I roll three dice and keep only the largest two I know that rolling two 6-sided dice I will get 7 as medium result. But what if I roll three dice and get only the largest two? How much is that going to increase the medium result? And what about with 4 dice, 5 etc...?
Thank you!
 A: Let's suppose we have 3 dice and pick the largest two. Then if $X$ is the value of the largest die result, and $Y$ is the value of the next-largest, we want $p(x,y)=P(X=x,Y=y)$. If we observe 3 distinct die results, there are 6 ways the dice could be assigned to those results. If we only observe 2 distinct die results, there are 3 ways the dice could be assigned to those results, and if we only observe 1 distinct die result, there is only 1 way for the dice to be assigned to those results. So we have $$p(x,y)=\begin{cases}\frac{6(y-1)+3}{6^3}&x>y\\\frac{3(y-1)+1}{6^3}&x=y\\0&x<y\end{cases}=\begin{cases}\frac{6y-3}{6^3}&x>y\\\frac{3y-2}{6^3}&x=y\\0&x<y\end{cases}.$$
And we note that $\sum_{x,y}p(x,y)=\sum_{y=1}^{6}(6-y)\frac{6y-3}{6^3}+\frac{3y-2}{6^3}=1$, as of course it should. Now if we let $T=X+Y$, we can find $p(t)=P(T=t)$, for $2\leq t\leq 12$. This gives $$p(t)=\begin{cases}\sum_{y=\min\{1,t-6\}}^{(t-1)/2}p(t-y,y)&t\text{ odd}\\\sum_{y=\min\{1,t-6\}}^{t/2}p(t-y,y)&t\text{ even}\end{cases}=\begin{cases}\sum_{y=\min\{1,t-6\}}^{(t-1)/2}\frac{6y-3}{6^{3}}&t\text{ odd}\\\left(\sum_{y=\min\{1,t-6\}}^{t/2-1}\frac{6y-3}{6^3}\right)+\frac{3(t/2)-2}{6^3}&t\text{ even}\end{cases}$$
This equals $$p(t)=\begin{cases}\frac{3(t-1)^{2}}{4\cdot 6^3}-\frac{3(t-7)^2}{6^3}\mathbf{1}_{\{>7\}}(t)&t\text{ odd}\\\frac{3(t-2)^{2}}{4\cdot 6^3}+\frac{3t-4}{2\cdot 6^3}-\frac{3(t-7)^2}{6^3}\mathbf{1}_{\{>7\}}(t)&t\text{ even}\end{cases}=\begin{cases}\frac{3(t-1)^{2}-12(t-7)^2\mathbf{1}_{\{>7\}}(t)}{4\cdot 6^3}&t\text{ odd}\\\frac{3(t-1)^{2}+1-12(t-7)^2\mathbf{1}_{\{>7\}}(t)}{4\cdot 6^3}&t\text{ even}\end{cases}.$$
Once again, it's easy to check that this defines a probability distribution, noting that $\mathbf{1}_{\{>7\}}(t)$ is the function which equals 1 if $t>7$ and is 0 otherwise.
Then the mean of this distribution is just $\sum_{t=2}^{12}t*p(t)=\frac{203}{24}\approx 8.46.$
I believe that the approach I've just described is probably the best for extending to more dice; one could, in the case of 3 dice, view the variable of interest as $X+Y+Z-M,$ where $M=\min\{X,Y,Z\},$ and $X,Y,Z$ are the three dice results, and use linearity of expectation to compute the mean. However, with more dice, this will get a bit more... dicey.
