## Roll That Die

Roll That Die
A fair six-sided die is rolled 6 times.  What is the probability of rolling a number greater than four at least five times?
Source: mathcontest.olemiss.edu 9/27/2010

Solution
Let’s start by working on a simpler problem. Let’s roll the die 3 times instead of 6 times. What is the probability of rolling a number greater than four at least two times?

STEP 1
We first have to understand the problem. In the case of rolling a six-sided die, a number greater than 4 is either 5 or 6. At least two times means either two times or three times. So we roll the die three times and look out for either the number 5 or 6 showing either two times or three times. The following are examples of desirable outcomes:

Example 1: $\begin{matrix}\underline{n}&\underline{5} &\underline{6}\end{matrix}$
Example 2: $\begin{matrix}\underline{5}&\underline{n} &\underline{5}\end{matrix}$
Example 3: $\begin{matrix}\underline{6}&\underline{6} &\underline{n}\end{matrix}$
Example 4: $\begin{matrix}\underline{6}&\underline{6} &\underline{6}\end{matrix}$

In the above examples, $\mathit{n}=1,2,3\: or\: 4$. The variable $\mathit{n}$ occurs in the first position in Example 1, second position in Example 2, and third position in Example 3. And in Example 4, $\mathit{n}$ does not show up. It is much simpler to work with this variable’s position than to worry about where 5 or 6 are located.

STEP 2
Example 1
can occur in $4\cdot 2\cdot 2=16$ outcomes when $\mathit{n}$ appears in the first position. Example 2 can occur in  $2\cdot 4\cdot 2=16$ outcomes when $\mathit{n}$ appears in the second position. Example 3 can occur in  $2\cdot 2\cdot 4=16$ outcomes when $\mathit{n}$ appears in the third position. Example 4 can occur in  $2\cdot 2\cdot 2=8$ outcomes when there is no place for $\mathit{n}$. Thus, the total number of desirable outcomes is $16+16+16+8=56$.
STEP 3
The total number of possible outcomes when rolling a six-sided die three times is $6\cdot 6\cdot 6=216$.

STEP 4
Thus, the probability of rolling a number greater than 4 at least two times is $\frac{56}{216}=0.2593$.

Now we can turn our attention to the original problem of rolling the die six times and calculate the probability of rolling a number greater than 4 at least five times.

STEP 5
Now the $\mathit{n}$ variable can occur in six different positions as follows:

Example 1$\begin{matrix}\underline{n}&\underline{5} &\underline{6}&\underline{6}&\underline{5}&\underline{5}\end{matrix}$ occurs in $4\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2=2^{7}$ outcomes.
Example 2$\begin{matrix}\underline{5}&\underline{n} &\underline{6}&\underline{6}&\underline{5}&\underline{5}\end{matrix}$ occurs in $2\cdot 4\cdot 2\cdot 2\cdot 2\cdot 2=2^{7}$ outcomes.
Example 3$\begin{matrix}\underline{6}&\underline{5} &\underline{n}&\underline{6}&\underline{5}&\underline{5}\end{matrix}$ occurs in $2\cdot 2\cdot 4\cdot 2\cdot 2\cdot 2=2^{7}$ outcomes.
Example 4$\begin{matrix}\underline{6}&\underline{5} &\underline{5}&\underline{n}&\underline{5}&\underline{5}\end{matrix}$ occurs in $2\cdot 2\cdot 2\cdot 4\cdot 2\cdot 2=2^{7}$ outcomes.
Example 5$\begin{matrix}\underline{6}&\underline{5} &\underline{5}&\underline{6}&\underline{n}&\underline{5}\end{matrix}$ occurs in $2\cdot 2\cdot 2\cdot 2\cdot 4\cdot 2=2^{7}$ outcomes.
Example 6$\begin{matrix}\underline{6}&\underline{5} &\underline{5}&\underline{6}&\underline{6}&\underline{n}\end{matrix}$ occurs in $2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 4=2^{7}$ outcomes.

And when $\mathit{n}$ disappears,
Example 7
$\begin{matrix}\underline{6}&\underline{5} &\underline{5}&\underline{6}&\underline{6}&\underline{5}\end{matrix}$ occurs in $2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2=2^{6}$ outcomes.

Thus, the total number of desirable outcomes is $6\cdot 2^{7}+2^{6}=2^{6}\left ( 2\cdot 6+1 \right )=2^{6}\left ( 13 \right )=64\cdot13=832$.
And the total number of possible outcomes is $6\cdot 6\cdot 6\cdot 6\cdot 6\cdot 6=46656$.

The probability of rolling a number greater than 4 at least five times is therefore
$\frac{832}{46656}=0.0178$.