# Symmetric 6 sided dice, exactly 5 sixes; chance of six > 0.95

• October 7th 2011, 01:49 AM
magicka
Symmetric 6 sided dice, exactly 5 sixes; chance of six > 0.95
Hello, I'm having problems with probability and I hope You can help me! (If this is more suitable for uni level, please move it there)

My problem is : A symmetric dice is rolled 30 times. What is the probability that a 6 turns up exactly 5 times?

My logic:
We define
$\Omega = \{ (a_1, a_2, ... , a_{30} ) | a_i \in {1,...,6}, i=1,..,30 \}$
Obviously cardinality of $\Omega$ is $k(\Omega)=6^{30}$.
Now we focus of cardinality of $A = \{"exactly \,5 \, sixes" \}$. In order to have exactly 5 sixes, we need to have 5 sixes in $(a_1,...,a_{30})$ and rest can be from 1 to 5. So we have $5^{25}$ choices for rest. When we selected numbers, we can "move" block of sixes in any way we want (since it doesn't matter if 5 sixes felt in first 5 throws or last), we can do that in 25 ways*. So $k(A)=25 \cdot 5^{25}=5^{27}$ and we get probability by dividing those two: $P(A)=\frac{5^{27}}{6^{30}}$

*The number of r-permutations of set with n elements is $\frac{n!}{(n-r)!}$, we can consider here a block of 5 sixes as one big block. By this logic we can choose block positions in $\frac{25!}{24!}$, which would be my proof for above selection.

Second problem: How many dice you need to throw in order to have more then 0.95 chance of six.

I do not have a clear idea how to solve this one, my first thought was calculating chance of complement (i.e. chances of numbers 1 to 5 < 0.5), so basically I'm looking at $(\frac{5}{6})^n = 0.05$ and I get n has to be over 16. My problem is how to solve this in more strict way.

This is binomial: $\binom{N}{k}\left( p\right)^k\left(1-p\right)^{N-k}$.
Now for this problem: $\binom{30}{5}\left(\frac{1}{6}\right)^5\left(\frac {5}{6}\right)^{25}$