## anonymous one year ago I have an unlimited supply of standard 6-sided dice. What's the fewest number of dice that I have to simultaneously roll to be at least 90% likely to roll at least one 6?

1. anonymous

My understanding is, I used complementary counting and count the probability of not rolling a 6, P(6)=1/6. So the probability of not rolling a six is 1-1/6=5/6. Now we are given that we need 90% to be our 'successes' so the remaining 10% are our 'losses'. This means that 10% = 5/6. What do I do from here?

2. phi

I would use the binomial distribution https://en.wikipedia.org/wiki/Binomial_distribution you want Pr(1 or more 6's) >= 0.9 which is the same as Pr(0 6's) < 0.1 using the binomial distribution, with Pr(6) =1/6 we have $\Pr( \text{k successes, n trials}) = \left(\begin{matrix}n \\ k\end{matrix}\right)p^k(1-p)^{n-k}\\ Pr(0)= \left(\begin{matrix}n \\0\end{matrix}\right)\left(\frac{1}{6}\right)^0\left(\frac{5}{6}\right)^n \le 0.1$ solve for n

3. phi

n > log(.1)/log(5/6) =12.6 so n= 13 is the min number of dice to roll to have a 90% chance of getting 1 or more 6's in one roll

4. anonymous

is there a simpler way to do this. I mean without using Binomial and log