Past exam of the MSc in advanced computer science course of the University of Oxford / 2023 / Computational game theory / 3 / b

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 $p=\mathrm{Pr}(D)$ and $1-p=\mathrm{Pr}(R)$; 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 isThis expression gives the unique global maximum on $[0,1]$:

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.