Formula and percentages I need help making a formula. Here is the basic concept: I have 1 twenty-sided die (from now on 1d20) and another 16-sided die (1d16). I roll each, and the die which displays the lowest number "wins". This math I can do, and it is obvious that the 1d16 will roll a lower number more often than the 1d20. However if I roll 2d20 vs. 1d16. Which has the better odds of winning (i.e. rolling a lower number)? Simultaneously, a separate but related question of mine is which has a better odds of rolling a 1. 
The reason I want to formulate a formula is because the above is part of a prototype game mechanic, and I am curious as to which die combinations have technically better chances of "beating" an opponent. Thank you for your time.
 A: Your second question is easy to answer: You are more likely to roll a $1$ with $2$ $D20$'s. The probability of that is $1$ minus the probability of not getting a $1$, and so:
$P(1|2D20) = 1-P(1') = 1-\frac{19}{20}\cdot\frac{19}{20}=0.0975$
While the probability of rolling a $1$ with one $D16$ is:
$P(1|1D16) = \frac{1}{16} = 0.0625$
For the other question: which has better odds of getting the lower number: that's a little nasty since you need to pick the minimum of the two... Intuitively I would think one of the $2$ $D20$ is likely to 'beat' the $D16$  ... but you'll have to do the math.
A: if you roll a d16 and a d20
$\frac {16}{320}$ the dice are equal
$\dfrac {\sum_\limits{n=1}^{15} n + 16\cdot 4}{320} = \frac {184}{320}$ the d16 wins
$\dfrac {\sum_\limits{n=1}^{15} n}{320} = \frac {120}{320}$ the d20 wins
1 d16 vs 2d20.
the d16 has a $\frac 1{16}$ chance to roll any number (call it $n$ in [1,16])
A win for the 2 d20 the first d20 is lower than $n$ is $\frac {n-1}{20}$
or the first d20 is greater than or equal to $n$ and the second die is less than n.
$\frac {20(n-1) + (21-n)(n-1)}{400} = \frac {-n^2 +42n - 41}{400}$
2 d 20 wins:
$\frac {1}{6400} \sum_\limits{n=1}^{16} -n^2 +42n - 41\\
\frac {1}{6400} (-\frac 16 (16)(17)(33) + \frac 12 42(16)(17) - 41*16 = \frac {3560}{6400}$
2 d 20 loses:
$\frac {1}{6400} \sum_\limits{n=1}^{16} (20-n)^2 = \frac {2456}{6400}$
Tie 
$\frac 1{6400} \sum_\limits{n=1}^{16} 2(21-n) - 1 = \frac{384}{6400}$
