Modelling the variance of dollar values of the rare/mythic slot in a Magic: the Gathering booster pack I'm trying to model the variance of a certain kind of card that is pulled from a Magic: the Gathering trading card pack. My simplified model is this:

*

*I open a booster pack containing exactly one card. In this instance, the probability of getting a mythic is $w_M=\frac{1}{8}$ and the probability of a rare is $1-w_M=\frac{7}{8}$, so:

*

*$\frac{7}{8}$ of the time, it is a rare card that is selected from a list of $i$ equally likely rare cards having dollar values $R = (r_1,...,r_i)$

*$\frac{1}{8}$ of the time, it is a mythic card that is selected from another list of $j$ equally likely mythic cards having dollar values $M = (m_1,...,m_j)$
A example set of values we can easily work with might be:

*

*$R = (0.1, 0.2, 0.5, 3.0, 3.5)$

*$M = (0.1, 5, 10)$
Calculating the population variance of each component of $R$ or $M$ would be straightforward:
$$\sigma^2_R=\frac{\sum^i_{k=1}{(r_i - \mu_R)^2}}{i} = \frac{(0.1-1.46)^2+(0.2-1.46)^2+(0.5-1.46)^2+(3.0-1.46)^2+(3.5-1.46)^2}{5} = 2.1784$$
$$\sigma^2_M=\frac{\sum^j_{k=1}{(r_i - \mu_M)^2}}{i} = \frac{(0.1-5.033..)^2+(5-5.033..)^2+(10-5.033..)^2}{3} = 16.3356$$
However, I'm not sure how to combine these two values into something that accurately reflects the entire model of the variance of dollar values from getting a rare card $\frac{7}{8}$ of the time and a mythic card $\frac{1}{8}$ of the time. I expect that naively averaging variances like this doesn't work, but I think this incorrect calculation hints at what I'm trying to calculate:
$$\sigma^2_{R\cap M}= \frac{7\sigma^2_{R}+\sigma^2_{M}}{8}=\frac{7\times2.1784+16.3356}{8}=3.9481$$

After doing some research of what I've modelled, this feels like trying to calculate the variance of combined weighted discrete independent uniform distributions - that is to say, each of $R$ and $M$ are discrete uniform distributions in themselves, and I'm trying to combine them together in a way that one distribution is seven times more likely to occur than the other.
When looking at combining continuous independent uniform distributions, I found the Irwin-Hall distribution, but that seems to be modelling equally likely continuous independent uniform distributions and likely more complex than what I'm looking for.

Some ideas I have to work around my inability to combine these distributions would be:

*

*Experimentally calculate variance by simulating a large number of pack openings

*Making some combined uniform discrete distribution $R_7M = R\cup R\cup R\cup R\cup R\cup R \cup R \cup M$ that has seven of each rare value and one of each mythic value (this doesn't work because there isn't typically the same number of mythic and rares)

*Doing something like above but with tuples of overall probability and dollar values like $RM = {(\frac{7}{7i+j},r_1), ..., (\frac{7}{7i+j},r_i), (\frac{1}{7i+j},m_1), ..., (\frac{1}{7i+j},m_j)}$ and then calculating the variance using these tuples somehow?


I'm looking for a way that for arbitrary $P_M, R, M$ I can either transform the problem into something that I can calculate the variance on, or a way to calculate the combined weighed variance of the two distributions. I'm sure that I'm just forgetting some basic concept of statistics and probability - this concept of combined variance feels like something that would have been covered in an introductory statistics class, but I'm just not finding the right words to describe it.
 A: This is a mixture distribution, which in my case is a combination of two distributions.
From the Wikipedia article's section on calculating moments of a mixture distribution, the mean and variance of a mixture distribution is:
$$
\mu = \sum^n_{i=1}w_i\mu_i \\
\sigma^2 = \sum^n_{i=1}w_i(\sigma^2_i + \mu^2_i) - \mu
$$
We can find the variance of the mixture distribution representing the combined distribution of rares and mythics from my question, represented by $R\cap M$. ​First, we calculate the overall mean of our mixture distribution using $w_R$, $\mu_R$, $w_M$, and $\mu_M$:
$$
w_R = \frac{7}{8}\\
w_M = 1 - w_R = \frac{1}{8}\\
$$
$$
\mu_{R\cap M} =w_R\mu_R+w_M\mu_M \\
= \frac{7}{8}\times1.46 + \frac{1}{8}\times5.0333 \\
= 1.9066 \\
$$
Then, we'll calculate the variance of the combined distributions using the variance and mean of both distributions as well as the mixture's mean:
$$
\sigma_{R\cap M} = w_R(\sigma_R^2+\mu_R^2) + w_M(\sigma_M^2 + \mu_M^2) - \mu_{R\cap M} \\
= \frac{7}{8}\times(2.1784+1.46^2) + \frac{1}{8}\times (16.3356 + 5.0333..^2) - 1.9066^2 \\
= 5.3448
$$
