Question

Three teams participate in a tournament in which each team plays both of the other two teams exactly once. The teams are evenly matched so that in each game, each team has a 50% chance of winning the game. No game can end in a tie. At the end of the tournament, if one team has more wins than both of the other two teams, that team is declared the unique winner of the tournament. Otherwise, the tournament ends in a tie. The probability that the tournament ends in a tie, is

• A. 1/8
• B. 1/4
• C. 3/8
• D. 1/2

Answer 1

Answer: (B)

$T_1$ and $T_2$ plays and
$\Vert lly{T}_2$ and $T_3$ and $T_3$ and $T_1$ Total games played is 3
$P$ (game ends in tie) i.e. every team wins exactly one game
$\text{ Case-1: }{T}_{1}v/s{T}_2\Rightarrow T_1\text{ wins }$
$T_{2}v/s{T}_3\Rightarrow T_2\text{ wins }$
$T_{3}v/s{T}_1\Rightarrow T_3\text{ wins }$
$P(\text{ ties })=\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{8}$
$\text{ Case-2: }{T}_{1}v/s{T}_2\Rightarrow T_2\text{ wins }$
$T_{2}v/s{T}_3\Rightarrow T_3\text{ wins }$
$T_3^{2}v/s{T}_1\Rightarrow T_1\text{ wins }$
$P(ties)=\frac{1}{2}\cdot \frac{1}{2}\cdot \frac{1}{2}=\frac{1}{8}$
Hence $P($ ties $)=\frac{1}{8}+\frac{1}{8}=\frac{1}{4}$