The system can be described by a 5 states Markov chain, with the state described by a number of consecutive six's accumulated so far. The transition matrix is:
$$
P = \begin{bmatrix}
1-p & p & 0 & 0 & 0 \\
1-p & 0 & p & 0 & 0 \\
1-p & 0 & 0 & p & 0 \\
1-p & 0 & 0 & 0 & p \\
0 & 0 & 0 & 0 & 1
\end{bmatrix}
$$
where for the case at hand $p=\frac{1}{6}$ is the probability to get a six at the next rolling of the die.
The classic way to solve this is to consider the expected number of rolls $k_i$ given an initial state $i$. We are interested in computing $k_0 = \mathbb{E}(X)$.
Conditioning on a first move, the following recurrence equation holds true:
$$
k_i = 1 + p k_{i+1} + (1-p) k_0 \qquad \text{for} \qquad i=0,1,2,3
$$
with boundary condition $k_4 = 0$. Solving this linear system yields:
$$
k_0 = \frac{p^3+p^2+p+1}{p^4} \quad
k_1 = \frac{p^2+p+1}{p^4} \quad
k_2 = \frac{p+1}{p^4} \quad
k_3 = \frac{1}{p^4} \quad
k_4 = 0
$$
In[24]:= Solve[{k0 == 1 + k1 p + (1 - p) k0,
k1 == 1 + k2 p + (1 - p) k0, k2 == 1 + p k3 + (1 - p) k0,
k3 == 1 + p k4 + (1 - p) k0, k4 == 0}, {k1, k2, k3, k0,
k4}] // Simplify
Out[24]= {{k1 -> (1 + p + p^2)/p^4, k2 -> (1 + p)/p^4, k3 -> 1/p^4,
k0 -> (1 + p + p^2 + p^3)/p^4, k4 -> 0}}
Substituting $p=\frac{1}{6}$ gives $k_0 = \mathbb{E}(X) = 1554$.
Added A good reference on the subject is a book by J.R. Norris, "Markov chain" (
Amazon). The chapter on discrete Markov chains is
available on-line for free from the author. Section 1.3 discusses finding mean hitting times $k_i$.
Let's calculate the probability, then convert that to odds.
On a fair die, half the numbers are even and half the numbers are odd. So, the probability for a single roll of getting an even number or an odd number is $\dfrac{1}{2}$.
The probability for a specific roll are unaffected by previous rolls, so we can apply the product principle and multiply probabilities for each roll. Each roll has probability of $\dfrac 1 2$ of obtaining the desired result. So, we have:
$$P(E,E,O) = \dfrac 1 2 \cdot \dfrac 1 2 \cdot \dfrac 1 2 = \dfrac 1 8$$
Now, the probability of that not happening is $$1-\dfrac 1 8 = \dfrac 7 8$$
So, the odds are 7:1 against the desired outcome.
Best Answer
EDIT : Thanks to Andre, he gave me a slap in the face right when I needed it, i.e. before my applied analysis exam tomorrow. XD I'll make this answer right before I bring anymore shame on me.
My ex-approach for the case $Y=5$ is indeed making everything more complicated. The more easy and non-complicated way to do this is just use the definition of the expectation : $$ \mathbb E(X \, | \, Y = 5) = \sum_{n=1}^{\infty} \, n \, \mathbb P(X = n \, | \, Y = 5). $$ Now we just need to compute those probabilities. We know that $P(X = 5 \, | \, Y = 5) = 0$, so that removes this term. For the first four terms, note that $$ \mathbb P(X = n \, | \, Y = 5) = \left( \frac 45 \right)^{n-1} \left( \frac 15 \right) = \frac{4^{n-1}}{5^n}, n=1, 2, 3, 4. $$ because the condition $Y=5$ only means for $X$ that the first four rolls cannot take the value $5$, hence the value of those rolls become independent and uniformly distributed over $\{1,2,3,4,6\}$.
For the next rolls, $Y=5$ gives information on the first five rolls but no information on the ones after. Thus $$ \mathbb P(X = n \, | \, Y = 5) = \left( \frac 45 \right)^4 \left( \frac 56 \right)^{n-6} \left( \frac 16 \right). $$ The $n-6$ stands for the number of rolls after the first five rolls which are not a 6 before you actually get your first $6$ (which gives the $1/6$ term). Therefore, the series we look at in the expectation can be computed, after the first 5 terms, as the derivative of a geometric series in $\left( \frac 56 \right)$. I don't wanna compute it right now because I am going to mess it up, I am definitively too tired for this.