A fair die is rolled three times. What is the probability that
a) the second and third rolls are both larger than her first roll? [Ans:$P(A)=55/216$]
b) the result of her second roll is greater than that of her first roll and the result of her third roll is greater than the second? [Ans: $P(B)=20/216$]
I have the answers to these questions, but I'm more concerned about how to get to the answer or the idea behind it. All I know is that the sample space is $6^3 = 216$. Can someone please help me out? Thanks in advance.
 A: For part (a), first consider the simple case where we roll two dice. What is the probability that the second roll is greater than the first?
Assuming a six-sided dice, the total number of outcomes $(i,j)$ is $6^2$. We need to count the outcomes in which the second roll is greater than the first; i.e. the number of pairs where $j > i$. 
Any set of two distinct integers $\{a,b\}$ must have one smaller and one larger element; let us agree that the smaller number indicates the result of the first roll, and the larger number is the result of the second roll. 
It follows that the number of such pairs is $6\choose2$, and our probability is $$\frac{6\choose2}{6^2}$$ Can you use this result to reason about three rolls? Or use independence of events?
A similar argument holds for part (b). Any set of three distinct integers $\{a,b,c\}$ must have a smallest, middle, and largest element. Let the convention now be the smallest is roll 1, the middle is roll 2, and the largest is roll 3. 
The total number of outcomes $(i,j,k)$ is $6^3$, and we have $6\choose3$ outcomes  for which $i<j<k$, so the result is $$\frac{6\choose3}{6^3} = \frac{20}{216}$$ Can you generalize this result to an increasing sequence of rolls of length $m$?
Edit: Further clarification for (a)
The set of outcomes $(i,j,k)$ for which $j > i$ and $k > i$ can be partitioned as follows (letting $i$ = roll 1, $j$ = roll 2, $k$ = roll 3):
roll 3 > roll 2 > roll 1
roll 2 > roll 3 > roll 1
(roll 2 = roll 3) > roll 1
We can simply count the number of events in each partition. We just argued from (b) that the number of outcomes where roll 3 > roll 2 > roll 1 is $20$. By symmetry, the number of outcomes where roll 2 > roll 3 > roll 1 is also $20$. 
Finally, we need to count the outcomes where (roll 2 = roll 3) > roll 1. This is the same as the number of outcomes for which roll 2 > roll 1 in the 2-dice case (hence the hint), which we established is ${6\choose2} = 15$.
Tallying these up, we find the total is $20+20+15=55$, as desired.
A: a) If the first die is a 6, then there is no way the other two can be larger. If the first is a 5, then there is only one way both the other dice can be larger (I.e. a 6 and a 6). If the first is a 4, then there are are 4 ways for the others to be larger -- ${(5,5),(5,6),(6,5),(6,6)}$. If it's a three, then we have 9 ways, ${(4,4),(4,5),(4,6),(5,4),(5,5),(5,6),(6,4),(6,5),(6,6)}$. Continuing this process will get you the answer -- $1+4+9+16+25=55$ possible ways.
b) Again, if the first roll is a 6, then no possible ways. If the first roll is a 5, again, no possibilities. If the first is a 4, then there's only one way -- $(5,6)$. If it's a three, then there are 3 ways, ${(4,5),(4,6),(5,6)}$. If it's a 2, there are 6 ways, ${(3,4),(3,5),(3,6),(4,5),(4,6),(5,6)}$. If it's a one, there's 10 ways. This gives us $1+3+6+10=20$ ways.
Sorry if the answer is a bit crude, I'm on the phone :P hope this helps!
