Discrete Mathematics for Computer Scientists and Mathematicians , Chapter- 2 , Exercise- 2.1, Question-36

Question

Twenty athletes compete in a contest. Each of 3 judges assign 20 different ratings to the 20 athletes. For an athlete to be named winner he must be given the highest rating by at least 2 judges. Compute the fraction of cases for which a winner is named.

Comments

Consider the complement of the required event where no athlete gets highest rating by at least 2 judges. This will happen when all the three judges select 3 different athletes as the highest rated. For this, select 3 athletes out of 20 in $\binom{20}{3}$ ways, permute them in $3!$ ways and then arrange the remaining 19 athletes in $19!$ ways for each judge. So total ways $= \binom{20}{3}3!(19!)^3$.

So final probability $= 1 - \frac{\binom{20}{3}3!(19!)^3}{(20!)^3} = 1 - \frac{6840}{(20)^3} = 1 - 0.855 = 0.145$

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