I have two fair four-sided dice. The dice are both numbered 3, 4, 5, and 6. I am going to roll both dice and multiply the scores. What is the probability that the product is a multiple of 3?

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01-11-2026
Mathematics

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I have two fair four-sided dice. The dice are both numbered 3, 4, 5, and 6. I am going to roll both dice and multiply the scores. What is the probability that the product is a multiple of 3?

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GinnyAnswer GinnyAnswer · Answer 1 · 2026-11-01T15:25:51-05:00

Each die has four sides: $\{3, 4, 5, 6\}$, resulting in $16$ total possible outcomes when two are rolled.
A product is a multiple of $3$ if at least one factor is a multiple of $3$ ($\{3, 6\}$).
Using the complement rule, identify outcomes where the product is not a multiple of $3$: these are pairs from $\{4, 5\}$, giving $2 \times 2 = 4$ outcomes.
The probability of not getting a multiple of $3$ is $\frac{4}{16} = \frac{1}{4}$, so the probability of getting a multiple is $1 - \frac{1}{4} = \boxed{\frac{3}{4}}$.

Explanation

Analyze the problem and identify total outcomes

We are rolling two fair four-sided dice, each labeled with the numbers $\{3, 4, 5, 6\}$. We need to determine the probability that the product of the two scores obtained is a multiple of $3$. Total outcomes are calculated by multiplying the number of faces on the first die by the number of faces on the second die, which gives $4 \times 4 = 16$ possible outcomes.

Identify the condition and use the complement rule

A number is a multiple of $3$ if it is divisible by $3$. In the set $\{3, 4, 5, 6\}$, the multiples of $3$ are $\{3, 6\}$, and the non-multiples are $\{4, 5\}$. For a product of two numbers $D_1 \times D_2$ to be a multiple of $3$, at least one of the factors must be a multiple of $3$. Instead of counting the favorable outcomes directly, it is easier to use the complement rule: the product is NOT a multiple of $3$ if and only if both factors are NOT multiples of $3$.

Calculate the probability of the complement event

The outcomes where the product is NOT a multiple of $3$ occur only if both dice show values from the set $\{4, 5\}$. Since there are $2$ such values for each die, the total number of combinations where neither die shows a multiple of $3$ is $2 \times 2 = 4$. These pairs are $(4, 4)$, $(4, 5)$, $(5, 4)$, and $(5, 5)$. The probability of this event is $\frac{4}{16} = \frac{1}{4}$.

Determine the final probability

Since the total probability of all possible outcomes is $1$, the probability that the product IS a multiple of $3$ is given by:
\[1 - P(\text{product is not a multiple of } 3) = 1 - \frac{1}{4} = \frac{3}{4}\]
Thus, the probability that the product is a multiple of $3$ is $\frac{3}{4}$.

Examples

This type of probability problem is foundational for understanding games of chance. For instance, in board games using custom dice to decide game events (like "all players gain 3 points if the product is a multiple of 3"), knowing these odds helps in balancing game mechanics so that certain rewards occur at the desired frequency.