<p>Forty teams play a tournament. Each team plays every other team just once. Each game results in a win for one team. If each team has a 50% chance of winning each game, the probability that at the end of the tournament, every team has won a different number of games is</p>
Step-by-Step Solution
Key Concept: With 40 teams in round-robin, each team plays 39 games. For all teams to have different win counts, they must have exactly 0, 1, 2, ..., 39 wins respectively. The total wins must equal total games played: C(40,2) = 780. The probability is the number of valid tournament outcomes divided by 2^780.
<p><strong>Step 1:</strong> Verify the constraint. In a round-robin tournament with 40 teams, total games = C(40,2) = 780. Total wins distributed = 780.</p><p><strong>Step 2:</strong> For all teams to have different win counts, the teams must have exactly {0, 1, 2, 3, ..., 39} wins in some order. Check: 0+1+2+...+39 = 39×40/2 = 780 ✓</p><p><strong>Step 3:</strong> Count favorable outcomes. If we assign team i to have exactly i wins (a specific ordering), there is exactly ONE tournament outcome that produces this result. By symmetry, there are 40! ways to assign which team gets which win-count.</p><p><strong>Step 4:</strong> Count total outcomes. Each of the 780 games can go either way → 2^780 total possible tournament outcomes.</p><p><strong>Step 5:</strong> Calculate probability = 40!/2^780</p><p>∴ Answer: B</p>
Correct Answer: B