Question

Two teams are playing a series of soccer games, each of which is independent. Team 1 has probability p of winning each game, and team 2 has probability 1 − p of winning each game. The winner of the series is the first team to win two soccer games. Find the expected number of games played. This will be a function of p. Let Y be the total number of soccer games played in the series and first determine the probability for each possible value of Y. Show that this expected number of games played is maximized when p = 1/2.

Answer 1

Here team 1 winning probability = p

Team 2 winning probability = 1 - p

Here the winner of the series of the first team to win two soccer games. So here if  Y be the total number of soccer games played in the series.

so here there is either two games to be played or 3 games is payed if there is an tie after first game.

so here sample space S(Y) = {2,3}

so here Y = 2 when either team 1 wins both game of team 2 wins both games

P(Y = 2) = Pr(Team 1 wins both games) + Pr(Team 2 wins both games)

P(Y=2) = p* p + (1-p) * (1-p) = p² + 1 -2p + p² = 2p² - 2p + 1

now for Y = 3 there will be one game won each by both teams

P(Y= 3) = 2 * p * (1-p) = 2p - 2p²

so

E(Y) = 2 * p(2) + 3 * p(3)

E(Y) = 2 * (2p²-2p +1) + 3 *(2p - 2p²)

= 4p² -4p + 2 + 6p - 6p² = -2p² + 2p + 2 = -2(p² - p + 1)

so here we have to find for what p it would be maximum

dE(Y)/dp = -2 (2p - 1) = 0

2p - 1 = 0

p = 1/2

so here for p = 1/2 the expected value would be highest.