I have the following question bugging me and I'm not sure how to get to the correct answer. It's about the probability in a video game:
A) Let's say I play 10 levels of a game, where each level has the chance to spawn a loot-box with a probability of 1%. The Prob. to get 3 loot-boxes in 10 runs would than be solved with the binomial formula P(k=3)=0.0001118478
B) Now if each box has an itempool of 12 items, what is the probability that I get exactly three of them (e.g item A, B, and C). Can I just multiply the result of the binomial formula with (1/12*1/11*1/10) or not?
Are the probabilites of both events considered dependent or independent? And, if I would go to simulate this problem to get to P by brut-force would I choose a hidden-markov model?
Thank you for your help :)
The events "get exactly 3 boxes" and "get exactly boxes with items $A, B, C$" are not independent: if the latter has happened, then the former necessarily has happened. If we call the first event $X$, and the second event $Y$, then $$ \mathbb P(X \cap Y) = \mathbb P(Y) \neq \mathbb P(X)\mathbb P(Y). $$ However, you don't want independence here: you are trying to calculate $\mathbb P(Y)$, which is the same as $\mathbb P(X \cap Y)$, and the calculation you were actually trying to do is as follows: $$ \mathbb P(Y) = \mathbb P(X \cap Y) = \mathbb P(X)\mathbb P(Y \mid X). $$ So, you start with the calculation for $X$, getting exactly three items. Then, you calculate the probability of getting exactly $A, B, C$ under the assumption that you got three items. You can multiply these, as you were trying to do. However, your computation for $\mathbb P(Y \mid X)$ was not quite correct: assuming that the boxes can't contain repeats, you can still get $A, B, C$ in any order, so the result is actually $3! \times \tfrac1{12} \times \tfrac1{11} \times \tfrac1{10}$. If the items in the boxes can repeat, then it's $3! \times \left(\tfrac1{12}\right)^3$.