Tuesday Birthday Problem
Below is a classic series of probability questions about children. They begin gently and grow steadily more difficult, to the point of defying intuition. These problems have confounded enough people over the years to earn the title of “Boy or Girl Paradox”.
First, let’s state some assumptions that are used throughout the problems. Much like real life, a child is born either a boy or girl with equal probability, 1/2. The sex of each child is independent of the other. A child is equally likely to be born any day of the week, and the day is again independent from others and from the sex. In statistical notation, the sex of the child ~iid Bernoulli(1/2) and day of week born is uniform over the seven days.
1. What is the probability of having a boy?
Answer: 1/2
Without any additional information, we refer to the given assumptions, namely, that a boy is born with 1/2 chance.
2. A family has two children. The older one is a boy. What is the probability that the other one is also a boy?
Answer: 1/2
That the older one is a boy has no bearing on the sex of the younger one, since each child’s sex is independent.
3. A family has two children. At least one child is a boy. What is the probability that the other one is also a boy?
Answer: 1/3
Surprised that it isn’t 1/2?
We can solve this from first principles. Enumerating the four equally probable events for a two-child family in birth order gives {GG, GB, BG, BB}, where B stands for boy and G for girl. Since the problem states that the family has at least one boy, the GG case is excluded, leaving only 3 equally probable events {GB, BG, BB}. Of these, 1 has two boys, giving the answer of 1/3.
This question is similar to the previous one, except we no longer know which child is the boy mentioned. If he’s the younger child, the two possible events are {GB, BB}. If he’s the older one, two possibilities also exist: {BG, BB}. It might seem that, taken together, there are 4 possibilities, of which 2 have two boys, but in fact the BB case is duplicated between the sets. When combining them, we must be careful to not double-count it. This is an insight that will prove useful in the next question.
Another way to solve the problem is by applying the definition of conditional probability: P(A|B) = P(A and B)|P(B), where P(A|B) is the probability A occurs given that B has occurred. If B is the count of boys, the problem can be expressed as:
P(B=2|B≥1)
= P(B=2 and B≥1)/P(B≥1)
= P(B=2)/P(B≥1)
= P(B=2)/(P(B=1) + P(B=2))
= ((1/2)²)/(2 * (1/2)² + (1/2)²)
= (1/4) / (3/4)
= 1/34. A family has two children. At least one child is a boy born on a Tuesday. What is the probability that the other one is also a boy?
Answer: 13/27
Why should the day of the week that the child is born matter? The given information alters the sample space, which has grown by a factor of 7 * 7, now that the birth day of the week for each child is a factor.
Let ‘w’ represent a birthday that falls on any day of the week and ‘t’ be Tuesday.
If the boy born on a Tuesday is the younger child there are two possibilities permitted by the given condition: {GwBt, BwBt}. If he’s the older one, then there there are also two possibilities: {BtBw, BtGw}. In combining these sets, we must be careful to remove the duplicated BtBt, which is included in both the BwBt and BtBw cases. The combined set is thus: {GwBt, BwBt, BtBw, BtGw} - {BtBt}. The set has 7 (GwBt) + 7 (BwBt) + 7 (BtBw) + 7 (BtGw) - 1 (BtBt) = 27 elements. Of those, 7 (BwBt) + 7 (BtBw) - 1 (BtBt) =13 have two boys, giving 13/27.
13/27 ~ .48, which is closer to 1/2 than to 1/3, the answer from the previous problem.
5. Generalization: A family has two children. At least one child is a boy with a condition that occurs with probability p. What is the probability that the other one is also a boy?
Answer: (2-p)/(4-p)
Let B’ be the count of boys born with the condition.
If p = 1, that means the condition applies to every boy and girl, and it thus gives no information. We recover the “at least one is a boy” problem listed in part 3, as well as its probability (2-1)/(4-1) = 1/3.
In part 4, p = 1/7, where the condition is that the birth day of the week is a Tuesday. Its probability is (2-(1/7)) / (4-(1/7)) = 13/27.
As p, the probability of having the condition, ranges from 0 to 1, the probability that a family has two boys given it has at least one boy with the condition curves from 1/2 toward 1/3.
These problems illustrate the counterintuitive phenomenon whereby two variables, which are usually independent, become conditionally dependent given certain other information.
