###Analysis
Let's answer all the questions at once by supposing you are independently throwing an $m$-sided die (call it $X$) and an $n$-sided die (call it $Y$), where each die has faces numbered 1, 2, 3, *etc.* and all faces have equal probabilities of occurring.
The independence assumption means that any particular outcome where the first die shows the value $i$ (between $1$ and $m$) and the second die shows the value $j$ (between $1$ and $n$) is $1/(nm).$
The *Law of Total Probability* says you can break down the event "$Y$ is greater than $X$" into separate events corresponding to the value shown by $X$ and sum the probabilities of each separate event. Here goes:
1. **$X$ shows a 1.** For $Y$ to exceed $X,$ it must therefore show a 2, 3, *etc*, up to $n.$ There are $n-1$ ways in which this can happen, each of which with probability $1/(nm),$ so the chance is $(n-1)/(nm).$
2. **$X$ shows a 2.** For $Y$ to exceed $X,$ it must therefore show a 3, 4, *etc*, up to $n.$ There are $n-2$ ways in which this can happen, each of which with probability $1/(nm),$ so the chance is $(n-2)/(nm).$
3. **The pattern potentially continues** up to the case where $X$ shows the value $n,$ in which case there's no way $Y$ can exceed $X.$
4. Obviously we need to stop with the case $X=m,$ because it cannot result in any larger number. The probability that $Y$ exceeds $X$ in this case is $(n-m)/(nm).$
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###Solution
Summing all these possibilities *and assuming $n \ge m$* (as in the question) gives
> When $n \ge m,$ $$\eqalign{\Pr(Y \gt X) &= \frac{n-1}{nm} + \frac{n-2}{nm} + \cdots + \frac{n-m}{nm} \\&= \frac{n(n-1) - (n-m)(n-m-1)}{2} \times \frac{1}{nm}.}$$
The left hand factor counts the number of outcomes where $Y$ exceeds $X$ while the right hand factor of $1/(nm)$ converts that to a probability. (The count can be simplified to $m(2n-m-1)$ for computational purposes, but that simpler formula obscures the idea that led to the answer.)
In case $m$ is the larger number, the sum stops with $n-m=0$ giving a simpler value:
> When $n \lt m,$ $$\Pr(Y \gt X) = \frac{n(n-1)}{2} \times \frac{1}{nm}.$$
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###Illustration
For example, the $4(2(6)-4-1)/2 = 14$ possible ways in which a d6 can exceed a d4 are
d4 (X): 1 1 1 1 1 | 2 2 2 2 | 3 3 3 | 4 4
d6 (Y): 2 3 4 5 6 | 3 4 5 6 | 4 5 6 | 5 6
and indeed the count is $5+4+3+2 = 14.$ **The chance therefore is $14/(4\times 6)= 7/12.$** **Similarly, the answers in the other two cases are $27/48$ (d8 *vs* d6) and $44/80=11/20$ (d10 *vs* d8).**
If you're interested in the chance that a d4 will exceed a d6 (an example of where $n$ is less than $m,$ here are the possibilities:
d6 (X): 1 1 1 | 2 2 | 3
d4 (Y): 2 3 4 | 3 4 | 4
and indeed the count is $3+2+1 = 6 = 4(4-1)/2.$
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###Comments
Notice that $14/24$ and $6/24$ do not sum to $1,$ because there's a third possibility of a tie. Evidently there are four ways a tie can occur in the case, giving a chance of $4/(4\times6).$ The chances to add to $1$: $14/24 + 6/24 + 4/24 = 24/24=1,$ as they ought. This is a helpful check of the formulas and the arithmetic.
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###Brute force calculation
Finally, since you have a good SO reputation, you must have an aptitude for programming. **With smallish values of $m$ and $n$ you can just generate all the possibilities and count them up.** Here's an `R` implementation to illustrate:
f <- function(m, n) nrow(subset(expand.grid(Y=1:n, X=1:m), Y > X))
This does not do the division by $nm$ so that you can obtain the *count* of possibilities directly. Examples:
> f(4,6)
[1] 14
> f(6,8)
[1] 27
> f(8,10)
[1] 44
> f(6,4)
[1] 6