Ro-Sham-Bo-Fu Game Theory

[PairTheBoard]

At the "Moola" website where they give you a penny and let you play games with a chance to double it up 30 times, one of the games is a variation of Rock-Paper-Scissors called Ro-Sham-Bo-Fu.

It's the same Rock-Paper-Scissors except you play for a number of points, N, and you are also given 2 bonus points if you don't use the specified tainted object. For example, you might be playing for N=6 points and, win or lose, you get 2 bonus points if you avoid use of the Rock. Depending on N, what's the optimum frequencies with which you should be choosing your objects?

For example, if the Rock is tainted and you are playing for N=1 point then it's clear you would always choose scissors. Let's call Scissors the Superior non-tainted object and Paper the Inferior non-tainted object in this case. So you always have a Tainted Object, Superior Object, and Inferior Object.

I think as N gets larger the optimum frequency for the Tainted Object goes up along with the optimum frequency for the Inferior Object. Letting "Op" denote optimum frequency I'm guessing it's always true that,

Op(Superior) > Op(Tainted} > Op(Inferior)

and that these approach equality as N gets large.

What does Game Theory have to say about this game?


You know, if this Moola thing got popular enough and you could achieve a consistent edge, playing a Kelly Criteria could turn into some real money.

The Hi-Lo game there could use some analysis as well.

PairTheBoard