Let's roll a pair of dice two times. What is the probabiblity of **not** getting a 7 or 11 total on either of two tosses?

We'll set

We'll set

`N`

equal to the number of rolls we'd like to make. Next, we'll set up a for-loop that will roll each die `N`

times, add their numbers and see if the sum is $7$ or $11$.
`N`

, then fix the `if`

statement to look for a sum of `7`

or `11`

.
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Here's a way of assessing this answer. This table represents all 36 ways two die can be rolled.

(1,6) | (2,6) | (3,6) | (4,6) | (5,6) | (6,6) |

(1,5) | (2,5) | (3,5) | (4,5) | (5,5) | (6,5) |

(1,4) | (2,4) | (3,4) | (4,4) | (5,4) | (6,4) |

(1,3) | (2,3) | (3,3) | (4,3) | (5,3) | (6,3) |

(1,2) | (2,2) | (3,2) | (4,2) | (5,2) | (6,2) |

(1,1) | (2,1) | (3,1) | (4,1) | (5,1) | (6,1) |

For example, $(5,2)$ means die one ended up with a $5$ and die 2 ended up with a $2$. If you look carefully, you'll count $8$ combinations (of 36 total) that result in a $7$ or $11$ (see the diagonal elements and the $(5,6)$ and $(6,5)$ in the upper right).

Now, $8$ in $36$ is a probability of $8/36$ or $2/9$. Thus, the chance of NOT getting a $7$ or $11$ is $1-2/9=7/9$. This is after the dice are rolled just once. (If you don't like the "1-" bit, count all pairs that do not sum to 7 or 11. There are 28 of them, or a $28/36=7/9$ (same answer) chance of NOT rolling a 7 or 11.)

Since the second toss of the dice is independent of the first roll, we use the "and" idea again: "no 7 or 11 after the first toss, **AND** no 7 or 11 after the second toss." This means we multiply $7/9\times 7/9$ and get $49/81=0.6$.