Four fair dice are thrown simultaneously. What is the probability that exactly two of them show the same face (forming one pair) and the remaining two dice show two different faces, both different from the pair and from each other?

Introduction / Context:
This question is a more advanced dice probability problem. You are asked for the probability that when four dice are rolled, the pattern of faces is exactly one pair and two distinct singletons, with all three values different from one another. This means the outcome looks like a,a,b,c with a, b, and c distinct. Counting such patterns correctly requires careful combinatorial reasoning.

Given Data / Assumptions:

• Four fair six sided dice are thrown at the same time.
• Each die has faces 1, 2, 3, 4, 5, 6.
• All 6^4 = 1296 ordered outcomes are equally likely.
• We want outcomes with exactly one pair and two other distinct values.
• Thus the multiset of values must be of the form {a, a, b, c} with a, b, c all different.

Concept / Approach:
To count favorable outcomes:

• First choose the value that will appear twice (the pair).
• Then choose the two distinct values that will appear once each.
• Finally, arrange these four dice so that the multiset a, a, b, c is assigned to positions in all possible ways.

After counting favorable ordered outcomes, we divide by the total 6^4 to obtain the probability. This is a standard counting pattern in problems about dice and repeated values.

Step-by-Step Solution:
Step 1: Choose the value for the pair. There are 6 possible face values, so 6 choices for a. Step 2: From the remaining 5 distinct face values, choose 2 values for the singletons b and c. Number of ways = 5C2 = 10. Step 3: For a fixed choice of a, b, and c, we must count the permutations of the multiset {a, a, b, c} on four dice. Step 4: The number of distinct permutations of a multiset with 4 items where one value repeats twice is 4! / 2!. Step 5: 4! / 2! = 24 / 2 = 12 ordered arrangements. Step 6: Therefore, the total number of favorable outcomes = 6 * 10 * 12 = 720. Step 7: The total number of ordered outcomes when four dice are rolled is 6^4 = 1296. Step 8: Probability = favorable / total = 720 / 1296. Step 9: Simplify 720 / 1296 by dividing numerator and denominator by 72 to obtain 10 / 18, and then by 2 to get 5 / 9.

Verification / Alternative check:
To verify, consider that all possible value patterns for four dice can be categorized by the type of repetition: four of a kind, three of a kind plus a distinct singleton, two disjoint pairs, exactly one pair and two distinct singletons, and all four distinct. You can compute the counts for all categories using standard formulas and confirm that the sum is 1296. The count 720 for the exactly one pair and two distinct singles category fits consistently in this breakdown, and the resulting probability 5/9 is therefore trustworthy.

Why Other Options Are Wrong:
Option 4/9 would correspond to 4 * 1296 / 9 = 576 favorable outcomes, which is too small. Option 7/18 is approximately 0.388 and does not arise from any natural counting split for this pattern. Option 1/3 and option 2/9 are both significantly smaller probabilities and would imply far fewer favorable cases than we actually count. None of these incorrect options match the careful enumeration of 6 choices for the pair value, 10 ways to pick the two singletons, and 12 permutations of each multiset.

Common Pitfalls:
A common mistake is to forget that b and c must be distinct from each other and from a, which leads to counting patterns like a, a, a, b or a, a, b, b that do not satisfy the condition. Another frequent error is to assume the dice are indistinguishable and use combinations instead of permutations, which ignores the fact that each die is a separate position. Some students also double count permutations when dealing with repeated values. Always fix the value counts first, then count permutations with the correct division for identical items, and finally multiply by the number of ways to choose the distinct values themselves.

Final Answer:
The probability that exactly one pair appears and the other two dice show different faces is 5/9.