Suppose I keep rolling a die and I stop once I got two consecutive 6. What is the expected number of time to roll the die under the condition that all the rolls are even number?
So, the sample space are sequences like: 262246442266 or 2262466 etc. Note that we cannot just think of the problem as rolling a 3-sided die, see this blog post to understand why.
A hint (from Gil Kalai's page):
The game ends immediately when a number $\notin\{2,4\}$ is thrown.
You need the probability of reaching double-six after $k$ single sixes and $n+k$ twos and fours.
There are ${n+1\choose k}$ ways to place the single sixes, and the probability of each one is $\left(\frac16\right)^k\left(\frac13\right)^{n+k}$.
I get the probability of reaching double-six with no single-six is $\frac1{6^2}\frac32$; with one single-six is $\frac1{6^3}\frac13\left(\frac32\right)^2$, and so on, with an extra factor of $\frac16\frac13\frac32=\frac1{12}$ for each extra six. That gives a total probability of $\frac1{6^2}\frac32\frac1{1-1/12}=\frac1{22}$.
When weighted by the number of rolls, I get $15/121$, so the expected number of rolls, conditioned on all numbers being even, would be $30/11$
Perhaps I'm missing something, but if all of the rolls are even, the probability of rolling a 6 is $\frac{1}{3}$, so rolling two in a row would have a probability of $\big(\frac{1}{3}\big)^2=\frac{1}{9}$. If your chance of success is $\frac{1}{9}$, that means you'd expect to get your result in 9 rolls.
inspired by my colleagues, I think one can approach this problem by Markov chain as follow:
Let the game ends immediately once odd number or two consecutive 6 show up. Let $X$ be the random variable which is defined as the number of rolls when the game ends with 2 consecutive 6 and $0$ otherwise.
Let $e_1=E[X|X_0=2\text{ or }4]$, $e_2=E[X|X_0=6]$, $p_1=P(\text{game ends with two consecutive six}|X_0=2\text{ or }4)$ and $p_2=P(\text{game ends with two consecutive six}|X_0=6)$.
From property of Markov chain, we have: $p_1=\frac{2}{6}p_1+\frac{1}{6}p_2$ and $p_2=\frac{1}{6}+\frac{2}{6}p_1$ so we get $p_1=\frac{1}{22}$ and $p_2=\frac{2}{11}$.
$e_1=\frac{3}{6}\cdot 0+\frac{2}{6}(e_1+p_1)+\frac{1}{6}(e_2+p_2)$ and $e_2=\frac{3}{6}\cdot 0+\frac{2}{6}(e_1+p_1)+\frac{1}{6}\cdot 1$ since $E[1|X_0=2\text{ or }4]=p_1$ and $E[1|X_0=6]=p_2$.
After solving these equations, one can get $e_1=\frac{15}{121}$.
$P(\text{game ends with two consecutive six})=p_1=\frac{1}{22}$.
Denote $A=\text{game ends with two consecutive six}$. Then we have $E[X]=P(A)E[X|A]+P(\text{game ends at odd numbers})E[X|\text{game ends at odd numbers}]=P(A)E[X|A]$.
As a result, the expected number of time to roll the die under the condition that all the rolls are even number$=E[X|A]=\frac{E[X]}{P(A)}=\frac{30}{11}$.
