xxii. Boy or Girl and Bertrand's Box

Conditional-probability puzzles where the intuitive answer $\frac{1}{2}$ is wrong, or right only for a different way of learning the information. Rule: write down the sample space and the protocol that produced the information, then apply Bayes.

Puzzle	Information	Answer
boy or girl (Gardner)	“at least one of my two children is a boy"	$P(\text{two boys})=\frac{1}{3}$
	"the older child is a boy"	$\frac{1}{2}$
	"at least one is a boy born on a Tuesday”	$\frac{13}{27}$
	general: “at least one is a boy with a trait of probability $q$”	$\frac{2-q}{4-q}$ (from $\frac{1}{3}$ at $q=1$ to $\frac{1}{2}$ as $q\to 0$)
	you meet a random one of the two children and it is a boy	$\frac{1}{2}$
	the parent picks a random child and says “I have a boy born on Tuesday”	$\frac{1}{2}$
Bertrand’s box	boxes GG, SS, GS; random box, random coin, it is gold; is the other gold?	$\frac{2}{3}$
three prisoners	one of A, B, C is pardoned at random; the warden tells A “B will be executed”	A still $\frac{1}{3}$, C now $\frac{2}{3}$

// Two children; sex (0 = boy) and a trait uniform on t values (t = 7: weekday) are independent.
// P(both boys | at least one is a boy with trait 0), by enumerating the 4t^2 equally likely families.
pair<ll, ll> boyWithTrait(int t) {
    ll cond = 0, both = 0;
    for (int s1 = 0; s1 < 2; s1++) for (int d1 = 0; d1 < t; d1++)
        for (int s2 = 0; s2 < 2; s2++) for (int d2 = 0; d2 < t; d2++)
            if ((s1 == 0 && d1 == 0) || (s2 == 0 && d2 == 0)) { cond++; both += s1 == 0 && s2 == 0; }
    ll g = gcd(both, cond); return {both / g, cond / g};
}
 
// Same families, but the parent reports a uniformly random child: "a boy with trait 0". P(both boys | that report).
double boyWithTraitRandomChild(int t) {
    double cond = 0, both = 0;
    for (int s1 = 0; s1 < 2; s1++) for (int d1 = 0; d1 < t; d1++)
        for (int s2 = 0; s2 < 2; s2++) for (int d2 = 0; d2 < t; d2++)
            for (int who = 0; who < 2; who++) {                  // each report has weight 1/2
                int s = who ? s2 : s1, d = who ? d2 : d1;
                if (s == 0 && d == 0) { cond += 0.5; both += 0.5 * (s1 == 0 && s2 == 0); }
            }
    return both / cond;
}

Proof: the boy-or-girl answers

The four families BB, BG, GB, GG (older first) are equally likely. “At least one boy” removes GG, leaving BB among three equally likely families: $\frac{1}{3}$. “The older is a boy” leaves BB, BG: $\frac{1}{2}$.

Trait of probability $q$. A child is “a boy with the trait” with probability $\frac{q}{2}$, independently for the two children. So $P(\text{at least one})=1-(1-\frac{q}{2}{)}^2=q-\frac{q^2}{4}$, and $P(\text{both boys and at least one has the trait})=\frac{1}{4}(1-(1-q{)}^2)=\frac{2q-q^2}{4}$. The ratio is $\frac{q(2-q)/4}{q(4-q)/4}=\frac{2-q}{4-q}$; for $q=\frac{1}{7}$ it is $\frac{13/7}{27/7}=\frac{13}{27}$. Intuition: the more specific the trait, the less likely that two boys both qualify, so the condition nearly identifies one particular child, as in “the older is a boy”.

Random child. If a uniformly random child is reported, $P(\text{report "boy with trait"}\mid \text{family})=\frac{1}{2}[\text{child 1 qualifies}]+\frac{1}{2}[\text{child 2 qualifies}]$. Taking expectations, $P(\text{report})=\frac{1}{2}\cdot \frac{q}{2}+\frac{1}{2}\cdot \frac{q}{2}=\frac{q}{2}$; in a BB family each child qualifies with probability $q$, so $P(\text{report and BB})=\frac{1}{4}\cdot q$. The posterior is $\frac{q/4}{q/2}=\frac{1}{2}$ for every $q$, including $q=1$ (“I met a random child, a boy”). The same words carry different information under different protocols. Both functions were checked against the closed forms for $t=\mathrm{1..30}$.

Proof: Bertrand's box and the three prisoners

Box. The six coins are equally likely to be drawn. Three are gold, and two of those lie in the GG box. Given a gold coin, the box is GG with probability $\frac{2}{3}$, and then the other coin is gold. (The tempting answer $\frac{1}{2}$ treats “GG or GS” as equally likely, ignoring that GG produces a gold coin twice as often.)

Prisoners. If A is pardoned, the warden names B or C with probability $\frac{1}{2}$ each; if C is pardoned he must name B; if B is pardoned he cannot name B. So $P(\text{"B"})=\frac{1}{3}\cdot \frac{1}{2}+\frac{1}{3}\cdot 1=\frac{1}{2}$, $P(A\mid \text{"B"})=\frac{1/6}{1/2}=\frac{1}{3}$, $P(C\mid \text{"B"})=\frac{1/3}{1/2}=\frac{2}{3}$. This is Monty Hall in disguise: A’s chance cannot change because the warden can always name someone other than A.

Related notes

• Probability and Expectation
• Monty Hall
• Two Envelopes Paradox