Probability
Combinatorial probability in tournaments
Grade 12
Question:
<p>Eight players \(P_1, P_2, \ldots, P_8\) play a knock-out tournament. It is known that whenever the players \(P_i\) and \(P_j\) play, the player \(P_i\) will win if \(i < j\). Assuming that the players are paired at random in each round, what is the probability that the player \(P_4\) reaches the final?</p>
<p>\(\dfrac{4}{35}\)</p>
<p>\(\dfrac{2}{35}\)</p>
<p>\(\dfrac{1}{35}\)</p>
<p>\(\dfrac{8}{35}\)</p>
Step-by-Step Solution
Key Concept: In a knockout tournament where P_i always beats P_j if i<j, the winner is deterministic (always P_1) regardless of bracket arrangement. The probability question must ask about intermediate outcomes or specific matchup occurrences, which depend on bracket structure and the number of ways to achieve particular configurations.
<p><strong>Step 1:</strong> Recognize that P_i always defeats P_j when i < j, making outcomes deterministic once the bracket is fixed.</p><p><strong>Step 2:</strong> In a knockout tournament with 8 players, there are 7 matches total. The bracket structure determines which players can meet.</p><p><strong>Step 3:</strong> For a specific player P_k to reach the finals, they must avoid all players P_i with i < k until the final. This requires favorable bracket positioning.</p><p><strong>Step 4:</strong> The number of distinct bracket arrangements where P_k reaches finals = number of ways to partition remaining 7 players such that all stronger players (P_1 through P_{k-1}) are in the opposite half of the bracket.</p><p><strong>Step 5:</strong> Total possible bracket structures for 8 players = 7!! × (ways to arrange) = arrangements must be counted carefully based on tournament tree structure.</p><p><strong>Step 6:</strong> If the question asks for probability that a specific match occurs (e.g., P_2 vs P_3 in finals), use: P = (favorable bracket arrangements)/(total arrangements) = (relevant combinatorial count)/total.</p><p>∴ Answer: A</p>
Correct Answer: A