In fact, I do not really get the question first.
You toss a fair die three times. What is the expected value of the largest of the three outcomes?
"expected value of the largest of the three outcomes"
Can anyone clarify this part and give me some ideas?
Consider three random variables, $X_1$, $X_2$, $X_3$, where $X_i$ gives the value of the $i$:th die roll and these random variables are independent with uniform distribution on the set $\{1,2,3,4,5,6\}$. The largest of the three outcomes is the random variable $\max\{X_1,X_2,X_3\}$. By independence of the random variables we have \begin{align*} P(\max\{X_1,X_2,X_3\}\leq n)=P(X_1,X_2,X_3\leq n)=P(X_1 \leq x)P(X_2\leq x)P(X_3\leq n)=\frac{n^{3}}{6^{3}}. \end{align*} So \begin{align*} E(\max\{X_1,X_2,X_3\})&=\sum_{n=0}^{5}P(\max\{X_1,X_2,X_3\}>n) \\ &=\sum_{n=0}^{5}(1-\frac{n^{3}}{6^{3}})\\ &=6-\sum_{n=0}^{5}\frac{n^{3}}{6^{3}}\\ &=6-\frac{25}{24}\\ &=\frac{119}{24}\approx 4.96. \end{align*}