ARTIFICIAL INTELLIGENCE/Student Engineering Notebook
Page 06·notes
ADVERSARIAL SEARCH & GAME THEORY—~24 MIN READ

Stochastic Games & Expectiminimax Decision Trees

“Chance nodes, expected value propagation, and evaluation function linearity constraints.”

1.Stochastic Games & Expectiminimax Trees

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

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.

WORKED EXAM ANALYSIS · STEP-BY-STEP DERIVATION2 Marks

3. GATE Worked Numerical: Expectiminimax Root Decision

⚠️ Pitfall Warning:

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.

SOURCE TRANSPARENCY & CITATIONS

Primary Academic References & Exam Sources

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Curated University Lecture Series
Artificial Intelligence: A Modern Approach (4th Edition) ·
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