Two Trips with At Least One International in December

Probability · Medium · Free problem

Two trips occurred last year, and each trip independently has probability $p$ of being international. Each trip occurs in a uniformly random month. You learn that at least one of the two trips was an international trip that happened in December.

What is the probability that both trips were international?

Work it out for general $p$, then compute the answer for $p = \frac{1}{2}$. Compare your result to the simpler version where you only know "at least one trip was international" (without the December detail). Why does the extra information change the answer?

Hints

  1. Think about what you are actually conditioning on. The December detail is not cosmetic -- it changes the event space. Consider how the Boy or Girl paradox applies here.
  2. Define $D_i$ as the event that trip $i$ is both international and in December, so $P(D_i) = p/12$. You are conditioning on $D_1 \cup D_2$. Use inclusion-exclusion for the denominator, and note that if both trips are international, $A$ fails only when neither lands in December.
  3. The numerator is $p^{2}[1 - (11/12)^{2}] = 23p^{2}/144$ and the denominator is $p(24 - p)/144$. Divide to get
3p/(24 - p)$. For $p = 1/2$ this yields 3/47$, which is much larger than the
/3$ you get without the month detail.

Worked Solution

How to Think About It: This is the "Tuesday Boy" paradox wearing a travel disguise. The December detail feels irrelevant -- surely knowing *when* the international trip happened shouldn't change the probability that *both* are international? But it does, and understanding why is one of the deepest lessons in conditional probability. The key: "at least one is international in December" is a *more specific* event than "at least one is international," so conditioning on it shifts the posterior differently.

Before doing any math, think about extremes. If December were replaced by a very rare event ("international trip on February 29th in a leap year"), then learning that at least one trip satisfies this extremely specific criterion makes it very unlikely that *both* trips independently satisfy it. The posterior should be close to $0$ for both being international-on-that-rare-day, but the question asks about both being international (in any month), which is a weaker condition. In the limit where the month specification becomes infinitely precise, the answer converges to $p$.

Quick Estimate: With $p = 1/2$ and 12 months, each trip has probability $p/12 = 1/24$ of being an international December trip. The event "at least one international in December" is rare -- probability roughly

\times 1/24 = 1/12$ by inclusion-exclusion (the overlap term is negligible). Given this event occurred, most of the time exactly one trip is the international-December one. The other trip is international with probability $p = 1/2$, so we expect the answer is close to
/2$. More precisely, the slight double-counting correction should pull it just below
/2$. Gut estimate: around $0.48$ to $0.49$.

Approach: Condition on the event $A$ = "at least one trip is international and in December" using the multiplication rule.

Formal Solution:

Let each trip independently be international with probability $p$, and independently fall in each of the 12 months with equal probability

/12$. Define:

  • $D_i$ = event that trip $i$ is international AND occurs in December. We have $P(D_i) = p/12$.

The conditioning event is $A = D_1 \cup D_2$, i.e., at least one trip is an international December trip.

By inclusion-exclusion:

$P(A) = 2 \cdot \frac{p}{12} - \left(\frac{p}{12}\right)^{2} = \frac{p(24 - p)}{144}$

Now we need $P(\text{both international} \cap A)$. If both trips are international (probability $p^{2}$), then $A$ fails only if *neither* trip is in December:

$P(\text{both international} \cap A) = p^{2} \left[1 - \left(\frac{11}{12}\right)^{2}\right] = p^{2} \cdot \frac{23}{144}$

Verification: For $p = 1/2$ this gives

3 \cdot (1/2) / (24 - 1/2) = 11.5/23.5 = 23/47 \approx 0.489$, which lies in $(0, 1)$ and matches our estimate.

So:

$P(\text{both international} \mid A) = \frac{p^{2} \cdot 23/144}{p(24 - p)/144} = \frac{23p}{24 - p}$

Comparison with the simpler version: If we only know "at least one is international" (no December detail):

$P(\text{both international} \mid \text{at least one}) = \frac{p^{2}}{2p - p^{2}} = \frac{p}{2 - p}$

For $p = 1/2$: this gives

/3 \approx 0.333$.

The December version gives

3/47 \approx 0.489$ -- substantially higher. The December detail makes the answer closer to $p$ itself (which is
/2$). In the limit where we replace "December" with an event of probability $\epsilon \to 0$, the answer converges to $p$.

Answer: For general $p$ with 12 months:

$P(\text{both international} \mid \text{at least one international in December}) = \frac{23p}{24 - p}$

For $p = 1/2$: the answer is $\dfrac{23}{47} \approx 0.489$.

Without the December detail, the answer is $\dfrac{p}{2-p}$, which equals

/3$ for $p = 1/2$.

Intuition

This problem is a disguised version of the famous "Tuesday Boy" paradox from probability theory. The naive reaction is that the December detail is irrelevant -- if at least one trip is international, why should it matter *when* it happened? But "at least one international trip in December" is a much more specific event than "at least one international trip." When you condition on a more specific event, you are restricting to a smaller slice of the sample space, and the relative proportions within that slice shift.

Here is the deep intuition: if exactly one trip is international and it happens to be in December, there is no information about the second trip. If both are international, there are *two chances* for at least one to land in December, making the conditioning event more likely. This asymmetry inflates the posterior probability of both being international. As the identifying detail becomes rarer (replace December with a specific day), the answer approaches $p$ -- because in the limit, the rare event almost certainly identifies a *specific* trip, and the other trip is international independently with probability $p$. This is the same logic behind why "I have two children, one is a boy born on a Tuesday" gives a different answer than "I have two children, one is a boy." In quant interviews, this tests whether you truly understand conditional probability or just memorize formulas.

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