Loot box probability: Is it advantageous to pay twice the money for 2x the 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!
 A: 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
A: For the 300-gold version, the chance of not getting the top prize with one purchase is $1-\frac3{1000}$; therefore the chance of not getting the top prize with $20$ purchases (assuming the game makes each purchase independent from the others) is $(1-\frac3{1000})^{20}$, making the chance of getting the top prize at least once
$$
1 - (1-\tfrac3{1000})^{20} \approx 5.832\%.
$$
For the 600-gold version, the chance of not getting the top prize with one purchase is $1-\frac6{1000}$; therefore the chance of not getting the top prize with $10$ purchases is $(1-\frac6{1000})^{10}$, making the chance of getting the top prize at least once
$$
1 - (1-\tfrac6{1000})^{10} \approx 5.84\%.
$$
So this option gives a larger chance of getting the top prize at least once.
I think the reason the second option gives a larger chance of getting the top prize at least once is that the first situation makes it more likely to get the top prize more than once, which "siphons off a bit of the probability" so to speak.
