A method for simplifying manual random number generation? Is there a good way to take a more complicated dice-rolling mechanism and create a simpler dice-rolling mechanism that gives roughly similar results?
Example 1: Say you have a game where both sides roll a number of 6-sided dice (each side rolling a possibly different number of dice) looking for 6s. Unfortunately, other rules of the game frequently cause the number of dice rolled to be excessive (say, 40-60 or 80). The assorted statistics of the resulting distribution is fairly easy to calculate, but is there a way to  use that to create a simpler mechanism (using, say, not more than 10 dice) that produces approximately the same results?
Example 2: Say you have a game where $n$ 6-sided dice are rolled looking for values greater than $x$, to identify possible successes. These successes are then verified by rolling 1 die per possible success looking for a value greater than $y$. ($n, x,$ and $y$ are parameters of the mechanism.) Is there a way to create a mechanism that produces approximately the same results with only one step?

For example 1, as in Xi’an’s answer, you could use a 10 sided die 3 times to get the first 3 digits of a unif(0,1) rv. Then use inverse cdf. If the result is simply the number of 6’s out of 80 throws, this is approximately normal with mean 80/6 and variance 805/36. Find some other manual rv approximately normal and transform it. For example the number of 6’s out of 20 throws is approximately normal with mean 10/6 and variance 25/36. Call that X. Then, 80/6+sqrt(805/36) (X-10-6)/sqrt(25/36) has approximately the righ normal distribution.
For example 2, you may be able to do it in one step. Assume you start with 6 sided dice. Probability of success is $(6-x)(6-y)/36$. If this fraction can be written with a denominator equal to the number of sides of a die you have, then you can do it in one step. If you have a 36 sided die, you can always do it. I’ve never seen one, but it would be a trapezohedron similar to the 10 sided die. If not, you have to check for your specific situation. For example, $x=3$, $y=2$. The probability is $1/3$ which can be done with either 6 or 12 sided die. If you only have 6 sided die, you can do it if and only if  one of the parameters is 3 and the other is 2 or 4.
