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

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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?

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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*}

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Roll a die, each time we get an outcome value either 1,2,3,4,5 or 6.

Roll a die thrice, we get 3 outcome values.

For example the three outcomes could be first roll 3, second roll 5, third roll 1. Then the largest of the three outcomes is 5.

Indeed, the largest of the three coutcomes can be any of 1, 2, 3, 4, 5 and 6. Let's suppose they have probabilities $P_1$, $P_2$, $P_3$, $P_4$, $P_5$ and $P_6$.

So we want an expected value of this experiment.

The expected value is similar to the notion of average. To be accurate in our case it's

\begin{equation} \text{Expected value} = 1\cdot P_1 + 2\cdot P_2 + .. + 6\cdot P_6 \end{equation}