Below is a problem that I did. I have good reason to believe that the answer is around $12.244599$.
Problem:
Suppose a person rolls $4$ six sided dice. He does not count the die with the lowest value. He then adds the other three values up. Call that sum $s$. What is the expected value of $s$?
Answer:
Let $p_6$ be the probability that the lowest die has a value of $6$.
\begin{align*}
p_6 &= \left( \frac{1}{6} \right) ^4 = \frac{1}{1296} \\
\end{align*}
Let $p_{56}$ be the probability that the lowest die has a value of $5$ or $6$. Let $p_5$ be the probability that the lowest die has a value of $5$.
\begin{align*}
p_{56} &= \left( \frac{2}{6} \right) ^4 = \frac{16}{1296} = \frac{1}{81} \\
p_{56} &= p_5 + p_6 \\
p_5 &= p_{56} – p_{6} = \frac{16}{1296} – \frac{1}{1296} = \frac{15}{1296}
\end{align*}
Let $p_{46}$ be the probability that the lowest die has a value of $4$, $5$ or $6$. Let $p_4$ be the probability that the lowest die has a value of $4$.
\begin{align*}
p_{46} &= \left( \frac{3}{6} \right) ^4 = \frac{81}{1296} = \frac{9}{144} = \frac{1}{16}\\
p_{46} &= p_4 + p_5 + p_6 \\
p_4 &= p_{46} – p_{5} + p_{6} = \frac{81}{1296} – \frac{15}{1296} – \frac{1}{1296} = \frac{81 – 15 – 1}{1296} \\
p_4 &= \frac{65}{ 1296 } \\
\end{align*}
Let $p_{36}$ be the probability that the lowest die has a value between $3$ and $6$. Let $p_3$ be the probability that the lowest die has a value of $3$.
\begin{align*}
p_{36} &= \left( \frac{4}{6} \right) ^4 = \frac{256}{1296} \\
p_{36} &= p_3 + p_{46} \\
p_3 &= p_{36} – p_{46} = \frac{256}{1296} – \frac{81}{1296} = \frac{175}{1296} \\
\end{align*}
Let $p_{26}$ be the probability that the lowest die has a value between $2$ and $6$. Let $p_2$ be the probability that the lowest die has a value of $2$.
\begin{align*}
p_{26} &= \left( \frac{5}{6} \right) ^4 = \frac{625}{1296 } \\
p_{26} &= p_2 + p_{36} \\
p_2 &= p_{26} – p_{36} = \frac{625}{1296 } -\frac{256}{1296} = \frac{369 }{ 1296}
\end{align*}
Let $p_{16}$ be the probability that the lowest die has a value between $1$ and $6$. Let $p_1$ be the probability that the lowest die has a value of $1$.
\begin{align*}
p_{16} &= \left( \frac{6}{6} \right) ^4 = \frac{1296}{1296 } = 1 \\
p_{16} &= p_1 + p_{26} \\
p_1 &= p_{16} – p_{26} = \frac{1296}{1296 } – \frac{625}{1296 } = \frac{671}{1296 }
\end{align*}
Now, as a partial check of my work, I want to verify that $p_1 + p_2 + p_3 + p_4 + p_5 + p_6 = 1$.
\begin{align*}
p_1 + p_2 + p_3 + p_4 + p_5 + p_6 &=
\frac{ 671 }{ 1296} + \frac{ 369 }{ 1296} + \frac{175}{1296} + \frac{65}{ 1296 } + \frac{15}{1296} + \frac{1}{1296} \\
p_1 + p_2 + p_3 + p_4 + p_5 + p_6 &= \frac{ 1215 }{ 1296} + \frac{65}{ 1296 } + \frac{15}{1296} + \frac{1}{1296} \\
p_1 + p_2 + p_3 + p_4 + p_5 + p_6 &= \frac{ 1215 + 65 + 15 + 1 }{ 1296} = \frac{1296} {1296 } \\
p_1 + p_2 + p_3 + p_4 + p_5 + p_6 &= 1
\end{align*}
It passes this check. Now we can find $E(s)$.
