I am stucked in some of this question about sample space and probability 
*

*What is the size of the sample space for the following scenario:



Roll $3$ six-sided dice, and discard the highest roll

In this question I know that sample space is all possible outcomes so the set should be $\{1,2,3,4,5\}$, right? So the answer should be $5^3$? I am really confused about this so please help me 


*What is the probability that you will be dealt a 5-card hand that contains no Face Cards?


In this question I know that there is 2,598,960 possible hand but I am stuck in how to find a probability for this question 
Thanks for reading and helping me 
 A: Students are often asked to identify the sample space. However, in many cases there is more than one natural candidate for sample space. It can be convenient, when possible, to use a sample space in which all outcomes are equally likely. In our problem, it is convenient to imagine the dice to blue, white, and red, and to use as sample space all ordered triples $(b,w,r)$, where $b$ is the number showing on the blue, $w$ the number showing on the white, and $r$ is the number showing on the red. This sample space has $216$ elements. 
For the second question, there are $\binom{52}{5}$ ways to choose a $5$-card hand. With proper shuffling, these are close to equally likely.
We now count the "favourables." There are $12$ face cards, and therefore $40$ non-face cards. There are therefore $\binom{40}{5}$ to choose a hand all of whose cards are non-face.
The required probability is therefore $\frac{\binom{40}{5}}{\binom{52}{5}}$.
Now let us do this another way, that illustrates the fact that there is not necessarily only one suitable sample space.
Imagine picking the cards one at a time, and record what we got, in order. There are $(52)(51)(50)(49)(48)$ possibilities, all equally likely.
How many of these sequences are favourable? It is $(40)(39)(38)(37)(36)$. For the probability, divide.
A: The answer to question 1 (that's all I have time for, sorry) is as follows.  First, without "dropping the highest roll", the outcomes are triples
$$
(x_1, x_2, x_3)
$$
of scores from the three dice.  If, say, you roll the three dice, and die 1 yields score $x_1 = 3$, die 2 yields score $x_2 = 5$, and die 3 yields score $x_3 = 2$, then the outcome is $(3, 5, 2)$.  Dropping the highest of these scores would leave the pair $(3, 2)$.
So, your outcomes are pairs of scores, which came from triples of scores with a maximum dropped.
These pairs, however, are not equiprobable.
