Stochastic Games & Expectiminimax Decision Trees
“Chance nodes, expected value propagation, and evaluation function linearity constraints.”
1.Stochastic Games & Expectiminimax Trees
§ 3.1.Real-world environments and games such as Backgammon include elements of chance (dice rolls, card shuffling). In stochastic game trees, we introduce a third type of node: the Chance Node.
The Expectiminimax Formulation:
$Expectiminimax(s) = \begin{cases} Utility(s) & if s is terminal \\ \max_{a} Expectiminimax(Result(s, a)) & if Player(s) = MAX \\ \min_{a} Expectiminimax(Result(s, a)) & if Player(s) = MIN \\ \sum_{s'} P(s') × Expectiminimax(s') & if s is a Chance Node \end{cases}$
2.The Linearity Requirement for Evaluation Functions
§ 3.2.In standard deterministic Minimax, an evaluation function requires only an order-preserving (monotonic) scale: replacing evaluation v with v^3 or \log(v) does not alter which move is maximum or minimum.
However, in Expectiminimax, expected utility is a weighted linear sum: \sum P(i) v_i.
Applying a non-linear monotonic transformation alters the expectation! For example, expected value of outcomes (1, 9) with P = 0.5 is 0.5(1) + 0.5(9) = 5. If we apply non-linear transform f(x) = x^2, the expectation becomes 0.5(1) + 0.5(81) = 41, which can invert the preferred decision compared to an alternative move.
Theorem: An evaluation function in expectiminimax must be a positive affine transformation f(x) = a x + b (a > 0) of the true probability of winning to preserve optimal move choices.
3. GATE Worked Numerical: Expectiminimax Root Decision
Weight each child of a chance node by its explicit probability before computing the expected value!
Problem (Original Practice Problem · Modeled on GATE DA Pattern):
A MAX node A has two possible moves leading to chance nodes C1 and C2.
Chance node C1 has outcomes: Left (prob 0.7, utility 10) and Right (prob 0.3, utility 20).
Chance node C2 has outcomes: Left (prob 0.5, utility 5) and Right (prob 0.5, utility 25).
Calculate: (a) Expected value of C1, (b) Expected value of C2, and (c) The optimal decision at Root A.
Step 1: Compute Expected Value of C1:
V(C1) = P(Left) Utility(Left) + P(Right) Utility(Right)
V(C1) = 0.7 10 + 0.3 20 = 7 + 6 = 13.
Step 2: Compute Expected Value of C2:
V(C2) = P(Left) Utility(Left) + P(Right) Utility(Right)
V(C2) = 0.5 5 + 0.5 25 = 2.5 + 12.5 = 15.
Step 3: Root Decision at MAX Node A:
V(A) = max(V(C1), V(C2)) = max(13, 15) = 15.
Optimal Action: Choose move to Chance Node C2.
Conclusion: V(C1) = 13, V(C2) = 15; Root A selects C2.
Primary Academic References & Exam Sources
Every concept in Notes From a B.Tech Brain is synthesized from authoritative academic texts, seminal peer-reviewed research, verified exam archives, and renowned university lectures.