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A game tree evaluation rule using maxima at the controlled player’s decision nodes and expected values at chance nodes:respectively. Terminal utilities fix the boundary values. These rules exactly optimize a finite-horizon expected utility objective by backward induction. For cyclic infinite-horizon reachability problems, a finite Markov decision process formulation and value iteration replace naive expansion of an infinite game tree. Start with one at successful terminal states and zero elsewhere, keep terminal values fixed, and apply the maximizing/expectation backups repeatedly. The $t$th iterate is the greatest success probability achievable within $t$ transitions, by backward induction. These values are nondecreasing and bounded by one. For every policy, eventual success is the increasing set union of success within finite horizons, so its probability is the limit of those finite-horizon probabilities. Taking the supremum over policies and horizons in either order proves that the iterates converge to the optimal eventual-success probability. The limiting fixed-point equations can also admit larger solutions, so the initialization from terminal-success indicators matters.

This version of value iteration updates state values in place, allowing later states in a sweep to use values just computed for earlier states. In a finite discounted Markov decision process, a sweep in the order $1,\dots ,m$ isThis formula uses the old self-component; solving a self-dependent scalar equation would be a different update convention. The ordinary synchronous update uses only old components throughout.

For $0\le \gamma <1$, let $V_{\ast }$ solve the maximizing Bellman equation and let the initial supremum norm error be $E$. Each component update has error at most $\gamma E$, because its transition probabilities average components whose errors are at most $E$. After every state has been updated once, the whole sweep has error at most $\gamma E$. Repeated full sweeps therefore converge to the same fixed point. The order can affect intermediate values and speed, even though it leaves the fixed point unchanged.

The game tree alternates MAX nodes for joint robot choices and CHANCE nodes for joint monster outcomes. An interior MAX node has at most $25$ children; an interior CHANCE node at most $4^2=16$. A terminal safe meeting has utility one and capture has utility zero. With no specified preference for speed, the objective is to maximize the probability of eventual safe meeting.

Use expectimax: maximize at MAX nodes and take the expected value at CHANCE nodes. For nonterminal states, its finite-horizon backup isHere the horizon counts half-turns, terminal values stay fixed at zero or one, and $V_0$ is zero at nonterminal states. This computes the best chance of success within the horizon. Maximizing the worst monster outcome would solve a different problem from the stipulated random movement.

Because waiting and movement cycles can repeat states, the infinite game tree need not terminate. For the unbounded objective, use value iteration on the finite state graph, starting from terminal-success indicators as above: $V_t$ increases to the optimal eventual-success probability. A reachability-policy method or sufficiently long finite-horizon policies can extract the actions; arbitrary tie choices in the limiting equations must not select a policy that waits forever instead of realizing the value. Alternatively impose an explicit finite horizon and use depth-limited expectimax, stating that changed objective. No method guarantees success with probability one from every initial configuration.

For a finite discounted Markov decision process, value iteration repeatedly applies the Bellman operator:The operator is a contraction by factor $\gamma$ in the supremum norm, so the iterates converge to its unique fixed point, the optimal value. The proof bounds the change of each action value by $\gamma \Vert V-W{\Vert }_{\infty }$ and uses the same bound after maximization.

In a one-state system with one action giving reward one, $V_{n+1}=1+\gamma V_n$ and the limit is $1/(1-\gamma )$. Undiscounted problems require different assumptions; this contraction proof fails at $\gamma =1$.