Rock Paper Scissors
You are going to participate in a rock-paper-scissors game. If you win the game, you get one dollar, if you lose, you have to pay one dollar, and if the game ends as a draw, you do not exchange any money.
Your opponent has already informed you that he never plays rock. What
is your best strategy to participate in this game?
Link to the problem on Twitter: https://twitter.com/Riazi_Cafe/status/1664986539582017537
As some friends mentioned, the correct strategy in this game is to
play scissors with probability \(2/3\)
and rock with probability \(1/3\).
The mathematical definition of the best strategy in this type of games
is the strategy that has the highest average utility in the worst case.
It is obvious that playing paper is of no use to us (since we either
lose or get a draw in that case). So we either play rock or scissors.
Let \(p\) be the probability that we
play scissors and \(1-p\) be the
probability that we play rock.
Based on this, if our opponent plays scissors, the game is a draw
with probability \(p\) and a win for us
with probability \(1-p\). If the
opponent plays paper, with probability \(p\) we win the game and with probability
\(1-p\) we lose the game. So we have to
choose the probability \(p\) in such a
way that our minimum utility is maximized in the above two cases. In
other words, \(p\) should be chosen in
a way that \[\min\{p-(1-p), 1-p\}\] is
maximized. The above expression is maximized for \(p = 2/3\).
Link to the solution on Twitter: https://twitter.com/Riazi_Cafe/status/1665301277142654976