# Expected number of rolls until all dice are removed

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.

• $\ X_i=1\$ with probability $\ \frac{1}{6}\$ and $\ X_i=0\$ with probability $\ \frac{5}{6}\$, so $\ \mathbb{E}\big(X_i\big)=1\cdot\frac{1}{6}+0\cdot\frac{5}{6}=\frac{1}{6}\$, and $\ \mathbb{E}(X)=\frac{n}{6}\$, not $\ 6n\$. But in any case, $\ \mathbb{E}(X)\$ is the expected number of dice removed on each throw, and it's not clear how this is related to the expected number of throws needed to remove all the dice. Jan 14 at 12:09
• This is equivalent to math.stackexchange.com/questions/26167/…. For each die, the number of rolls it takes for 3 to occur is a geometric random variable with $p=1/6$. The total number of rolls is the maximum of these geometric random variables, hence the linked question. Jan 14 at 17:25

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 $$f$$?

• I see $$f(r_0,r_1,\ldots,r_n) = \sum_{i=0}^{n} \binom{n}{i}\left(\frac{5}{6}\right)^{i}\left(\frac{1}{6}\right)^{n-i} (1+r_i)$$ Jan 14 at 12:45

$$P\text{(one particular dice remains after first k tosses)}=\left(\frac{5}{6}\right)^k$$

$$P\text{(one particular dice is removed within first k tosses)}=1-\left(\frac{5}{6}\right)^k$$

$$P\text{(all dice are removed within first k tosses)}=\left(1-\left(\frac{5}{6}\right)^k\right)^n$$

$$P\text{(all dice are removed within first k-1 tosses)}=\left(1-\left(\frac{5}{6}\right)^{k-1}\right)^n$$

$$P\text{(last dice is removed exactly at the k^{th} toss)}=\left(1-\left(\frac{5}{6}\right)^{k}\right)^n-\left(1-\left(\frac{5}{6}\right)^{k-1}\right)^n$$

Now let $$X=\text{number of tosses until all dice are removed}$$.

$$E(X)=\sum_{k=1}^\infty{k\left(\left(1-\left(\frac{5}{6}\right)^{k}\right)^n-\left(1-\left(\frac{5}{6}\right)^{k-1}\right)^n\right)}$$

And here is a way to approximate $$E(X)$$ for large $$n$$.

Let $$Y=\text{number of tosses until one particular dice is removed}$$.

Consider the curve of the probability density function of $$Y$$. The $$n$$ dice are uniformly distributed in the region under this curve, and the right-most dice has a $$Y$$-value of $$E(X)$$.

$$\left(\frac{5}{6}\right)^{E(X)}=P\left(Y>{E(X)}\right)=(\text{area under the curve to the right of {E(X)}})\approx\frac{1}{n+1}\approx{\frac{1}{n}}$$

$${E(X)}\approx\frac{\ln{n}}{\ln{\left(\frac{6}{5}\right)}}$$

Technology (e.g. desmos) verifies that

$$\lim_{n\to\infty}\frac{E(X)}{\frac{\ln{n}}{\ln{\left(\frac{6}{5}\right)}}}=1$$