Claudia and Adela are betting to see which one of them will ask the boy they like for his telephone number. To decide they roll dice. If none of the dice are a multiple of 3, Claudia will do it. If exactly one die is a multiple of 3, Adela will do it. If 2 or more of the dice are a multiple of 3 neither one of them will do it. How many dice should be rolled so that the risk is the same for both Claudia and Adela?
1. **Define the probabilities:**
   - The probability that a single die is a multiple of 3 is \(\frac{2}{6} = \frac{1}{3}\).
   - The probability that a single die is not a multiple of 3 is \(\frac{4}{6} = \frac{2}{3}\).

2. **Calculate the probability that none of the \(x\) dice are multiples of 3:**
   - The probability that none of the \(x\) dice are multiples of 3 is \(\left(\frac{2}{3}\right)^x\).

3. **Calculate the probability that exactly one of the \(x\) dice is a multiple of 3:**
   - The probability that exactly one die is a multiple of 3 and the rest are not is given by:
     \[
     \binom{x}{1} \left(\frac{1}{3}\right)^1 \left(\frac{2}{3}\right)^{x-1} = x \cdot \frac{1}{3} \cdot \left(\frac{2}{3}\right)^{x-1}
     \]

4. **Set the probabilities equal to each other:**
   - We need the probability that none of the dice are multiples of 3 to be equal to the probability that exactly one die is a multiple of 3:
     \[
     \left(\frac{2}{3}\right)^x = x \cdot \frac{1}{3} \cdot \left(\frac{2}{3}\right)^{x-1}
     \]

5. **Simplify the equation:**
   - Divide both sides by \(\left(\frac{2}{3}\right)^{x-1}\):
     \[
     \frac{2}{3} = x \cdot \frac{1}{3}
     \]
   - Multiply both sides by 3:
     \[
     2 = x
     \]

6. **Verify the solution:**
   - If \(x = 2\), then:
     - The probability that none of the dice are multiples of 3 is \(\left(\frac{2}{3}\right)^2 = \frac{4}{9}\).
     - The probability that exactly one die is a multiple of 3 is \(2 \cdot \frac{1}{3} \cdot \left(\frac{2}{3}\right) = 2 \cdot \frac{1}{3} \cdot \frac{2}{3} = \frac{4}{9}\).

Since both probabilities are equal, the solution is correct.

The final answer is \( \boxed{ x = 2 } \).