2 x 1/7)
= 13/27
Additional scenarios Having considered the Two Boys classical problem, and the Tuesday Boy problem, several variations can be presented for students’ practice, for example: …
Rina Zazkis and Chanakya Wijeratne, Simon Fraser University, Canada
• Mr. Smith has two children. The older child is a boy born on Tuesday. What is the probability that the other child is also a boy?
References
• Mr. Smith has two children. One of his children was born on Tuesday. What is the probability that the other child was also born on Tuesday? • Mr. Smith has three children. One of the three is a boy. What is the probability that he has 3 boys? • Mr. Smith has three children. One of the three is a boy born in Spring. What is the probability that he has 3 boys? Alternative calculation The probability in scenario (c ) above, that is, the probability of Mr. Smith having two boys, when it is known that one of his children is a Tuesday Boy can be determined directly.
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Ejersbo, L.R. & Leron, U. (2014). Revisiting the medical diagnosis problem: Reconciling intuitive and analytical thinking. In Chernoff, E.J. & Sriraman, B., Probabalistic Thinking: Presenting Plural Perspectives. Springer. Peled, O.N., Peled, I., & Peled J.U. (2013). Uncommon knowledge of a common phenomenon: intuitions and statistical thinking about gender birth ratio. International Journal of Mathematical Education in Science and Technology, 44(1), 59-69. Rehmeyer, J. (2010). When intuition and math probably look wrong. Science News. Available: http://www. sciencenews.org/view/generic/id/60598/description/When_ intuition_and_math_probably_look_wrong
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