Each of five, standard, six-sided dice is rolled once. What is the probability that there is at least one pair but not a three-of-a-kind (that is, there are two dice showing the same value, but no three dice show the same value)?

To solve this problem, we need to calculate the probability that when five standard six-sided dice are rolled, there is at least one pair, but no three-of-a-kind. Here's the step-by-step approach:

### Step 1: Understand the requirements
- We need at least one pair, but no three-of-a-kind.
- This implies there could be configurations like:
  - One pair and three other different numbers (e.g., 2, 2, 3, 4, 5)
  - Two pairs and one different number (e.g., 2, 2, 3, 3, 4)

### Step 2: Calculate the total number of possible outcomes
Each die can roll any of the six numbers, so the total number of possible outcomes is:
\[
6^5 = 7776
\]

### Step 3: Count the favorable outcomes

#### Case 1: One pair and three other different numbers
- Choose 1 number to appear twice: \(\binom{6}{1}\) ways.
- Choose 3 different numbers for the other three dice: \(\binom{5}{3}\) ways.
- Arrange these numbers: \(\frac{5!}{2!1!1!1!} = \frac{120}{2} = 60\) ways.
- The 2 repeated numbers can show up in \(6^2\) ways.

Total for this case:
\[
\binom{6}{1} \times \binom{5}{3} \times \frac{5!}{2!1!1!1!} \times 6^3 = 6 \times 10 \times 60 = 3600 \text{ ways}
\]

#### Case 2: Two pairs and one different number
- Choose 2 numbers to appear as pairs: \(\binom{6}{2}\) ways.
- Choose 1 different number for the fifth die: \(\binom{4}{1}\) way.
- Arrange these numbers: \(\frac{5!}{2!2!1!} = \frac{120}{4} = 30\) ways.
- The repeated numbers can show up in \(6^4\) ways.

Total for this case:
\[
\binom{6}{2} \times \binom{4}{1} \times \frac{5!}{2!2!1!} \times 6 = 15 \times 4 \times 30 = 1800 \text{ ways}
\]

### Step 4: Probability Calculation
The total number of favorable outcomes is:
\[
3600 + 1800 = 5400
\]

The probability is:
\[
\frac{5400}{7776} \approx 0.6944
\]

### Conclusion
The probability that there is at least one pair but not a three-of-a-kind when rolling five six-sided dice is approximately **0.6944**.

---

Would you like further details or have any questions?

Here are some related questions you might find interesting:
1. What is the probability of getting exactly two pairs in five dice rolls?
2. How do we calculate the probability of getting three-of-a-kind in five dice rolls?
3. Can you determine the probability of rolling a full house with five dice?
4. What is the probability of rolling five of a kind with five dice?
5. How does the probability change with four dice instead of five?
6. What is the expected value of the sum when rolling five six-sided dice?
7. What is the probability of rolling a straight (all different values) with five dice?
8. How do probabilities change if the dice are not fair?

**Tip:** When dealing with probability problems involving combinations, it helps to break down the problem into distinct cases and consider each one separately.

Explore the probability of rolling five standard six-sided dice and obtaining at least one pair but no three-of-a-kind. Detailed step-by-step solution includes combinatorial calculations and probability formula applications. Suitable for Grades 10-12 students studying advanced probability and combinatorics.

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