Probability of a chess game

Question

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.)

Answer 1

Vlad has three decisions, which leads to eight possible global strategies:

1. What strategy to use in the first game
2. What strategy to use if he loses first
3. 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.

Answer 2

If Vlad plays the first game timidly he has a $.1$ chance of losing the first game and then he can not win the match at all. He has a $.9$ chance of tying and if so he must win the second game to win. He can't win a game if he plays timidly so he must play the second game boldly, in which case he has a $4/9$ chance of winning. So his chances of winning are $.1 * 4/9 = .4$ with this strategy. (The strategy of playing the first game timidly, and the second game boldly if he ties the first.)

If Vlad play the first game boldly he has a $5/9$ chance of losing, in which case he can not win the match at all. He has a $4/9$ chance of winning the first game. If he plays the second game timidly he will have have a $.9$ chance of tying the second game and will win the match and a $.1$ of losing and tying the match. That is a $4/9*.9=.4$ chance of winning. This strategy (playing the first boldy, and the second timidly if he wins) is exactly the same as the first..

If he plays the second game boldly he will have a $4/9$ of winning and winning the match but a $5/9$ of losing and tying the match. $\frac 49*\frac 49 = 16/81$ is obviously not as good as strategy 2.

That's if his goal is to maximize winning. If his goal is to maximize his likely score... that's another question.

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All right expected score:

After the first game he has a score of $k$ his expected score if he plays the second game timidly is $k + .1*0 + .9*.5 = k + .45$. His expected score if he plays the second game boldly is $k + 4/9*1 + 5/9*0 = k + .44444444$ So he should always play the second game timidly.

As for the first game, if he plays timidly $k$ is expected to be $.45$ but if he plays boldly $k$ is expected to be $.444444$ so his best strategy is to play both games timidly. However that is not the best strategy to maximize his ability to win (which is impossible). His expected winnings are $.9$ with a $.01$ probability of losing $0-2$, a $.9*.1 + .1*.9 = .18$ probability of losing $.5 - 1.5$ and $.81$ probabality of tying $1-1$. (So the expected score is $.01*0 + .18*.5 + .81*1 $ to $.01*2 + .18*1.5 + .81*1 = .9$ to $1.1$.)

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If his goal is to maximize his probability of winning and if failing that, the have a maximum score his best strategy (from the first part) is to either play timidly the first game then play boldly if he ties or to play boldly the first game and then play timidly if he wins.

If he plays timidly the first game and loses he should play timidly the second game to maximize his expected score (as he will not win). Thus his probalities are $.1*.1=.01$ that he loses $0-2$. The probability is $.1*.9=.09$ that he loses the first game and ties the second game to lose $.5 - 1.5$. The probability is $.9*5/9 = .5$ that he ties the first game and loses the second to lose $.5 - 1.5$ and the probability is $.9*4/9 = .4$ that he ties the first and wins the second to win. So a probability of winning at $.4$ and a probability of losing at $.6$ and an expected score of $.09*0 + (.09 + .5)*.5 + .4*1.5$ to $.09*2 + (.09+.5)*1.5 + .4*.5 = .895 $ to $1.105$.

If he plays boldly the first game the probailities are $\frac 59*.1 = .05555...$ that he loses $0-2$. The probability is $\frac 59*.9 = .5$ that he loses $.5-1.5$. The probability is $\frac 49*.1 = .044444....$ that hee ties $1-1$ and the probability is $\frac 49*.9 = .4$ that he wins $1.5-5$. With the expected score of $.05555*0 + .5*.5 + .0444444*1 + .4*1.5$ to $.055555*2 + .5*1.5 + .0444444*1 = $ .89444444....$ to $1.10555555....$.

So best strategy is timid first, then bold if he ties or timid if he loses.

......

Meanwhile we can figure out other strategies if the goal is to minimize probability of losing, or winning first, not losing second, maximizing score third, etc.

Answer 3

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 $.1975 \cdot 1 + .4938\cdot .5 + .3086\cdot 0= 0.4444$.