A chess game can result in 3 outcomes, either player can win, or the game can be a tie. The game is played best out of 2.
George can play in 2 styles, bold style with probability $p_b$ of winning, and loses otherwise.
Cautious in which he draws with probability $p_c$, and loses otherwise.
If a game is tied after 2 matches, it keeps going until someone wins. George will play boldly in such tie-breakers.
Find the probability player George wins both games given bold style ? This is $p_b^2$ correct?
Find probability George wins playing cautious in both games 1 and 2 ? This can only be done such that Tie game 1, Tie game 2 and wins game three using bold. $p_c \cdot p_c \cdot pb$.
Assume that $p_b$ is $<0.5$, (so George is the worse player). Show that if George adopts strategy C. above, that for some values of $p_b$ and $p_c$ , he can win more often than not. How do you explain his being able to end up at an advantage?
I am having trouble with part 3.
if we assume for game 1 George plays bold then we have the outcomes (win,tie) which counts as a win. George can have (lose (by bold playing), win (by bold)). but this is a Lose,Win which results in a draw outcome. If he starts playing cautious, then he can win using the outcome Tie,Tie, Win (by bold). so the total outcomes i think are $p_b*p_c$ + $p_c*p_c*p_b$.
A bit stuck here on the 3rd part. any help appreciated.