You play a game where you roll a die up to three times and keep the highest value as the result of the game. Find the expected value of the result.
I calculated expected values of increasing numbers of die rolls by using probability of outcomes multiplied by the expected value of the previous roll, where I stop rolling if I get a 6 but keep going otherwise.
$$E_1 = 3.5$$
$$E_2 = \frac{1}{6}(6) + \frac{5}{6}(E_1)$$
$$E_3 = \frac{1}{6}(6) + \frac{5}{6}(E_2)$$
I get $E_3 = \frac{307}{72}$ but the answer is $\frac{119}{24}$. Why is my method wrong?
We could use distribution of maximums directly for this problem. Let $X_{M}$ be the maximum of roll die $X$. Then, $P(X_{M}\leq x)=F(x)^{n}$. Here $n=3$ because of rolling times. So we have a cdf $F(X_{M})=\frac{x^{3}}{216}$ then the density would be $p({X^{j}_{M}})=dF(X^{j}_{M})=F^{j}(X_{M})-F^{j-1}(X_{M})$ due to discrete sample space. We numerically calculate these, $dF(X^{6}_{M})=\frac{91}{216}$, $dF(X^{5}_{M})=\frac{61}{216}$, $dF(X^{4}_{M})=\frac{37}{216}$, $dF(X^{3}_{M})=\frac{19}{216}$, $dF(X^{2}_{M})=\frac{7}{216}$, $dF(X^{1}_{M})=\frac{1}{216}$. Then the expected value will be, $\sum_{j=1}^{6}x_{j}p({x^{j}_{M}})=\frac{546}{216}+\frac{305}{216}+\frac{148}{216}+\frac{57}{216}+\frac{14}{216}+\frac{1}{216}=\frac{1071}{216}$. After simplification by $9$ both numerator and denominator, this equals $\frac{119}{24}$.