70 6-sided dice Question re: Gaming I play a game where a d6 (a regular, fair 6 sided die) is used to determine whether or not a weapon hits an opponent's vehicle.  There is one possible scenario where the attacker can roll 70d6 in one attack step.  A "hit" is scored on a 4, 5 or 6 on each die.  In other words, you roll all dice, look at each and count up the 4's, 5's and 6's.  A 4 counts as 1 hit and a 5 counts as 1 hit while a 6 counts as 2 hits.
Question 1 - What is the probability of scoring at least 20 hits when rolling this 70d6 attack?
Question 2 - How does the probability change if a 3 is also 1 hit?
Thanks!
 A: Here is an approximation by simulation:
In a million sessions of 70 rolls of the die according to Question 1,
I found that it is almost sure to score at least 20 hits. [Notice I have labeled die faces with the appropriate numbers of hits; R treats the three 0's as different faces.]
set.seed(926); m = 10^6
die = c(0,0,0,1,1,2); rolls=70
hits = replicate(m, sum(sample(die,rolls,rep=T)))
mean(hits >= 20)
[1] 0.999998

If you intended 30 rolls of the die (as mentioned in a comment), then
its a little over 50%.
set.seed(926); m = 10^6
die = c(0,0,0,1,1,2); rolls=30
hits = replicate(m, sum(sample(die,rolls,rep=T)))
mean(hits >= 20)
[1] 0.540681

Either way, I suggest you try the CLT method proposed by @awkward.
Seventy rolls, or even 30, should be enough to get a reasonably good
approximation. Then you can compare results from the CLT with results from
simulation.
Note: Here is a histogram (blue) for the number of hits in 70 rolls, along with the
best-fitting normal curve (orange). The fit is not perfect, but is surely good
enough to show that there is almost no probability below 20.
# for rolls=70
mu = mean(hits);  sg = sd(hits);  mu;  sg
[1] 46.66131
[1] 6.235426
hist(hits, prob=T, col="skyblue2", ylim=c(0,.07))
curve(dnorm(x, mu, sg), add=T, col="orange", lwd=2)


A: The expected number of hits is just $70$ times the expected number for one die.  For $4,5,6$ hitting, you expect $\frac 46$ hit per die or $\frac {4\cdot 70}6=41\frac23$ hits on average per roll.  If a $3$ hits as well, you now expect $\frac 56$ from one die, for a total of $58\frac 13$ per roll.  The chance of scoring at least $20$ is very high in either case.  We can approximate the distribution with a normal one.  For the first case the variance of one die is $\frac 16\cdot (4+1+1)-(\frac 23)^2=\frac 59$ so the variance of the whole roll is $70\cdot \frac 59=39\frac 19$ and the standard deviation is about $6$.  The first is over $3$ standard deviations low, which happens less than once in $600$ tries.  The other is even more likely to yield at least $20$ hits.
