Dice Game: n Throws
You roll a fair six‐sided die up to \(n\) times. After each roll you may stop and take that face value, or discard it and continue (losing the current value). What is the optimal stopping rule and expected value \(E_n\)?
Answer
The optimal rule is a threshold which is based on the EV of the remaining throws, i.e. \(\,t_n=\lceil E_{n-1}\rceil\); stop on any roll \(X\ge t_n\), otherwise continue. One shows by the law of total expectation that
\[
E_n
=\frac{\lfloor E_{n-1}\rfloor}{6}\,E_{n-1}
\;+\;\Bigl(1 - \frac{\lfloor E_{n-1}\rfloor}{6}\Bigr)\,
\frac{\lceil E_{n-1}\rceil + 6}{2}.
\]
With \(E_0=0\), this yields:
\[
\begin{aligned}
E_1&=\frac{0}{6}\cdot0 + \Bigl(1-\tfrac{0}{6}\Bigr)\frac{1+6}{2}=3.5,\\
E_2&=\frac{3}{6}\cdot3.5 + \Bigl(1-\tfrac{3}{6}\Bigr)\frac{4+6}{2}=4.25,\\
E_3&=\frac{4}{6}\cdot4.25 + \Bigl(1-\tfrac{4}{6}\Bigr)\frac{5+6}{2}\approx4.6667,\\
E_4&=\frac{4}{6}\cdot4.6667 + \Bigl(1-\tfrac{4}{6}\Bigr)\frac{5+6}{2}\approx4.9444,\\
E_5&=\frac{4}{6}\cdot4.9444 + \Bigl(1-\tfrac{4}{6}\Bigr)\frac{5+6}{2}\approx5.1296,\\
E_6&=\frac{5}{6}\cdot5.1296 + \Bigl(1-\tfrac{5}{6}\Bigr)\frac{6+6}{2}\approx5.2747.
\end{aligned}
\]
The corresponding stop‐thresholds are \(t_1=1\), \(t_2=4\), \(t_3=5\), \(t_4=5\), \(t_5=5\), and \(t_6=6\).