<s>[INST] 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.
 [/INST] 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:


*

*$C_1=1^3=1$

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

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

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

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

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

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