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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!

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  • $\begingroup$ Note that not everyone is familiar with die game notation. 30d6 means you throw 30 standard (assumed fair) six-sided dice. $\endgroup$ – Arthur Sep 26 '18 at 11:29
  • $\begingroup$ Generating functions work as well as anything here. You are interested in the coefficients of $\frac {(4+2x+x^2)^{30}}{6^{30}}$. You'll want to use a computer $\endgroup$ – lulu Sep 26 '18 at 11:52
  • $\begingroup$ Would anyone be able to explain why this post got voted down? I would like to make sure I meet the criteria expected of this forum next time I post. Thank you. $\endgroup$ – MattN Sep 26 '18 at 11:59
  • $\begingroup$ @Lulu - THank you. By coefficients, do you mean solving for X when the equation = 0? $\endgroup$ – MattN Sep 26 '18 at 12:02
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    $\begingroup$ Note: I mistyped in my first comment. It should have been $\frac {(3+2x+x^2)^{30}}{6^{30}}$. If you really want $70$ dice then replace $30$ with $70$, but with that many dice the probability of getting a score of at least $20$ is essentially $1$. $\endgroup$ – lulu Sep 26 '18 at 12:09
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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.

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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)

enter image description here

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