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.
This may help: the chance of getting a 1 is $\frac{1}{6}$, so the chance of getting seven 1s is $(\frac{1}{6})^7$.
However, we also want to count getting seven 2s, seven 3s, etc. There are six numbers, so now we have a chance of $6\cdot(\frac{1}{6})^7$ which is $(\frac{1}{6})^6$
Best Answer
On a D-20 dice there are 20 possible outcomes. The probability of rolling any particular face of the dice (assuming that it is a fair dice) is ${1}\over{20}$. These events are independent of each other, this means that the outcomes of the dice do not influence each other. To find a particular two dice combination we have to take the probability of the first outcome and multiply it by the probability of the second outcome. $$\frac{1}{20} * \frac{1}{20} = \frac{1}{400}$$ The chance of rolling a one and a 20 is $\frac{1}{400}$. This is actually the same probability as any particular combination of those two dice. For example, rolling a 3 and a 16 would also be a $\frac{1}{400}$ probability.
Since there are two possible ways to get the combination of one and 20, first dice is one and second is 20, or first dice is 20 and second is one. You add the two probabilities together $$\frac{1}{400} + \frac{1}{400} = \frac{2}{400} = \frac{1}{200}$$
This is the final probability of this event.