= Solution
Let $v_j$ be the optimal <expected value> before seeing the next offer when $j$ rounds remain. With one round left the offer must be accepted, so $v_1=1/2$. For $j>1$, observing $x$ gives a choice between $x$ now and the continuation value $v_{j-1}$. <Independence> of future <uniform distributions> makes that continuation value independent of past offers. Thus the <uniform-offer stopping recursion> is
$$
v_j=\int_0^1\max(x,v_{j-1})\,dx=\frac{1+v_{j-1}^2}{2}.
$$
It gives $v_2=5/8$ and $v_3=89/128$. The <Snell envelope> rule therefore yields the explicit strategy
$$
\boxed{\begin{array}{c|c}
\text{round}&\text{accept when}\\ \hline
1&\xi_1\ge89/128\\
2&\xi_2\ge5/8\\
3&\xi_3\ge1/2\\
4&\text{always}
\end{array}}
$$
Only rounds actually reached are played. At a threshold the two actions have identical continuation <expected value>; either convention is optimal, and exact equality has probability zero. The optimal expected payout before the first offer is
$$
\boxed{v_4=\frac{1+(89/128)^2}{2}=\frac{24305}{32768}.}
$$
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