6 teams play a round robin in which every pair meets exactly once. Each game is won by either team with probability 1/2, independently, with no draws. What is the expected number of teams that win all of their games? Give a fraction a/b in lowest terms.

Learn / Library / Probability

Undefeated in round robin

ProbabilityEasy

$6$ teams play a round robin in which every pair meets exactly once. Each game is won by either team with probability $\frac{1}{2}$ , independently, with no draws. What is the expected number of teams that win all of their games? Give a fraction a/b in lowest terms.

Checking an answer needs JavaScript. The hint and the solution open without it.

Hint

Use linearity of expectation over teams.

Solution

Each team plays $5$ games.
A given team wins all of them with probability $(\frac{1}{2})^{5} = \frac{1}{32}$ .
By linearity of expectation, the expected number of undefeated teams is $6 \times \frac{1}{32}$ .
This equals $\frac{6}{32} = \frac{3}{16}$ .
The answer is $\frac{3}{16}$ .

Answer: $\frac{3}{16}$

Next question Back to the library