Probability of a chess game I have been asked to solve this problem and have literally NO idea where to begin. Any insight would be great!

Vlad is to play a 2-game chess match with Gary and wishes to maximize his chances of winning. To do this, he may select a strategy right before he plays each game: timid or bold. (Note: this means that Vlad may choose his strategy in the second game after he knows the result of the first game.) 

(A few remarks on chess matches: a 2-game match means that the players play exactly two games, with a win counting 1, a draw counting 0.5, and a loss counting 0. The player with more points wins the match. Note that the match may end in a tie, if, for example, each player wins one game.) 

Unfortunately, Gary is the superior player. If Vlad plays timidly, Gary will still win 10 percent of those games, and the rest will be draws. If Vlad plays boldly, Gary will win 5/9 of those games, and lose the rest. 

Assume Vlad selects his strategies optimally, and find (a) the probability that he wins the match, (b) the probability that the match ends in a tie, and (c) the probability that Gary wins the match. (N.B. The problem requires that you determine Vlad's optimal approach to the match.)
Vlad has three decisions, which leads to eight possible global strategies:  


What strategy to use in the first game  
What strategy to use if he loses first   
What strategy to use if he doesn't lose first.  


One of the eight global strategies is to play timidly in all three situations.
Run through the eight global strategies.
In at least one of them, Vlad has a better chance of winning than Gary does.
The expected score for the bold strategy is higher, so the optimal strategy is for Vlad to play bold.

Under the timid strategy, each match ends D (for draw) or L (for Vlad loses) with probability $.9$ and $.1$, respectively. So the 2-game match ends DD (tie) with probability $.9^2 = .81$ and LL, DL, or LD (loss) with total probability $1-.81=.19$. The expected score is $.81\cdot .5 + .19\cdot 0= 0.405$.

Under the bold strategy, each match ends W (for Vlad wins) or L (for Vlad loses) with probability $4/9$ and $5/9$, respectively. So the 2-game match ends WW (win) with probability $(4/9)^2 =.1975$, ends WL or LW (tie) with total probability $40/81 \approx .4938$, and LL (loss) with probability $(5/9)^2 \approx .3086$. The expected score is about $.1975 \cdot 1 + .4938\cdot .5 + .3086\cdot 0= 0.4444$.