# Finding the expected value of the sum of four dice with only the largest three counting

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 $$s$$?

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*}

• Suppose the person rolls e.g. 3,3,4,5. Do we have $s=3+4+5$ in that situation (then see my answer) or do we have $s=4+5$? Commented May 24, 2020 at 14:56
• @drhab Based on how the question is stated, and how Bob carried out the calculations in the post, it seems like you always keep three die. So your answer seems to the be the correct interpretation. Commented May 24, 2020 at 15:07
• @paulinho Thank you (and so is yours). Commented May 24, 2020 at 15:33
• @drhab Correct you only keep the largest three die rolls.
– Bob
Commented May 24, 2020 at 15:50

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!
Let $$M:=\min\left(D_{1},D_{2},D_{3},D_{4}\right)$$ where the $$D_{i}$$ denote the outcomes of the dice rolls.
Then $$s=\sum_{i=1}^{4}D_{i}-M$$.
Here: $$\mathbb{E}M=\sum_{k=1}^{6}P\left(M\geq k\right)=\sum_{k=1}^{6}\left(\frac{7-k}{6}\right)^{4}$$
So that: $$\mathbb{E}s=4\times\frac{7}{2}-\sum_{k=1}^{6}\left(\frac{7-k}{6}\right)^{4}$$