A game for two players, \(A\) and \(B\), can be won by player \(A\), with probability \(p_A\), won by player \(B\), with probability \(p_B\), where \(0 < p_A + p_B < 1\), or drawn. A match consists of a series of games and is won by the first player to win a game. Show that the probability that \(A\) wins the match is
\[
\frac{p_A}{p_A + p_B}.
\]
A second game for two players, \(A\) and \(B\), can be won by player \(A\), with probability~\(p\), or won by player \(B\), with probability \(q = 1 - p\). A match consists of a series of games and is won by the first player to have won two more games than the other. Show that the match is won after an even number of games, and that the probability that \(A\) wins the match is
\[
\frac{p^2}{p^2 + q^2}.
\]
A third game, for only one player, consists of a series of rounds. The player starts the game with one token, wins the game if they have four tokens at the end of a round and loses the game if they have no tokens at the end of a round. There are two versions of the game. In the cautious version, in each round where the player has any tokens, the player wins one token with probability \(p\) and loses one token with probability \(q = 1 - p\). In the bold version, in each round where the player has any tokens, the player's tokens are doubled in number with probability \(p\) and all lost with probability \(q = 1 - p\).
In each of the two versions of the game, find the probability that the player wins.
Hence show that the player is more likely to win in the cautious version if \(1 > p > \tfrac{1}{2}\) and more likely to win in the bold version if \(0 < p < \tfrac{1}{2}\).