We are to compute the probability that in 4 independent choices among 4 options (say labeled A, B, C, D, with equal likelihood), exactly one option appears twice, and two other distinct options appear once each, with the fourth position filled by a fourth distinct option — but wait: this would require 4 distinct options total, and one repeated. Since only 4 positions exist, and we want exactly one option repeated, and the other two being different, the only valid pattern is: one option appears t

["SEO Article: Computing the Probability of Exactly One Option Repetition in 4 Independent Choices Among 4 Equal-Likelihood Options", "When making independent choices among four equally likely options—say labeled A, B, C, and D—each with 25% chance, a fascinating combinatorial question arises: What is the probability that in 4 selections, exactly one option appears twice, and two other distinct options each appear once, with one option remaining completely unused?", "This pattern—commonly referred to as “exactly one repetition” or “one pair plus two singles” in probability theory—requires careful counting of feasible outcomes. Contrary to initial intuition, this configuration exactly fits a 4- choice sequence: for example, {A, A, B, C}, where A appears twice, B and C once each, and D not chosen at all.", "In this article, we will compute the probability of this specific distribution among the total number of possible 4-tuples, using combinatorics and probability principles to deliver a complete, clear explanation.", "---", "### Understanding the Scenario", "We select 4 items independently, each selected uniformly from {A, B, C, D}. All four options are equally probable, and choices are independent.", "The pattern we want counts ways to get:", "- Exactly one option appearing twice\n- Two distinct other options each appearing once\n- One option from the four total does not appear at all", "This totals 2 + 1 + 1 = 4 selections, consistent with the 4 spots — no invalid combinations.", "---", "### Step 1: Total Number of Possible Outcomes", "Since each of the 4 choices has 4 options, and selections are independent:", "[\n\ ext{Total outcomes} = 4^4 = 256\n]", "---", "### Step 2: Counting Favorable Outcomes", "To count outcomes matching the desired pattern:", "#### 1. Choose the repeated option\nWe pick one of the 4 options to appear twice.\nNumber of choices: ( \binom{4}{1} = 4 )", "#### 2. Choose two distinct options for the single appearances\nFrom the remaining 3 options, pick 2 to appear once each:\n[\n\binom{3}{2} = 3\n]", "The last option (the one unused) is automatically determined.", "#### 3. Count arrangements (permutations) of the multiset {X, X, Y, Z}", "We now count how many distinct sequences of 4 choices produce exactly two Xs, one Y, and one Z.", "This is a multiset permutation:", "[\n\ ext{Number of distinct arrangements} = \frac{4!}{2! \cdot 1! \cdot 1!} = \frac{24}{2} = 12\n]", "---", "### Step 3: Multiply to Get Total Favorable Outcomes", "[\n\ ext{Favorable outcomes} = \left( \binom{4}{1} \right) \ imes \left( \binom{3}{2} \right) \ imes 12 = 4 \ imes 3 \ imes 12 = 144\n]", "---", "### Step 4: Compute Probability", "[\nP = \frac{\ ext{Favorable outcomes}}{\ ext{Total outcomes}} = \frac{144}{256} = \frac{9}{16}\n]", "---", "### Additional Clarifications", "- Why never 3 or more of the same?\n The condition requires exactly one repetition, so no option can appear three or four times. Thus, the pattern AABB is excluded (two pairs), and AAAA, BBBB are excluded (full repetition).", "- Is it possible to have all distinct?\n Yes: A, B, C, D each appearing once — that gives ( 4! = 24 ) sequences. But this is different from our pattern.", "- What about D not chosen and two singles appearing?\n The unused option is fixed once we pick the repeated and singleton options, so the 12 arrangements apply only if D is unused and A appears twice, B and C once — matching our combinatorial model.", "---", "### Conclusion", "The probability of choosing, among four equally likely options in four independent picks, exactly one option repeated twice and two distinct options once each — with the fourth option absent — is:", "[\n\boxed{\frac{9}{16}}\n]", "This result highlights key principles in discrete probability: combinatorial counting, symmetry principles, and event-defined patterns. Understanding such distributions is essential in gaming theory, statistical sampling, and decision modeling.", "---", "Keywords: probability of repetition, independent choices, multinomial probability, combinatorics in probability, 4 choices, computing probability, combinatorics, statistics tutorial, probability theory, 4-choice scenarios, probability computation", "---", "Note: This article serves as both an educational resource and an SEO-optimized guide, using structured content to improve visibility while ensuring technical accuracy."]