However, since the better team is facing a team that is pretty good (wins more than it loses), it also must have a probability of victory below its historical rate of 0.9
A bad team can win more than it looses when it has very bad opponents.
For example: If I play chess against Magnus Carlsen, then I won't be as good as his regular opponents. No matter whether I have 60% win rate or not, Carlsen will have better chances against me than 'normally' (than against his historical opponents).
My naïve approach is to calculate the probabilities as the "proportion
of victory probability owned by team" (as I'm calling it).
$$ P(\text{team 1 wins while team 2 loses}) = \dfrac{0.6}{0.6 + 0.9} =
> 0.4\\ P(\text{team 2 wins while team 1 loses}) = \dfrac{0.9}{0.6 + 0.9} = 0.6 $$
How reasonable is this calculation?
Even when you ignore the idea that the two teams with 60% and 90% winrates may have different opponents of different strengths, then the percentages may not be compared without additional information. Say that the teams have opponents from the same pool, then still you can not say much about the probabilities for the wins in their match. An example is in the following model.
Consider a model game where two teams draw a random number from a normal distribution and the team with the highest number wins. A team might have different performance based on a different mean of their number. Say the draws are
$$X_A \sim N(\mu_A,\sigma^2) \\
X_B \sim N(\mu_B,\sigma^2) \\$$
and if $X_A>X_B$ then A wins and if $X_A< X_B$ then B wins.
To consider the performance against other teams we consider the performance of an other team $\mu_O$. Sometimes you face an opponent with a low $\mu_O$ and other times with a large $\mu_O$. Let's consider this to be distributed as normal
$$\mu_O \sim N(0,\sigma_O^2)$$
And the draw from another team is distributed as
$$X_O \sim N(0,\sigma_O^2+\sigma^2)$$
The difference in a match between the teams A and B and another team O will be distributed as
$$X_A-X_O \sim N(\mu_A,\sigma_O^2+2\sigma^2) \\
X_B-X_O \sim N(\mu_B,\sigma_O^2+2\sigma^2)$$
And we can determine $\mu_A$ and $\mu_B$ based on the 60-th and 90-th percentiles of the standard normal distribution.
$$\mu_A = q_{60} \sqrt{\sigma_O^2+2\sigma^2}\\
\mu_B = q_{90} \sqrt{\sigma_O^2+2\sigma^2}$$
Based on that we can compute the distribution of performance differences in a match between A and B
$$X_A-X_B \sim N(\mu_A-\mu_B,2\sigma^2)$$
and the probability for a win of A can be computed with the CDF of the standard normal distribution
$$P(X_A-X_B>0) = \Phi\left(\frac{\mu_A-\mu_B}{\sqrt{2\sigma^2}}\right) = \Phi\left((q_{60}-q_{90})\frac{\sqrt{\sigma_O+2\sigma^2}}{\sqrt{2\sigma^2}}\right)$$
and it depends on the ratio of within team performance variance $\sigma$ and between teams performance variance $\sigma_O$
sigma = 1
sigma_O = seq(0,4,0.01)
p = pnorm((qnorm(0.60)-qnorm(0.90))*sqrt(1+sigma_O^2/(2*sigma^2)))
plot(sigma_O/sigma, 1-p, main = "probability 90% team wins from 60% team", ylab = "probability")
The extreme situation of nearly 100% for large $\sigma_O/\sigma$ on the right side of the graph can be imagined intuitively with an athletics running duals/competition where players have little individual/personal variability and the randomness of winning depends on the variability in the opponents strengths. Say A always throws a ball around 30 meters and this makes them win about 60% of the duels, and B always throws a ball around 40 meters and this makes them win about 90% of the duels. The 40 meters is a lot more than 30 meters. Then, B will very likely win from A with nearly 100% probability.
