Ask The Wizard #383

120/47 = 2.553191 hours

2*sqrt(40) + 10*(π - cos^-1(3/7)) = 37.297725

• Play paper with probability 1/6
• Play rock with probability 1/3
• Play scissors with probability 1/2

Let’s outline some probabilities for the logician aiming to maximize his profits:

• r = Probability of playing rock
• p = Probability of playing paper
• s = Probability of playing scissors

It's clear that r+p+s = 1. Given r and p, s = 1-r-p.

In a scenario where both logicians are playing, they would be indifferent to the choices of their opponent, particularly if they randomize their selections.

The predicted gain from the opponent playing rock is calculated as: 3p - (1 - r - p) = 4p + r - 1.

For the opponent playing paper, the expected earnings would be: 2(1 - r - p) - 3r = 2 - 5r - 2p.

When the opponent chooses scissors, the anticipated win is: r - 2p.

These three equations must all yield the same result.

By equating the first and third equations, we find:

4p+r-1 = r - 2p

p = 1/6

By equating the first and second equations, we arrive at:

2 -5r - 2p = 4p + r -1

We already have that p=1/6, which conveniently allows us to deduce that r=1/3.

s=1-r-p = 1-(1/3)-(1/6) = 1/2.

Therefore, each player should choose paper with a probability of 1/6, rock with a probability of 1/3, and scissors with a probability of 1/2.

In a broader context, the strategy for this puzzle is to allocate each selection based on the potential monetary changes that would happen if the other two symbols were chosen.