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.
Best Answer
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 $$E[X_\min]=7-E[X_\max]=7-\frac{119}{24}=\boxed{\frac{49}{24}}.$$