# George chess strategy in best out of two games (bold, cautious style)

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.

1. Find the probability player George wins both games given bold style ? This is $p_b^2$ correct?

2. 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. $p_c \cdot p_c \cdot pb$.

3. 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.

• "Show that if George adopts strategy 3 above" -- what is strategy 3? Sep 9 '18 at 1:28
• it should be strategy C, where George will change to cautious if he is ahead. The question I've been giving is a bit ambiguous because you do not know if he starts playing bold, or cautious. in my solution, I consider he can play both styles in the first game. Sep 9 '18 at 2:08

It has to be pretty extreme, but it's possible. Let's say $p_c=.95$ and $p_b = .45.$ The strategy must be that they start bold, play cautious the middle round if they're ahead, else bold, then play bold on the tiebreaker if need be. Then the winning combinations are $(W,T,-),$ $(W,L,W),$ $(L,W,W).$ The probabilities of each of these are $$P(W,T,-)=p_bp_c=0.4275\\P(W,L,W)=p_b(1-p_c)p_b = 0.010125\\P(L,W,W)=(1-p_b)p_bp_b=0.11138.$$ The sum of these is $0.549 > 0.5.$
To make sure we haven't missed anything, the losing combinations are $$P(W,L,L) = p_b(1-p_c)(1-p_b) =0.012375\\P(L,W,L)= (1-p_b)p_b(1-p_b) = 0.13613\\P(L,L,-) = (1-p_b)(1-p_b) = 0.30250$$ which sum to $0.451.$
We cannot have a winning strategy where they start cautious and only play bold if we lose the first round. Then the only winning combinations are $(T,T,W)$ and $(L,W,W)$ which give winning probability $$p_c^2 p_b + (1-p_c)p_b^2 = p_b(p_c^2 + p_b(1-p_c)) \le p_b.$$ Similarly, the strategy of starting cautious then playing bold from then on has winning combinations $$P(T,W,-) = p_cp_b\\P(L,W,W) = (1-p_c)p_b^2$$ for a total probability of winning $$p_cp_b + (1-p_c)p_b^2 = p_b(p_c + (1-p_c)p_b) \le p_b$$
• It's interesting to observe what happens when the (minimum) length of the tournament increases. For "best of 4," George has 0.58 probability of winning if he "defends the advantage" (same $p_b$ and $p_c$ as yours). In the long run, however, his probability of winning declines. Sep 9 '18 at 15:06