Paradox of Stochastics

Conditional probability theory entails beautiful tasks with counterintuitive solutions. In addition to the well-known sibling problem, I will now briefly deal with another example: "I now have two children. One of them is a boy and was born on a Thursday. What is the probability that the other child is also a boy?"

We choose the result space

$$I = \{(J.MO,J.MO), (J.MO,J.DI), (J.MO,J.MI), \cdots, (M.SO,M.FR), (M.SO,M.SA), (M.SO,M.SO)\} $$ with $$|I| = 196.$$ $$|I| = 196.$$


$$ \begin{array}{rcl} B & = &\{(J.DO,J.MO), (J.DO,J.DI), (J.DO,J.MI), (J.DO,J.DO), (J.DO,J.FR), (J.DO,J.SA), (J.DO,J.SO),\\
& & (J.DO,M.MO), (J.DO,M.DI), (J.DO,M.MI), (J.DO,M.DO), (J.DO,M.FR), (J.DO,M.SA), (J.DO,M.SO),\\
& & (J.MO,J.DO), (J.DI,J.DO), (J.MI,J.DO), (J.FR,J.DO), (J.SA,J.DO), (J.SO,J.DO),\\
& & (M.MO,J.DO), (M.DI,J.DO), (M.MI,J.DO), (M.DO,J.DO), (M.FR,J.DO), (M.SA,J.DO), (M.SO,J.DO) \}\end{array}$$

with \(|B| = 27\) and

$$ \begin{array}{rcl} A & = &\{(J.DO,J.MO), (J.DO,J.DI), (J.DO,J.MI), (J.DO,J.DO), (J.DO,J.FR), (J.DO,J.SA), (J.DO,J.SO),\\
& &(J.MO,J.DO), (J.DI,J.DO), (J.MI,J.DO), (J.FR,J.DO), (J.SA,J.DO), (J.SO,J.DO)\}\end{array}$$

with \(|A| = 13\), so

  • \( P(A \cap B) = P(A) = \frac{13}{196} \),
  • \( P(B) = \frac{27}{196} \),
  • \( P_B(A) = \frac{P(A \cap B)}{P(B)} = \frac{13/196}{27/196} = \frac{13}{27} \neq \frac{1}{2} \).

In doing so, we neglect the fact that in some years there are more Thursdays than other days.