A: Condition on the outcome of the first roll.  Let the outcomes of the three rolls be $X_1, X_2, X_3$, in that order. For simplicity of notation, consider the events $A = (X_2 > X_1)$, and $B = (X_3 > X_1)$.  Then by the law of total probability, $$\Pr[A \cap B] = \sum_{k=1}^6 \Pr[(A \cap B) \mid (X_1 = k)] \Pr[X_1 = k].$$  But observe that $A \mid (X_1 = k)$ and $B \mid (X_1 = k)$ are independent events, because $X_2$ and $X_3$ are independent.  Furthermore, we easily see that $$\Pr[X_2 > k] = 1 - \Pr[X_2 \le k] = 1 - \frac{k}{6}.$$  Therefore, $$\Pr[(A \cap B) \mid (X_1 = k)] = \Pr[X_2 > k]\Pr[X_3 > k] = \Pr[X_2 > k]^2,$$ hence $$\Pr[A \cap B] = \sum_{k=1}^6 \Bigl(1 - \frac{k}{6}\Bigr)^2 \frac{1}{6} = \frac{55}{216}.$$
For the second part, an argument by counting is probably easiest, and was described in another answer.  The idea is to regard the outcomes of the three die rolls as an ordered triplet $(X_1, X_2, X_3)$ where each $X_i \in \{1, 2, 3, 4, 5, 6\}$ for $i = 1, 2, 3$.  The desired probability is simply the ratio of the number of desired outcomes divided by the total number of possible outcomes.  Of course, the latter is easy:  the total number of possible outcomes is simply $6^3 = 216$, since each outcome $X_i$ is independent of the others.  The numerator is slightly harder:  we observe that a desired outcome is one in which $1 \le X_1 < X_2 < X_3 \le 6$.  The key observation is to note that because the inequalities between the outcomes are strict, we cannot have any two die rolls equal in value.  Thus, if we count the number of ways to choose three distinct numbers from the set $\{1, 2, 3, 4, 5, 6\}$, this uniquely corresponds to a desired outcome of the three die rolls.  So for example, $\{2, 5, 6\} \iff (X_1, X_2, X_3) = (2, 5, 6)$.  Since there are $\binom{6}{3} = 20$ such ways to choose $3$ distinct numbers from $6$ distinct integers, we find that the desired probability is $$\Pr[X_1 < X_2 < X_3] = \frac{20}{216},$$ as claimed.
To do the above calculation using conditional probabilities is also possible, and requires a slight modification.  In particular, we must consider the probability $$\Pr[(X_3 > X_2 > X_1) \mid X_1 = k].$$  One way to compute this is to see that either $X_3 > X_2$, $X_3 = X_2$, or $X_3 < X_2$, given that both are greater than $X_1 = k$.  That is, $$\Pr[(X_3 > k) \cap (X_2 > k)] = \Pr[X_3 > X_2 > k] + \Pr[X_2 > X_3 > k] + \Pr[(X_3 = X_2) > k],$$ and we already know the LHS probability.  A simple symmetry argument shows that the first two terms on the RHS are equal, so all that remains is to find the last term.  But this is easy:  Since $X_2$ and $X_3$ are independent, we have $$\Pr[(X_3 = X_2) > k] = \sum_{j = k+1}^6 \Pr[X_3 = j]\Pr[X_2 = j] = \sum_{j=k+1}^6 \frac{1}{6^2} = \frac{6-k}{36}.$$  Thus $$\Pr[X_3 > X_2 > k] = \frac{1}{2}\left(\Bigl(1 - \frac{k}{6}\Bigr)^2 - \frac{6-k}{36}\right) = \frac{(5-k)(6-k)}{72},$$ and we have $$\Pr[X_3 > X_2 > X_1] = \sum_{k=1}^6 \frac{(5-k)(6-k)}{72} \cdot \frac{1}{6} = \frac{5}{54} = \frac{20}{216}.$$  As you can see, this is probably not the kind of solution you want to deal with on a first glance at the question.
A: Consider in 1st throw we got number 1
So in second throw we have to get higher number than 1,so there are 5 chances for getting higher number. 
Now consider in 2 Nd throw we got number 2
Now in third throw we have to get higher number than 2, so there are total 4 chances for getting higher number than 2
Hence probability is 
5*4/216 = 5/54
