When you roll 2 dice there are 36 equally likely options, starting at double one, $(1,1)$, and going all the way to double six, $(6,6)$. Of these, there are 11 equally likely ways to roll a $5$, $(1,5), (2,5), (3,5), (4,5), (5,5), (6,5), (5,1), (5,2), (5,3),(5,4),(5,6)$, so the odds of rolling a 5 is $\frac{11}{36}$.
Similarly there are 18 equally likely ways to roll an odd number (I'll let you think about which combinations these are), so the odds of rolling an odd number are $\frac{18}{36}=\frac{1}{2}$.
At this point you may think the odds of rolling a $5$ or and odd number should be $\frac{11}{36}+\frac{18}{36}=\frac{29}{36}$, however this is not the case. Note that six of the rolls contain both a $5$ and are odd:
$(2,5), (4,5), (6,5), (5,2), (5,4),(5,6)$
If we just add the two probabilities, we'll count these situations twice, even though they're no more likely to occur than any other combination! In reality there are only $11+18-6=23$ dice rolls which contain a $5$ or are odd (see if you can list them), and so the odds of rolling either a $5$ or an odd combination is $\frac{11}{36}+\frac{18}{36}-\frac{6}{36}=\frac{23}{36}$.
This is a simple application of the "principle of inclusion and exclusion", which you may want to look up.
Rephrase the question:
What is the probability of not seeing all $6$ values when rolling a die $7$ times?
Find the probability of the complementary event:
What is the probability of seeing all $6$ values when rolling a die $7$ times?
Use the inclusion/exclusion principle in order to count the number of ways:
- Include the number of ways to roll a die $7$ times and see up to $6$ different values: $\binom66\cdot6^7$
- Exclude the number of ways to roll a die $7$ times and see up to $5$ different values: $\binom65\cdot5^7$
- Include the number of ways to roll a die $7$ times and see up to $4$ different values: $\binom64\cdot4^7$
- Exclude the number of ways to roll a die $7$ times and see up to $3$ different values: $\binom63\cdot3^7$
- Include the number of ways to roll a die $7$ times and see up to $2$ different values: $\binom62\cdot2^7$
- Exclude the number of ways to roll a die $7$ times and see up to $1$ different values: $\binom61\cdot1^7$
Divide the result by the total number of ways, which is $6^7$:
$$\frac{\binom66\cdot6^7-\binom65\cdot5^7+\binom64\cdot4^7-\binom63\cdot3^7+\binom62\cdot2^7-\binom61\cdot1^7}{6^7}=\frac{35}{648}$$
Calculate the probability of the original event by subtracting the result from $1$:
$$1-\frac{35}{648}=\frac{613}{648}\approx94.6\%$$
Best Answer
The easiest way to think about it is that the first roll can be any of the $6$ possible outcomes, and then subsequent rolls can be any outcome except the previous outcome, hence any of $5$ possible outcomes. Therefore there are $6 \cdot 5^9$ such permuations.