There are $n$ fair dice. They are all tossed every time except the dice that are removed. A dice is removed if $3$ is rolled. What is the expected number of rolls? Any help would be appreciated.
My attempt: Consider the case of two dice. The process ends on step one with probability $\frac{1}{36}$, if one of the dice rolls up $6$ and other doesn't the on average $\frac{10}{36}(1+6)$ more steps are needed. Finally if $6$ doesn't roll up on both of the rolls then $\frac{25}{36}(z+1)$ rolls will be needed where $z$ is the expected number of rolls until both dice are removed. Thus
$$z= \frac{1}{36}+\frac{10}{36}(1+6)+\frac{25}{36}(z+1) $$
Is this correct? If it is I can extend it recursively for $n$ dice.
Best Answer
I have a hard time understanding what your event $X_i$ means? Die $i$ removed by what time? Because in the end all of the dice are removed and $X_i=1$ with $p=1$.
If you want to use this approach, you say that you have $n+1$ states (meaning the number of dice left in the game) and you trying to find $r_k=\mathbb E R(k)$, where $R(k)$ is the number or rolls if you start with $k$ dice. You know that $r_0 = 0$ and you can write recursive relations:
$$ \begin{aligned} r_1 &= 1+\frac56r_1 + \frac16r_0 \\ r_2 &= 1+\left(\frac56\right)^2r_2 + 2\left(\frac56\right)\left(\frac16\right)r_1+\left(\frac16\right)^2r_0\\ \ldots \end{aligned} $$
Or you can take a direct approach and calculate the distribution $R(n)$ directly. You can start to calculate the distribution of $R(1)$ by thinking what is the probability that 1 die wasn't remove on first $x$ rolls. And then see that the survival time of at least 1 die from $n$ is a function of survival times of each individual die: $R(n)=f(R_1(1), R_2(1)...,R_n(1))$. Can you see what function is $f$?