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?
[Math] Rock Paper Scissor – Probability Game
probability
Best Answer
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 $p>1/3.$