Average value of smallest dice rolled?

Question

We roll a dice until we get a $5$. What is the expected value of the smallest dice rolled?

I am looking for a thought process/an idea. Seeing a full solution would be great as well.

Calculating the probability of obtaining $1,\dots, 5$ is what I seem to struggle with. I thought it is simply a Bernoulli distribution, namely that the probability of getting $k$ in the first $i$ throws is simply $1-\left (\frac{5}{6}\right)^{i}$. Intuitively, the more I throw, the more likely I am to get a $1$ or $2$ so these numbers should have a greater weight than $3,4,5$. I can't seem to express this mathematically though.

Answer 1

Let $p_i$ be the relevant probability. So $p_i$ is the probability that the minimal roll is an $i$.

I'll compute $p_1$, the others are similar.

Case I: First roll is a $6$. $6$ can not be the minimum rolled, so that roll changes nothing.

Case II: First roll is a $5$. Then $5$ is the minimal roll.

Case III. First roll is a $2,3,4$. Then the probability that the minimal roll is $1$ is unchanged.

Case IV. First roll is a $1$. Then the minimal roll is guaranteed to be $1$.

Thus $$p_1=\frac 16\times (p_1+0+3p_1+1)\implies p_1=\frac 12$$

Sanity check: If we imagine that we keep on rolling after we get the $5$ then either $1$ or $5$ comes first and the probability of either is $\frac 12$. I expect this argument could be strengthened to compute all the cases.

Answer 2

Some thoughts to get you started: Obviously all the rolls before the first five cannot be a five.   However, each of those rolls is otherwise independent and unbiased.   Yet last roll will be five, thus if all dice rolled before the first five do show sixes, the least value will be five.

Let $X$ count the rolls before the first five, and let $Y$ measure the least result rolled (which can be conditioned over $X$).   Find $\mathsf E(Y)$.

Answer 3

Consider the following procedure

1. Set $t\leftarrow a$
2. Roll a die with result $X$.
3. If $X<t$, set $t\leftarrow X$
4. If $X\ne 5$, go back to step 2. Otherwise terminate

Let $f(a)$ be the expected value of $t$ upon termination of the above. We want to find $f(5)$ (or $f(a)$ for any $a\ge5$). Because the first roll can be either $\color{red}{<a}$ or $\color{green}{=5}$ or ($\color{blue}{\ge a\text{ and }\ne 5}$), we have $$\begin{align}f(1)&=\color{green}{\frac16\cdot 1}+\color{blue}{\frac56\cdot f(1)}\\ f(2)&=\color{red}{\frac16\cdot f(1)}+\color{green}{\frac16\cdot 2}+\color{blue}{\frac 46\cdot f(2)}\\ f(3)&=\color{red}{\frac16\cdot f(1)+\frac16\cdot f(2)}+\color{green}{\frac16\cdot 3}+\color{blue}{\frac 36\cdot f(3)}\\ f(4)&=\color{red}{\frac16\cdot f(1)+\frac16\cdot f(2)+\frac16\cdot f(3)}+\color{green}{\frac16\cdot 4}+\color{blue}{\frac 26\cdot f(4)}\\ f(5)&=\color{red}{\frac16\cdot f(1)+\frac16\cdot f(2)+\frac16\cdot f(3)+\frac16\cdot f(4)}+\color{green}{\frac16\cdot 5}+\color{blue}{\frac 16\cdot f(5)}\\ \end{align} $$ This can be solved from top to bottom, $$f(1)=1, f(2)=\frac32, f(3)=\frac{11}6,f(4)=\frac{25}{12}, $$ and finally $$ f(5)=\frac{137}{60}.$$