Calculating the probability that the sum of the 2 highest values of n dice is equal to x I'm trying to calculate the probability of the 2 highest values of n dice sum.
In these calculations i'm currently assuming a 7-sided die that is thrown 3 times.
With raw number i've found out through some form of logic and trial and error is that the sum of 2-5 can be calculated as
$$\text{sum to 2 = }(\frac{1}{7^3})$$
$$\text{sum to 3 = }(\frac{3}{7^3})$$
$$\text{sum to 4 = }(\frac{7}{7^3})$$
$$\text{sum to 5 = }(\frac{12}{7^3})$$
If i then increase the number of throws to 4 i get that
$$\text{sum to 2 = }(\frac{1}{7^3})$$
$$\text{sum to 3 = }(\frac{4}{7^3})$$
$$\text{sum to 4 = }(\frac{15}{7^3})$$
$$\text{sum to 5 = }(\frac{32}{7^3})$$
The only reason i've been able to calculate these is because i knew what the answer was supposed to be based on a spreadsheet. 
My problem is that i can't figure out the logic to how i should derive the numerator in case of n throws
 A: Outline:

*

*Find the joint cdf $F(x,y)$ for the highest and second highest rolls.


*Use this to find the joint pmf for the highest and second highest rolls.


*Use the joint pmf to find the pmf for the sum of these two rolls.
Let $X$ denote the highest roll, and $Y$ denote the second highest roll.
Joint cdf for $X$ and $Y$.
Given $x\ge y$, let $F(x,y)$ be the probability that the highest roll is at most $x$, and that the second highest roll is at most $y$. There are two ways this can happen; either exactly one of the rolls is at most $x$ yet greater than $y$, and the rest are at most $y$, or all of the rolls are at most $y$. The probability is then
$$
F(x,y)=\frac{n(x-y)y^{n-1}+y^{n}}{7^n}
$$
If instead $x<y$, then $F(x,y)=(x/7)^n$.
Joint pmf for $X$ and $Y$.
The probability that the highest roll is exactly $x$ and the lowest roll is exactly $y$ is
$$
f(x,y) = F(x,y)-F(x-1,y)-F(x,y-1)+F(x-1,y-1)
$$
Pmf of $X+Y$.
To find the probability $p(s)$ that the sum of the highest two rolls is $s$, just add up the probabilities $f(x,y)$ over all pairs $(x,y)$ for which $x+y=s$. For example,
$$
p(5)=f(4,1)+f(3,2)
$$
$$
\hspace{3.5cm}p(8)=f(7,1)+f(6,2)+f(5,3)+f(4,4)
$$
$$
p(12)= f(7,5)+f(6,6)
$$
Here is this solution implemented in Mathematica. You can verify the results are correct for the values you already know.
F[x_,y_] := If[x >= y, (n(x-y)y^(n-1)+y^n)/7^n, (x/7)^n];
f[x_,y_] := F[x,y] - F[x-1,y] - F[x,y-1] + F[x-1,y-1];
P[s_]    := Sum[f[x,s-x],{x,Max[1,s-7],Min[7,s-1]}];

For[s=2, s<= 14, s++,
 Print["P(",s,") = ",Simplify[P[s], Assumptions -> n>1]]
]

Output:

P(2) = 7^(-n)
P(3) = n/7^n
P(4) = (-1 + 2^n)/7^n
P(5) = (2^(-1 + n)*n)/7^n
P(6) = -(2/7)^n + (3/7)^n
P(7) = (3^(-1 + n)*n)/7^n
P(8) = -(3/7)^n + (4/7)^n
P(9) = ((-4 + 4^n)*n)/(4*7^n)
P(10) = -(2*(4^n - 5^n) + 2^n*n)/(2*7^n)
P(11) = ((-5*3^n + 3*5^n)*n)/(15*7^n)
P(12) = (4*(-5^n + 6^n) - 4^n*n)/(4*7^n)
P(13) = ((-6*5^n + 5*6^n)*n)/(30*7^n)
P(14) = (6 - (6/7)^n*(6 + n))/6

