What is the expected value of the largest of the three dice rolls? 
You toss a fair die three times. What is the expected value of the largest of the three outcomes?

My approach is the following:
calculate the probability of outcome when $\max=6$, which is
$$P(\text{at least one $6$ of the three rolls}) = 1 - P(\text{no }6) = 1 - (5/6)^3$$
and then calculate the probability of outcome when $\max=5$, which is
$$P(\text{at least one $5$ of the three rolls & $5$ is max}) = 1 - P(\text{no $5$ & $5$ is max}) = 1 - (4/6)^3.$$
I wonder if this approach is right.
 A: Let X denote the largest value, then:


*

*$P(X=1)=\frac{ 1}{216}$

*$P(X=2)=\frac{ 7}{216}$

*$P(X=3)=\frac{19}{216}$

*$P(X=4)=\frac{37}{216}$

*$P(X=5)=\frac{61}{216}$

*$P(X=6)=\frac{91}{216}$


Hence the expected value is:
$$1\cdot\frac{1}{216}+2\cdot\frac{7}{216}+3\cdot\frac{19}{216}+4\cdot\frac{37}{216}+5\cdot\frac{61}{216}+6\cdot\frac{91}{216}=\frac{119}{24}$$

Details:
Let $C_n$ denote the number of combinations with largest value $n$.
Observe that $C_n=n^3-\sum\limits_{k=1}^{n-1}C_{n-1}$, therefore:


*

*$C_1=1^3=1$

*$C_2=2^3-1=7$

*$C_3=3^3-1-7=19$

*$C_4=4^3-1-7-19=37$

*$C_5=5^3-1-7-19-37=61$

*$C_6=6^3-1-7-19-37-61=91$

A: More generally let's find $E[X_\max]$ and $E[X_\min]$ where $X_\max$ and $X_\min$ are the largest and smallest outcomes among $n$ independent rolls of an $s$-sided die with sides numbered from $1$ to $s$.
Define $X_i$ to be the random variable whose value is $1$ if $X_\max\ge i$ and $0$ otherwise; then
$$X_\max=\sum_{i=1}^sX_i$$
and
$$E[X_\max]=E\left[\sum_{i=1}^sX_i\right]=\sum_{i=1}^sE[X_i]=\sum_{i=1}^sP(X_i=1)=\sum_{i=1}^sP(X_\max\ge i)$$
$$=\sum_{i=1}^s[1-P(X_\max\lt i)]=\sum_{i=1}^s\left[1-\left(\frac{i-1}s\right)^n\right]=s-s^{-n}\sum_{i=1}^s(i-1)^n=\boxed{s-s^{-n}\sum_{i=1}^{s-1}i^n}.$$
Since the outcome of a single roll is distributed symmetrically about its mean $\frac{1+s}2$, we have $E[X_\min]+E[X_\max]=1+s$, so that
$$E[X_\min]=1+s-E[X_\max]=\boxed{1+s^{-n}\sum_{i=1}^{s-1}i^n}.$$
When $s=6$ and $n=3$ we have
$$E[X_\max]=6-6^{-3}(1^3+2^3+3^3+4^3+5^3)=6-6^{-3}\binom62^2=6-\frac{25}{24}=\boxed{\frac{119}{24}}$$
and
$$E[X_\min]=7-E[X_\max]=7-\frac{119}{24}=\boxed{\frac{49}{24}}.$$
A: Let $X_{i}$ be the value from individual die
Let Z be the max value: $Z = max(X_1, X_2, X_3)$
The CDF of Z is $F_Z(x) = P(Z\le x) = P(X_1\le x)P(X_2\le x) P(X_3\le x)  = F_{X_1}(x)^3 = (x/6)^3 $
Therefore PMF of Z: $f_Z(x) = F_Z(x) - F_Z(x-1) = (x^3-(x-1)^3)/6^3 $
$ E(Z) = \sum_{x=1}^6 xf_Z(x) = 119/24 $
A: Picture the cube of possible outcomes. The cells that represent a maximum of $6$ lie in a greedy half of the outer layer, which has $6^3-5^3=216-125=91$ cells in it.
The next layer represents max $5$, and has $5^3-4^3=125-64=61$ cells in it.
We can proceed in a similar manner and arrive at the sum of the whole cube:
$$6\cdot(6^3-5^3)+5\cdot(5^3-4^3)+4\cdot(4^3-3^3)+3\cdot(3^3-2^3)+2\cdot(2^3-1^3)+1\cdot(1^3-0^3)$$
$$=6^3(6)-5^3(6-5)-4^3(5-4)-3^3(4-3)-2^3(3-2)-1^3(2-1)-0^3(1)$$
$$=6\cdot6^3-5^3-4^3-3^3-2^3-1^3-0^3$$
$$=1296-225$$
$$=1071$$
Divide by the number of cells in the cube $6^3=216$, and the answer is:
$$\frac{1071}{216}=\frac{119}{24}\approx4.96$$
