Probability That the Product of Three Dice Is Prime
Three standard 6-sided dice are rolled. What is the probability that their product is a prime number?
Hints
- A prime has exactly two divisors: 1 and itself. What does that force the three dice values to look like?
- The product $d_1 \cdot d_2 \cdot d_3$ is prime only if exactly one die shows a prime and the other two show 1.
- Count: 3 choices for which die shows the prime, 3 primes available on a die ($2, 3, 5$), giving $9$ favorable outcomes out of $216$.
Worked Solution
How to Think About It: A prime number has exactly two positive divisors: 1 and itself. For a product of three positive integers to be prime, exactly one of them must equal that prime and the other two must equal 1. There is no other way -- if two or more factors are $\ge 2$, the product is composite (or at least 4). So the question reduces to: how many ways can one die show a prime and the other two show 1?
Quick Estimate: The primes on a die are $\{2, 3, 5\}$ -- three values. One die shows a prime (3 choices), the other two must each show 1 (probability $1/6$ each). There are 3 ways to pick which die shows the prime. So the count is $3 \times 3 = 9$ favorable outcomes out of $6^3 = 216$ total. That gives $9/216 = 1/24 \approx 4.2\%$.
Approach: Enumerate the favorable outcomes using the factorization constraint.
Formal Solution:
Each die shows a value in $\{1, 2, 3, 4, 5, 6\}$. Let the three results be $(d_1, d_2, d_3)$.
For $d_1 \cdot d_2 \cdot d_3$ to be prime, the product must have the form $p \cdot 1 \cdot 1$ for some prime $p$. The only primes achievable on a standard die are $2, 3, 5$.
The favorable outcomes are all permutations of $(p, 1, 1)$ where $p \in \{2, 3, 5\}$: - Choose which die shows the prime: $\binom{3}{1} = 3$ ways - Choose which prime: $3$ choices - The remaining two dice must each show $1$: $1$ way each
Total favorable outcomes: $3 \times 3 = 9$.
Total outcomes: $6^3 = 216$.
$$P(\text{product is prime}) = \frac{9}{216} = \frac{1}{24}$$
Answer: $\dfrac{1}{24} \approx 4.17\%$.
Intuition
The key insight is that primality is an extremely restrictive condition on a product. Since primes cannot be factored nontrivially, two of the three dice are forced to show 1 -- the least likely value on a die. This makes the probability very small. Note that 4 and 6 on a die are composite and can never contribute to a prime product, and even showing two primes (like $2 \times 3 \times 1 = 6$) gives a composite number.
This type of problem tests whether you can quickly translate a number-theoretic condition (primality) into a combinatorial constraint (exactly one non-one factor). The same reasoning extends: for the product of $k$ dice to be prime, exactly one die shows a prime and the remaining $k-1$ show 1, giving probability $(3k)/6^k$.