Question

there are 60 teams attend a chess tournament. every team will play with every other team exactly once. Supposed each team has a 50% chances of winning any games it plays and no ties occue which is the probability that no two teams win the same number of games

Answer 1

The probability that no two teams win the same number of games, P ≈ 2.084 × 10⁻⁴⁵³

The reason for arriving at the above probability is as follows:

The given parameters are;

The number of teams in the tournament, n = 60

The chance of a team winning a game = 50% = 0.5

The number of ties = No ties

The required parameter:

The probability that no two teams win the same number of games

Method:

Calculate the number of ways no two teams win the same number of games, and divide the result by the total number of possible outcomes

Solution:

The number of matches played, n = $\dbinom {60} {2}$ = 1,770

The possible outcomes = 2; Winning or losing

The total number of possible outcomes, $n_p$ = 2¹⁷⁷⁰

The number of games won by each team is between 0 and 59

The ways in which no two teams won the same number of games is given by the games won by the teams to be 0, 1, 2,..., 57, 58, 59

Therefore, the number of ways no two teams won the same number of games, the required outcomes, $n_k$ = 59!

$Probability = \dfrac{Number \ of \ possible \ outcomes}{Number \ of \ required\ outcomes}$

The probability that no two teams win the same number of games is given as follows;

$\mathbf{P(No \ two \ teams \ won \ the \ same \ number \ of \ games)} = \dfrac{n_k}{n_p}$

Therefore;

$P(No \ two \ teams \ won \ the \ same \ number \ of \ games) = \dfrac{59!}{2^{1,770}} \approx \mathbf{2.084 \times 10^{-453}}$

The probability that no two teams win the same number of games, P ≈ 2.084 × 10⁻⁴⁵³

Learn more about probability theory here:

brainly.com/question/1539712