Dice Game 1

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Problem Statement

Suppose that you roll a dice. For each roll, you are paid the face value. If a roll gives 4, 5 or 6, you can roll the dice again. Once you get 1, 2 or 3, the game stops. What is the expected payoff of this game?

Solution

This problem demonstrates the Law of Total Expectation (also called the Tower Rule) in action. The key insight is that the game has a recursive structure: if you roll high, you continue playing, effectively restarting the game.

Step 1: Define the Random Variable

Let X be the total payoff from the game. We want to find E[X].

Step 2: Condition on the First Roll

Let Y be the outcome of the first roll. We condition on Y to break down the problem.

The first roll can result in two mutually exclusive scenarios:

Scenario A: Y \in \{1, 2, 3\} (Game Stops)

  • Probability: P(Y \in \{1,2,3\}) = \frac{3}{6} = \frac{1}{2}

  • If this occurs, the game stops immediately

  • The payoff is simply the face value of the die

  • Expected payoff given this scenario: E[X \mid Y \in \{1,2,3\}] = \frac{1 + 2 + 3}{3} = 2

Scenario B: Y \in \{4, 5, 6\} (Game Continues)

  • Probability: P(Y \in \{4,5,6\}) = \frac{3}{6} = \frac{1}{2}

  • If this occurs, you receive the face value and get to roll again

  • Expected face value: \frac{4 + 5 + 6}{3} = 5

  • After receiving this payoff, you roll again, which is equivalent to starting a new game

  • Therefore: E[X \mid Y \in \{4,5,6\}] = 5 + E[X]

Step 3: Apply the Law of Total Expectation

The Law of Total Expectation states:

E[X] = E[E[X \mid Y]] = \sum_y E[X \mid Y = y] \cdot P(Y = y)

In our case, we condition on the two scenarios:

E[X] = P(Y \in \{1,2,3\}) \cdot E[X \mid Y \in \{1,2,3\}] + P(Y \in \{4,5,6\}) \cdot E[X \mid Y \in \{4,5,6\}]

Substituting our values:

E[X] = \frac{1}{2} \cdot 2 + \frac{1}{2} \cdot (5 + E[X])

Step 4: Solve for E[X]

Expanding the equation:

E[X] = \frac{1}{2} \cdot 2 + \frac{1}{2} \cdot 5 + \frac{1}{2} \cdot E[X]
E[X] = 1 + 2.5 + 0.5E[X]
E[X] = 3.5 + 0.5E[X]

Rearranging:

E[X] - 0.5E[X] = 3.5
0.5E[X] = 3.5
E[X] = \frac{3.5}{0.5} = 7

Final Answer

The expected payoff of this game is 7.

Verification

We can verify this result by considering the infinite series expansion:

  • With probability \frac{1}{2}, we get 2 and stop: contribution = \frac{1}{2} \cdot 2 = 1

  • With probability \frac{1}{2}, we get 5 and continue:

  • With probability \frac{1}{2}, we get 2 more and stop: contribution = \frac{1}{2} \cdot \frac{1}{2} \cdot 2 = 0.5

  • With probability \frac{1}{2}, we get 5 more and continue: contribution = \frac{1}{2} \cdot \frac{1}{2} \cdot 5 = 1.25

  • And so on…

The infinite series sums to 7, confirming our result.

Key Insights

  1. Law of Total Expectation: This problem is a classic application of E[X] = E[E[X \mid Y]], where conditioning on the first roll breaks the problem into manageable pieces.

  2. Recursive Structure: When you roll 4, 5, or 6, the game effectively restarts, creating a self-referential equation: E[X] = \text{immediate payoff} + \text{probability of continuing} \times E[X].

  3. Conditional Expectation: The expected payoff depends on the first roll’s outcome, so we compute E[X \mid Y] for each possible value of Y.

  4. Stopping Condition: The game stops when rolling low (1, 2, 3), which provides a natural base case for the recursion.

  5. Intuition: The expected payoff of 7 makes sense: you have a 50% chance of getting an average of 2 (low rolls), and a 50% chance of getting 5 plus continuing, which creates the recursive structure.

Alternative Approach: Direct Calculation

We can also solve this by explicitly writing out the infinite series:

E[X] = \frac{1}{2} \cdot 2 + \frac{1}{2} \cdot \left(5 + \frac{1}{2} \cdot 2 + \frac{1}{2} \cdot \left(5 + \cdots\right)\right)

This is a geometric series that converges to 7.

Generalization

Problem: Roll a die. If you get values in set A, you stop and receive that value. If you get values in set B, you receive that value and continue.

Let:

  • p_A = P(\text{roll} \in A)

  • p_B = P(\text{roll} \in B)

  • \mu_A = E[\text{roll} \mid \text{roll} \in A]

  • \mu_B = E[\text{roll} \mid \text{roll} \in B]

Then:

E[X] = p_A \cdot \mu_A + p_B \cdot (\mu_B + E[X])

Solving:

E[X] = \frac{p_A \mu_A + p_B \mu_B}{1 - p_B} = \frac{p_A \mu_A + p_B \mu_B}{p_A}

Applications

  1. Gambling Theory: Understanding expected value in games with continuation rules

  2. Decision Making: Evaluating strategies with recursive decision points

  3. Stochastic Processes: Modeling processes with stopping conditions

  4. Finance: Valuing options or contracts with continuation clauses

  5. Game Theory: Analyzing games with recursive payoff structures

Common Pitfalls

  1. Forgetting the recursive structure: It’s easy to miss that rolling high means restarting the game, not just adding a fixed amount.

  2. Incorrect conditioning: Make sure to condition on the first roll properly and account for all outcomes.

  3. Algebraic errors: When solving E[X] = a + bE[X], be careful with the algebra: E[X] - bE[X] = a, so E[X](1-b) = a.

  4. Confusing with independence: The rolls are independent, but the total payoff depends on the sequence, creating the recursive relationship.

Summary

The dice game demonstrates how the Law of Total Expectation elegantly solves problems with recursive structure. By conditioning on the first roll, we derive the equation E[X] = \frac{1}{2} \cdot 2 + \frac{1}{2} \cdot (5 + E[X]), which yields an expected payoff of 7. This problem showcases how conditional expectation and recursive thinking combine to solve complex expected value problems.