I have kind of an interesting and fun question I encountered today and I'm unsure how the derive the answer.
So let's say I have a mystery box that's worth $120 for 10 items. You are able to get repeats and each item is independent of the next.
There are 9 common items, 7 rare items, and 4 epic items in total.
Individually, each item has a 57.4%, 29.7%, and 12.9% chance to drop a common, rare, or epic item, respectively.
What is the value for each individual item (i.e. each individual common, rare, and epic item)?
So I tried an approach finding the weighted value of each item relative to the draw by doing:
$\sum_ {}= \frac{\frac{1}{P(x)}}{\sum_{x=1}^n\frac{1}{P(x)}}$
So it's taking each weighted inverse probability to get the odds of each item. (i.e. for the commons, $\frac{1}{.574} = 1.742$, for the rares, $\frac{1}{.297} = 3.367$, and for the epics, $\frac{1}{.129} = 7.752$, so $ \sum_{x=1}^n\frac{1}{P(x)} $ = 12.861, meaning each individual weight is .1355, .2618, and .6027, for commons, rares, and epics, respectively.
Then I said the value for each individual item (averaged out to $12/item, since 120 for 10), would be
12*.1355 = 1.625
12*.2618 = 3.142
12*.6027 = 7.233
for commons, rares, and epics, respectively. And this is where I'm kind of stuck. This definitely doesn't make sense, as each item is worth less than the average cost per item. I think I just need a different formula to use and was just kind of brainstorming ideas.
What's the correct way to approach this? And how does (or does) there being x amount of each common, rare, or epic item affect the average value per item?
I think I just need a fresh set of ideas or someone who knows a little more about statistics to help me answer this question, thanks :)