Calculating the Minimum Tries "n" required to Have a "Guranteed" Compound Event 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.
 A: 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 $275$.
