Dice Game Optimal Stopping Strategy

A player is playing a dice game where he/she can roll a 6-sided fair die up to three times (1, 2, or 3 times). The payoff of the game is the value of the last roll. Assuming the player is rational with a goal to maximize the expected payoff, what is the optimal stopping strategy?

We will start with the last roll and work backwards to the first roll by breaking the problem down to three components: roll once, roll up to two times and roll up to three times.

Scenario #1: Roll once

If the player rolls the die once, the last roll is the first roll. The expected payoff of the last role is (1+2+3+4+5+6)/6 = 3.5

Scenario #2: Roll up to two times

If the player rolls the die twice, the last roll is the second roll. The expected payoff of the last role is again 3.5 ((1+2+3+4+5+6)/6). To beat this payoff, the player has to roll a 4, 5, or 6 on the first roll for us to stop the game early. The optimal strategy is: if the first roll is 4, 5, or 6, stop the game and walk away with the value on the first roll; if the first roll is 1, 2, or 3, roll the die the second time and take home the value on the second roll. The expected payoff for this strategy is 1/2*(4+5+6)/3+1/2*3.5 = 4.25

Scenario #3: Roll up to three times

If the player rolls the die three times, the last roll is the third roll. The expected payoff of the last roll is 4.25 (details are explained in scenario #2). To beat this payoff, the first roll has to be a 5 or 6 for us to stop the game early. The optimal strategy is: if the first roll is 5 or 6, stop the game and walk away with the value on the first roll; if the first roll is 1, 2, 3, or 4, roll the die the second time; if the second roll is 1, 2, or 3, roll the die the third time and take home the value of the third roll; if the second roll is 4, 5, or 6, stop the game and walk away with the value on the second roll. The expected payoff for this strategy is 2/6*((5+6)/2)+4/6*4.25 = 14/3

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