\begin{align*}
E(s) &= p_1(3)\left( \frac{7}{2}\right) + p_2(3)\left( 4 \right) + p_3(3)\left( \frac{9}{2}\right)
+ p_4(3)\left( 5 \right) + p_5(3)\left( \frac{11}{2}\right) + p_6(3)\left( 6 \right) \\
%
E(s) &= p_1 \left( \frac{21}{2}\right) + p_2 (12) + p_3 \left( \frac{27}{2}\right)
+ p_4(15) + p_5 \left( \frac{33}{2}\right) + p_6 \left(18 \right) \\
%
E(s) &= \left( \frac{671}{1296 } \right) \left( \frac{21}{2}\right) + p_2 (12) + p_3 \left( \frac{27}{2}\right)
+ p_4(15) + p_5 \left( \frac{33}{2}\right) + p_6 \left(18 \right) \\
E(s) &= \left( \frac{14091}{1296(2) } \right) + \left( \frac{369 }{ 1296} \right) (12) + p_3 \left( \frac{27}{2}\right)
+ p_4(15) + p_5 \left( \frac{33}{2}\right) + p_6 \left(18 \right) \\
%
E(s) &= \left( \frac{14091}{2592 } \right) + \left( \frac{ 2(369)(12) }{ 1296(2)} \right) + p_3 \left( \frac{27}{2}\right)
+ p_4(15) + p_5 \left( \frac{33}{2}\right) + p_6 \left(18 \right) \\
%
E(s) &= \left( \frac{14091+ 2(369)(12)}{2592 } \right) + p_3 \left( \frac{27}{2}\right)
+ p_4(15) + p_5 \left( \frac{33}{2}\right) + p_6 \left(18 \right) \\
%
E(s) &= \left( \frac{22947}{2592 } \right) + \left( \frac{175}{1296} \right) \left( \frac{27}{2}\right)
+ \left( \frac{65}{ 1296 } \right) (15) + \left( \frac{15}{1296} \right) \left( \frac{33}{2}\right) + p_6 \left(18 \right) \\
%
E(s) &= \left( \frac{22947}{2592 } \right) + \left( \frac{175(27)}{1296(2)} \right)
+ \left( \frac{65(15)}{ 1296 } \right) + \left( \frac{15}{1296} \right) \left( \frac{33}{2}\right)
+ \left( \frac{1}{1296} \right) \left(18 \right) \\
%
E(s) &= \left( \frac{22947}{2592 } \right) + \left( \frac{175(27)}{2592} \right)
+ \left( \frac{65(15)(2)}{ 2592 } \right) + \left( \frac{15(33)}{1296(2)} \right) + \left( \frac{18}{1296} \right) \\
%
E(s) &= \left( \frac{22947}{2592 } \right) + \left( \frac{175(27)}{2592} \right)
+ \left( \frac{1950}{ 2592 } \right) + \left( \frac{15(33)}{2592} \right) + \left( \frac{36}{2592} \right) \\
%
E(s) &= \left( \frac{22947}{2592 } \right) + \left( \frac{4725}{2592} \right)
+ \left( \frac{1950}{ 2592 } \right) + \left( \frac{495}{2592} \right) + \left( \frac{36}{2592} \right) \\
E(s) &= \frac{22947 + 4725 + 1950 + 495 + 36}{2592 } = \frac{30153}{2592 } \\
E(s) &= \frac{10051 }{864} \\
E(s) &\doteq 11.633101851851
\end{align*}
Best Answer
Here is a simpler way to go about this question. Let $X_i$ be the $i$th dice roll, and note that $X_i$ is uniformly distributed between values $1$ thru $6$. You are interested in finding $$\mathbb{E}[X_1 + X_2 + X_3 + X_4 - \min(X_1,X_2, X_3, X_4)] = 4 \mathbb{E}[X_1] - \mathbb{E}[\min(X_1,X_2, X_3, X_4)]$$ Here we have used linearity of expectation. The first quantity is just $4 \cdot 3.5 = 14$. We proceed to find the distribution of the random variable $Y = \min(X_1,X_2, X_3, X_4)$. Here it is easiest to calculate the tail-CDF: $$\mathbb{P}(Y > y) = \left(\frac{6 - y}{6}\right)^4$$ Using the fact that $\mathbb{E}[X] = \sum_{x = 0}^\infty \mathbb{P}(X > x)$ for any random variable taking on values only in the natural numbers, we have that $$\mathbb{E}[Y] = \sum_{y = 0}^5 \mathbb{P}(Y > y) = \frac{1}{6^4}(1^4 + 2^4 + 3^4 + 4^4 + 5^4 + 6^4) = \frac{1}{6^4} \cdot \frac{1}{30}\cdot 6 \cdot 7 \cdot 13 \cdot 125 \approx 1.755$$ and hence the expectation we want is approximately $14 - 1.755 \approx \boxed{12.245}$, which indeed seems to be what you get!