We roll a fair die until a $5$ appears. What is the expected value of the minimum value rolled?

Question

Question: Given a fair dice, we roll until we get a $5.$ What is the expected value of the minimum value rolled?

Answer is $\frac{137}{60}.$

There is a similar question asked in MSE but I do not understand the method used by Henry.

In particular, if we let $X$ be the minimum value rolled up to and including $5$, then $$E(X) = \sum_{x=1}^5 xP(X=x) = 1 \times \frac12 + 2 \times \frac16 + 3 \times \frac1{12}+4 \times \frac{1}{20}+5 \times \frac15 = \frac{137}{60}.$$ It seems that we are using the fact that $$P(X=x) = \frac{1}{x(x+1)}.$$

I do not understand how to obtain the equation above.

Answer 1

First, $X$ is not the minimum value rolled before obtaining a $5$, it is the minimum value rolled up to and including the first roll that comes up $5$, so that $X=5$ is possible.

The event $X=5$ means that $5$ comes up before any of $1$, $2$, $3$, or $4$ (we don't care about $6$). Since each of the five numbers is equally likely to come up first, $$P(X=5)=\frac15.$$

Now suppose $1\le x\le4$. Now the event $X=x$ means that, among the $x+1$ numbers $1,\dots,x,5$, the number $x$ comes up first, and $5$ second. Thus we have $$P(X=x)=\frac{(x-1)!}{(x+1)!}=\frac1{(x+1)x}\text{ for }1\le x\le4.$$

Answer 2

A start:

If you get a $5$ on the first roll, then

$${\cal E}[x] = 5$$

If you get a $5$ on the second roll, then

$${\cal E}[x] = \frac{\left( \sum\limits_{i=1}^4 i\right) + 6}{5} = \frac{16}{5}$$

If you get a $5$ on the third roll, you write out all $25$ possible die outcomes: $11$, $12$, ... , $43$, $44$,..., $64$, $66$ and count the number of such outcomes in which $1$ is the minimum value appearing; count the number of outcomes where a $2$ is the minimum value appearing, up to $6$ being the minimum value

You have to get a general formula for ${\cal E}[X]$ for a given number of rolls (before a $5$ appears). This value is defined for arbitrary number of rolls... even 10000000.

Then you find the probability of getting each number of rolls before a $5$ appears. The probability it appears on the first roll is of course $1/6$. The probability it appears on the second roll is the probability

Answer 3

The probability that the minimum is $X\in\{1,2,3,4,5\}$ can be found as follows:

The probability of rolling a sequence of length $k$ using numbers from the set $\{X,X+1,\dots,6\}-\{5\}$ is $\left(\frac{6-X}{6}\right)^k$. Similarly, the probability of getting a sequence with values in $\{X+1,\dots,6\}-\{5\}$ is $\left(\frac{5-X}{6}\right)^k$. So the probability of a sequence of length $k$ with all elements in $\{X,\dots,6\}-\{5\}$ and at least one instance of $X$ is $\left(\frac{6-X}{6}\right)^k-\left(\frac{5-X}{6}\right)^k$. The probability of rolling a $5$ right after this sequence is $1/6$.

We sum over $k$ to find the probability of getting a sequence of any length with minimum $X$. This works out to $$P(X)=\frac{1}{6}\sum_{k=1}^\infty \left(\frac{6-X}{6}\right)^k-\left(\frac{5-X}{6}\right)^k = \left\{ \begin{array}{ll} \frac{1}{X} - \frac{1}{1+X} = \frac{1}{X(X+1)} & \text{for } X=1,2,3,4\\ \frac{1}{X} = \frac{1}{5} & \text{for } X=5\\ \end{array}\right. $$ Which is what we wanted.

Answer 4

Assume that a $5$ is first seen on roll $n$.
$5$ is the lowest seen with $n-1$ $6$s then one $5$.
$4$ is the lowest seen with $n-1$ $4$s and $6$s, but not all $6$s then one $5$.
$3$ is the lowest seen with $n-1$ $3$s, $4$s, and $6$s, but not all $4$s and $6$s then one $5$.
$2$ is the lowest seen with $n-1$ $2$s, $3$s, $4$s, and $6$s, but not all $3$s, $4$s, and $6$s then one $5$.
$1$ is the lowest seen with $n-1$ $1$s, $2$s, $3$s, $4$s, and $6$s, but not all $2$s, $3$s, $4$s, and $6$s then one $5$.
$$ \begin{array}{c|l|l} \text{lowest}&\text{chance with $n$ rolls}&\text{sum over $n$}\\ \hline 5&\,\left(\frac16\right)^{n-1}\frac16&\frac15\\ 4&\,\left[\left(\frac26\right)^{n-1}-\left(\frac16\right)^{n-1}\right]\frac16&\frac1{20}\\ 3&\,\left[\left(\frac36\right)^{n-1}-\left(\frac26\right)^{n-1}\right]\frac16&\frac1{12}\\ 2&\,\left[\left(\frac46\right)^{n-1}-\left(\frac36\right)^{n-1}\right]\frac16&\frac16\\ 1&\,\left[\left(\frac56\right)^{n-1}-\left(\frac46\right)^{n-1}\right]\frac16&\frac12 \end{array} $$ Expected value $=5\cdot\frac15+4\cdot\frac1{20}+3\cdot\frac1{12}+2\cdot\frac16+1\cdot\frac12=\frac{137}{60}$