sorry for the vague title. I've got this problem I've been working on solving. It is as follows:
Imagine there are a an infinite amount of gumballs with 12 different colors with an equal chance of being drawn. You can either pay 30 cents to get a random gumball of any color, or 50 cents to ensure you get a unique color to the ones you already have. What's the cheapest way to get all 12 differently colored gumballs?
I clarified with my teacher that the idea is to say how many times should you go with the 30 cent option before switching to the 50 cent. Such as, you should do 6 (50% of the total) using the cheaper option before switching to the more expensive option to ensure you don't get any duplicates. Another way to think of it is "once you've gotten 6 unique pulls, then you should start paying extra to only get more unique pulls."
However, I don't believe it would be after 6, I'm thinking it should be after 8. Let me explain:
Every time you pull a unique gumball, your chance to get a duplicate rises by 1/12. However, pulling a duplicate loses you 30 cents, whereas spending extra will "lose" you 20 cents (lose meaning that you could've theoretically gotten that gumball without paying extra). Therefore, for every three 50 cent plays, you could've gotten two duplicates and still "wasted" the same amount of money. However, I'm not really sure if my reasoning is sound here, since you would definitely be up two new unique gumballs.
In addition, it would be really nice if this could be explained with a simple algebraic formula. Such as , with if you could switch 12 out with x, 30 cents out with y, and 50 cents out with z. I think someone in the future looking to answer this problem might appreciate it if they have, say, 21 different items and the prices are 50 cents vs 90.
You should read about the coupon collector problem. If we already have $n$ distinct colors, the chance we get a new color if we do not specify it is $\frac {12-n}{12}=1-\frac n{12}$ The expected number of draws to get the next color is then $\frac {12}{12-n}$ and the expected cost of the next color is $0.3 \cdot \frac {12}{12-n}$. When this exceeds $0.50$ we should switch. That comes at $$\frac 53=\frac {12}{12-n}\\60-5n=36\\n=4.8$$ So you should take random balls until you have $5$ distinct colors