If a die is thrown thrice. Find the probability that the largest score is three times the smallest. I have no idea about the answer, but I'm viewing the question this way;
If the smallest score obtained from the any three throws of the die is $1$, then largest among the other two throws must be $3$, and likewise
if the smallest score obtained from any of the three throw of the die is  $2$ then a $6$ must also be scored from the other two throws.
This is how I think it should be done, but I'm not sure how get to the answer. Or if my logic about is correct. So I need to know how to arrive at the answer and with a good or easy to understand logic.
 A: Classify depending on the smallest die.
If the smallest dice is $1$ the largest is $3$ and the third can be $1,2$ or $3$, then the possible outcomes are:
$1,1,3$ ($3$ ways to order them)
$1,2,3$ ($6$ ways to order them)
$1,3,3$ ($3$ ways to order them)
If the smallest dice is $2$ then the largest must be $6$, then the possible outcomes are:
$2,2,6$ ($3$ ways to order them)
$2,3,6$ ($6$ ways to order them)
$2,4,6$ ($6$ ways to order them)
$2,5,6$ ($6$ ways to order them)
$2,6,6$ ($3$ ways to order them)
Adding we find there are $36$ ways in which this can happen, there are $6\cdot6\cdot6$ possible outcomes  when throwing a dice three times. Therefore the probability is $\frac{36}{6\cdot6\cdot6}=\frac{1}{6}\approx 0.167$
A: One can also solve the problem using the inclusion-exclusion principle to count the possible outcomes in each of two cases.
There are $3^3=27$ outcomes in which all of the dice are between $1$ and $3$ inclusive. $2^3=8$ of these have no $1$, $2^3=8$ of them have no $3$ and $1$ of them has neither a $1$ nor a $3$, so there are $$27-2\cdot8+1=12$$ that have a minimum of $1$ and a maximum of $3$.
Similarly, there are $5^3=125$ outcomes in which all of the dice are between $2$ and $6$ inclusive. $4^3=64$ of these have no $2$, $4^3=64$ of them have no $6$, and $3^3=27$ of them have neither a $2$ nor a $6$, so there are $$125-2\cdot64+27=24$$ that have a minimum of $2$ and a maximum of $6$.
Thus, there are altogether $12+24=36$ outcomes in which the maximum is three times the minimum, and the desired probability is therefore $$\frac{36}{6^3}=\frac16\;.$$
A: It is small enough to enumerate, but since you want formulas, divide into cases with subdivisions
Case 1: Extreme #s are 1 & 3
(a) 3rd # is a duplicate: 2 ways $\times \frac{3!}{2!}$ permutations = 6
(b) 3rd # is distinct: 1 way $\times 3!$ permutations = 6
Case 2: Extreme #s are 2 & 6
(a) 3rd # is a duplicate: 2 ways $\times \frac{3!}{2!}$ permutations = 6
(b) 3rd # is distinct: 3 ways $\times 3!$ permutations = 18
Favorable ways = 6 + 6 + 6 + 18 = 36 
Total ways = $6^3$
Pr = $\frac{1}{6}$
