# Loot box probability: Is it advantageous to pay twice the money for 2x the probability

probability

This question stems from loot boxes in a video game. The probabilities are independent.

If it costs 300 gold for a 3 in 1000 chance of winning the top prize, but you have the option to pay 600 gold for a 6 in 1000 chance of winning the same top prize, would you have a better chance of getting the top prize by buying twenty 300-gold boxes or ten 600 gold boxes?

Does the amount of gold you have to spend become a factor?
Does the math change when you compare a 400 gold box with 4 in 1000 chance vs. a 700 gold box with 7 in 1000 chance?

Would anyone be able to show me how to calculate this, please? I know I should show that I've attempted to solve this myself but I wouldn't know where to start. Just a lot of back of envelope trial and error. Thanks for your help!

Following along with the answer below, if you generalize the problem to spending a total of $$n$$ gold that can be broken down into the product of $$pq=n$$, what is better, going for $$p$$ vs $$q$$? Well, treating $$n$$ as a constant, we can say we for a percent chance of $$\frac {\frac n q} {1000}$$ we get $$q$$ chances, so the formula as a function of $$q$$ becomes
$$f(q)=1-(1-\frac n {1000q})^q$$ So taking derivatives, we get $$f'(q)=-(1-\frac n {1000q})^{q-1}\cdot (\frac n {1000q^2})$$
Note that this is strictly negative, so it will be maximized when $$q$$ is minimized. Thus it is always best to take the most expensive available options the least amount of times if you want at least 1 victory. As the other answer states, in exchange you are giving up the chances of multiple victories