= Solution
This is an <absent-minded decision problem>: the two decision nodes belong to the same <information set>. A <behavioural strategy> must therefore use the same <probability> at both visits. Put $p=\Pr(D)$ and $1-p=\Pr(R)$; fresh randomization occurs independently at each visit.
The terminal probabilities are $1-p$ for stopping immediately, $p(1-p)$ for going down once and then right, and $p^2$ for going down twice. Thus the <expected payoff> is
$$
U(p)=0(1-p)+8p(1-p)+p^2
=8p-7p^2
=\frac{16}{7}-7\left(p-\frac47\right)^2.
$$
This expression gives the unique global maximum on $[0,1]$:
$$
\boxed{\Pr(D)=\frac47,\qquad \Pr(R)=\frac37,\qquad
U_{\max}=\frac{16}{7}.}
$$
\Image[/past-exam-of-the-msc-in-advanced-computer-science-course-of-the-university-of-oxford/2023/computational-game-theory-behavioural-payoff.png]
{title=The optimal <behavioural strategy> improves on either fixed plan}
The independence of repeated randomizations is essential. A <mixed strategy> choosing once between the two deterministic plans “always $R$” and “always $D$” gives at most $1$, since their <payoffs> are $0$ and $1$. It cannot arrange “$D$ first, then $R$” as a deterministic plan respecting this <information set>. The failure of <perfect recall> is exactly why <mixed strategies> and <behavioural strategies> need not have the same attainable outcome distributions here.
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