Repeatedly rolling a six sided die four times and summing the highest three results gives you a distribution with what mean and standard deviation?

Question

Repeatedly rolling a six sided die four times and summing the highest three results gives you a distribution with what mean and standard deviation?

I've only taken AP statistics, but I would like to learn how to do this.

Answer 1

(D&D player?) It's easy enough to just enumerate the whole distribution. In R:

> a1 <- rep(1:6,times=216)
> a2 <- rep(1:6,times=36,each=6)
> a3 <- rep(1:6,times=6,each=36)
> a4 <- rep(1:6,each=216)
> tot <- a1+a2+a3+a4-pmin(a1,a2,a3,a4)

That's all it takes (there's a bunch of other ways to do it). Results:

> mean(tot)
[1] 12.2446
> n <- length(tot)
> sd(tot)*sqrt((n-1)/n)  # we have the whole distribution; we want n-denominator, not n-1
[1] 2.846844
> quantile(tot)
  0%  25%  50%  75% 100% 
   3   10   12   14   18 
> table(tot)  #frequencies
tot
  3   4   5   6   7   8   9  10  11  12  13  14  15  16  17  18 
  1   4  10  21  38  62  91 122 148 167 172 160 131  94  54  21 
> print(table(tot)/sum(table(tot))*100,d=2)  # as percentages
tot
     3      4      5      6      7      8      9     10     11     12     13 
 0.077  0.309  0.772  1.620  2.932  4.784  7.022  9.414 11.420 12.886 13.272 
    14     15     16     17     18 
12.346 10.108  7.253  4.167  1.620 

Answer 2

Each roll has a uniform distribution on $\{1, 2, 3, 4, 5, 6 \}$ and to get the distribution for the minimum you can use $P[m>k] = {\rm probability \ that \ all \ four \ rolls \ are \ } >k$ and by independence this is the probability that each roll is $\geq k$ raised to the fourth power where $m$ is the minimum of the 4 rolls. Now for each roll $P[X>k] =(6-k)/6 = 1 - k/6$ So $P[m>k] = (1-k/6)^4$ and therefore $P[m<=k]=1 - (1-k/6)^4$.

From this we get $P[m=k]=P[m \leq k]-P[m \leq k-1]$ - knowing that $m=k$ means that the smallest of the three in the sum you calculate is at least $k$.

Let $S$ be the sum of the three highest rolls. You want $P[S=k]$ for $k=3,...,18$. You can calculate this as $P[S=k]=∑_{j=1}^{6} P[S=k|m=j] P[m=j]$. Note that for a variety of $k$s and $j$s $P[S=k|m=j]=0$. For example $P[S=3|m=k]=0$ for $k=2,3,4,5,6$. Of course once you determine $P[S=k]$ for $3 \leq k \leq 18$ you can calculate the mean and variance.