Rock Paper Scissor - Probability Game I heard this recently and it got me thinking.  Let's say you and an opponent are playing rock/paper/scissors.  You know that your opponent can't play rock.  What is the optimal strategy to win? 
My initial approach is to not play paper (since you can only lose/draw so its not optimal) and play rock with probability $$a$$ and scissor with probability $$1-a$$.  If your opponent plays paper w.p $$b$$ and scissor w.p $$1-b$$.  So the expected winning is 
$$E = -3*a*b + a + b$$.  Here is where im stuck.  Taking a derivative doesn't work since the first derivative tells us nothing about whether its maximized, minimized or saddle point.  How else do you approach this problem?
 A: Suppose your opponent plays paper w.p. p and scissors w.p. (1-p):
If you play rock, you will win w.p. (1-p) and lose w.p. p.
If you play scissors, you'll win w.p. p and draw w.p. p.
Your strategy depends on your 'loss function'. If it's symmetrical (i.e. you get equal utility from winning as from losing, just with an opposite sign, with 0 utility from drawing), the following holds. Assume you get 1 pt from winning and 1pt from losing, 0pts for a draw (the number '1' is arbitrary... and real number will produce the same result):
If you always play rock: $E(points)=1-p-p=1-2p.$
If you always play scissors: $E(points) = p + 0(1-p)=p.$
Play scissors iff $p>1-2p$, i.e. if $p>1/3.$
A: If you play rock, you can win or lose
If you play scissors, you can win or draw
If you play paper, you can lose or draw.
Seems, scissors is the best choice directly.
Going by your way, note that $b$ is a constant. You cannot control the opponent. You can only maximize your own chances. So, $$E=(1-3b)a+b$$
This is an increasing linear if $b<\frac 13$ and decreasing otherwise. 
If the function is increasing, you choose the maximum $a=1$, and if the function is decreasing, you choose the minimum $a=0$
