Three teams (Team 1, Team 2, and Team 3) are playing soccer championship. In eac

Question

Three teams (Team 1, Team 2, and Team 3) are playing soccer championship. In each game, two teams play each other and the third team does not play. The winner of any given game n plays again in game n+1 against the team that did not play in game n, and the loser of game n does not play in game n+1. The probability that Team 1 will beat Team 2 in any game they play against each other is 0.3. The probability that Team 1 will beat Team 3 in any game they play against each other is 0.6. The probability that Team 2 will beat Team 3 in any game they play against each other is 0.8. (Assume that game always results in either win or lose) (a) Determine the probability that the two teams who play against each other in the first game will play against each other in the fourth game. (b) Show that this probability does not depend on which two teams play in the first game.

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Explanation / Answer

we are denoting a team beats other team by ">" this symbol....

possible starting games are

a) team I and team II

b) team II and team III

c) team I and team III

the 4th game between will be same game who will start the ist game ...that means the same team will not win consecutive two times among three consecutive games ....so that the starting team can meet with each other again in the 4th game ....We will study this by case by case ....

case 1)

assuming team I and team II starting the game ....we have two sub cases to show how they will meet with each other again in game 4

II>I III>II I>III

I>II III>I II>III

hence the probability is =.7*.2*.6+.3*.4*.8=.18

case 2)

assuming team II and team III starting the game

II>III I>II III>I

III>II I>III II>I

hence the probability is .8*.3*.4+.2*.6*.7=.18

and if we start the game with I and III we eill get the same result ....So the required probability is .18 hence we can say it is not varying with the starting teams ...as the probability is constant with respect the teams who play the ist game ...

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