Say I'm buying a pack of cards for $4.
The pack contains 5 cards, though I don't think this really impacts the scenario.
Simple Scenario
In a simple scenario, a pack might have a single type of insert that has 1:2 odds for a 50/50 chance at opening a pack with an insert in it.
In this scenario I would value an insert at $8 because that's how much it would cost me on average to get an insert and I'm valuing the base cards at $0.
Complex Scenario
In the more complex scenario, the pack has the following odds:
Type A) 1:1 - Guaranteed to get one and only one of this type
Type B) 1:3
Type C) 1:5
Type Z) ~ The default base you get if it's not one of the above cards
You will never get 2 of a single type in a pack. It's possible to get types A, B, & C all in a single pack.
QUESTION: What is the value of each type of card?
If I bought 100 packs at $4, I spent $400 and would expect (on average) to have received:
100 Type A cards
33 Type B cards
20 Type C cards
But I don't know how to divide & assign that $400 to these 153 cards as they don't seem like they should have equal weight.
My napkin math says divide total money spent by number of types
400/3 = 133.33
Then divide that by number of cards
133.33/100 = 1.33
133.33/33 = 4.04
133.33/20 = 6.66
Not sure if that seems right though.
There isn't a unique answer to this question. If $a,b,c$ are the values of types A,B,C, respectively, all that we can say is that $$100a+33b+20c=400\tag1$$ There are infinitely many solutions to this equation. I didn't completely follow the logic behind the method you used, but the way I did it came up with the same answer you did.
I would think that type C is worth $5$ times as much as type A, and type B is worth $3$ times as much, so we would get $$\begin{align}300a&=400\tag2\\ a&=\frac43\approx1.33,\\b&=4\\c&=\frac{20}3\approx6.67\end{align}$$ Note that in $(1)$ I rounded $33.\overline{3}$ off to $33$ as you did, but in deriving $(2)$ I used the true value. That accounts for the slight differences from your values.
Others may find different valuations more appropriate.