These dice are loaded… mathematically.
Rolling 1 red you have a 17% chance of getting a 6 and an 83% chance of getting a 3.
Rolling 1 green, there’s a 50% chance of a 5 and a 50% chance of a 2
Rolling 1 blue, there’s an 83% chance of rolling a 4 and a 17% chance of rolling a 1.
To solve for the probabilities of red beating blue I added the products of the probabilities of higher numbers of red… in math terms that looks like: .17 * .83 + .83 * .17 = .28… (Probability of red being 6 * bluebeing 4 + Probability of red being 3 * blue being 1.)
Then I played with rolling two dice at a time, again first finding the probability of each potential outcome (this time multiplying the fraction of sides of the first number by the fraction of sides of the second number, e.g. for a red outcome of 9 I did 5/6 * 1/6 + 1/6 * 5/6), and then followed the same procedure as before to find the odds of winning with any given match up.
With two dice, Red beats Blue 52% of the time, Green beats Red 68% of the time, and Blue beats Green 68% of the time. These results are the reverse of those above.