Assume the following scenario:
I have a magic dice tower that destroys 90% of all dice that enter it, only allowing 10% of dice through the tower to have a readable outcome (this is independent for each die that enters the tower). If I have infinite dice to place into this tower, how many dice minimum do I have to put in the tower to have a "guranteed" (99% chance) of having at least one die come out the bottom of the dice tower reading a roll of "6"?
My Attempt: Let the chance of surviving the dice tower be event A and let the chance of a die roll being "6" be event B. Since I want one to occur after another and event B is dependent on event A, I multiply the chance of B occurring after event A to get:
$P(A and B) = P(A) * P(B) = (1/10) * (1/6) = (1/60)$ for any given die I place into the tower.
Now to calculate n, the # of dice it takes to get at least a 99% chance of this occurrence, I run into an issue. I've researched a bit and find I can use:
$(1 - p)^n >= (1 - D)$ , where $p = P(AandB) = (1/60)$, and D = desired probability (99% in this case)
to find n (my goal), but I don't understand why this formula works. I initially tried to rearrange the binomial probability formula to solve for n, but the factorials in the formula make that unclear to me.
I appreciate anyone willing to help explain where this formula comes from and how it applies here.
There are two possible interpretations of the question. In the first interpretation, the tower always destroys 90% of the dice put into it and rolls the rest (so if I put 10 dice in, I always get 1 die out). In the second interpretation, the machine takes each die independently and either destroys it (90% chance) or rolls it (10% chance). I will assume the latter.
In this case, each die has a $10\% \cdot \frac{1}{6} = \frac{1}{60}$ chance of displaying a 6, and these events are mutually independent. The probability that a given die does not display a 6 is thus $1 - \frac{1}{60} = \frac{59}{60}$.
The probability that all $n$ dice rolled do not display a 6 is thus $(\frac{59}{60})^n$ by mutual independence. Thus, the probability that at least one die displays a six is $1 - (\frac{59}{60})^n$.
Now, we must find the smallest $n$ such that $1 - (\frac{59}{60})^n \geq 99\%$: that is, such that $(\frac{59}{60})^n \leq \frac{1}{100}$: that is, such that $n \log(\frac{59}{60}) \leq \log(\frac{1}{100})$: that is, such that $n \geq \frac{\log(1/100)}{\log(59/60)}$ (note that we must reverse the inequality since $\log(59 / 60)$ is negative).
Now, we use our calculator to find that $\frac{\log(1/100)}{\log(59/60)}$ is approximately $274.001$. So the required $n$ is $275$.