Four fair six faced dice are thrown simultaneously. What is the probability that all four dice show exactly the same number on their upper faces?

Introduction / Context:
This question explores probability in a situation involving four dice instead of the more common one or two dice. We want the probability that all four dice show the same face. This reinforces the idea of independent events and counting total outcomes when multiple identical random devices are used.

Given Data / Assumptions:

• Four fair six sided dice are thrown at the same time.
• Each die can land showing any integer from 1 to 6.
• All 6^4 = 1296 ordered combinations of faces are equally likely.
• We are interested in outcomes where all four dice show exactly the same number.

Concept / Approach:
Each die acts independently, so the total number of possible outcomes is the product of the possibilities for each die. For the favorable outcomes, all dice must show the same face. Since that face could be 1, 2, 3, 4, 5 or 6, we count the outcomes where all four dice show 1, all show 2, and so on. The probability is the ratio of these favorable outcomes to the total number of outcomes.

Step-by-Step Solution:
Step 1: Each die has 6 possible faces, so for four dice the total number of ordered outcomes is 6^4 = 6 * 6 * 6 * 6 = 1296.Step 2: For an outcome where all dice are the same, the first die can show any of the 6 faces.Step 3: Once the face value is chosen, all four dice must show that same value, so there is exactly 1 such ordered outcome for each chosen face.Step 4: Therefore, the number of favorable outcomes is 6, one for each possible common face value.Step 5: Probability that all four dice show the same number is 6 / 1296.Step 6: Simplify 6 / 1296 by dividing both numerator and denominator by 6 to get 1 / 216.
Verification / Alternative check:
We can also reason from the perspective of conditional probabilities. Suppose the first die shows some value. The probability that the second die matches it is 1/6, and the same is true for the third and fourth dice. The probability that all three remaining dice match the first is (1/6) * (1/6) * (1/6) = 1/216. This matches the result obtained from counting favorable and total outcomes, confirming the correctness of our answer.

Why Other Options Are Wrong:
1/36 would be correct for two dice showing the same face but is far too large for four dice. 2/216 and 4/216 are partial counts that would correspond to only some of the possible common face values and fail to represent the full set of 6 favorable outcomes. 1/1296 would be the probability of one specific ordered outcome, such as all dice showing 1, not the event where they all show the same number regardless of which number that is.

Common Pitfalls:
Students sometimes forget to count all six possible identical outcomes and instead count only one specific case. Another issue is confusing ordered outcomes with unordered ones, but in this problem, the order of dice does matter when counting the total outcomes. Understanding that independence leads to multiplying the number of possibilities for each die is crucial for this type of question.

Final Answer:
The probability that all four dice show exactly the same number is 1/216.