What is the probability of throwing all the even numbers with a dice, before you throw any odd number? I came across this question via a friend, and I am still thinking about it.
So we need to throw all even numbers before we throw any odd number. The first throw can be even or odd, that is simple. However, after that, we can throw the same even number, another even number or an odd number. This infinite loop of possibly throwing the same even number an infinite number of times, gets me confused in finding the answer.
Could someone help me with finding the answer, using possibly recursion techniques?
Kind regards,
Clayton44
 A: 
The first throw can be even or odd, that is simple. However, after that, we can throw the same even number, another even number or an odd number.

You are only concerned that the next throw that is not the first number is also even, and the next throw that is neither of those numbers is the last even face.
Suppose, without loss of generality, that we throw 2 on the first roll.   After that, we don't care how often we may throw 2 until we throw something else, we only care about what that something else will be.   So what is the probability of that roll being even, when given that it is not 2?

An alternate approach.
Let $E_{2,4,6}$ be the event that we see all of those three faces before any other.   We can define $E_{1,2,3}, E_{1,2,4}, \ldots, E_{4,5,6}$ likewise, each for every way there is to select distinct three faces from six.
Now, I posit that each of these events has identical probability, and also that they are mutually exclusive and exhaustive (ie they partition the outcome space).
So...
A: Suppose that there are $k$ even numbers before the first odd. Then we want
$$\sum_{k=3}^{\infty}\left(\frac{f(k)}{6^k}\right)\frac{1}{2}$$
Where $f(k)$ is the number of ways to have $k$ even numbers such that all $3$ evens appear. So $f(k)$ is $3^k$ minus the number of ways that we could have a string consisting solely of $(2,4)$ minus the number of ways we could have a string with $(2,6)$, minus the strings with $(4,6)$, minus the strings with $2$, minus the strings with $4$, minus the strings with $6$. There is only $1$ strings consisting of exactly one number with length $k$. There are $2^k-2$ strings consisting of exactly two numbers with length $k$. Thus
$$f(k)=3^k-3(2^k-2)-3=3^k-3\cdot 2^k+3$$
The answer is then
$$\frac{1}{2}\sum_{k=3}^{\infty}\left(\frac{3}{6}\right)^k-3\left(\frac{2}{6}\right)^k+3\left(\frac{1}{6}\right)^k$$
$$\frac{1}{2}\left[\sum_{k=3}^{\infty}\left(\frac12\right)^k-3\sum_{k=3}^{\infty}\left(\frac13\right)^k+3\sum_{k=3}^{\infty}\left(\frac16\right)^k\right]$$
Which boils down to computing three geometric series sums -- shouldn't be too hard.
