I'm having trouble reconciling the approaches to calculating the expected value in two different but similar dice game scenarios. Here's the first scenario involving a standard 6-sided die:
Problem Statement: "A player rolls a standard fair 6-sided die. If the player rolls a 6, the game ends and the player receives no payout. Otherwise, the player may quit and receive $k$, where $k$ is the upface of the previous roll. Otherwise, the player may roll again under the same rules. Assuming optimal play from the player, what is their expected payout?"
Given Solution: "Intuitively, given that the player does not roll a 6, the conditional expectation of their die roll is 3. The sum of the rest of the faces of the die is 15, which are distributed over 5 faces. Therefore, the player should keep any value at least 3 and re-roll any value below 3. Let $e_3$ be the expected payout of this strategy. We have that $e_3 = \frac{1}{3} \cdot e_3 + \frac{1}{2} \cdot 4$. Solving for this, we get that $e_3 = 3$, which is our optimal strategy."
In a similar problem involving a 100-sided die discussed here, the optimal threshold isn't directly taken as the expected value of the die's outcomes. Instead, a calculus approach determines the point at which the expected value of rolling exceeds the immediate payoff. The threshold calculated significantly differs from the simple average of die faces.
Question: Shouldn't the approach to finding the optimal stop condition be consistent across similar types of games? Specifically, if using the average value as a threshold is optimal for a 6-sided die, why wouldn't it be the same for a 100-sided die? Could the discrepancy arise from the larger range of values and probabilities, or is there a flaw in how the 6-sided die game's solution is presented?
I'm looking to understand whether I'm missing something fundamental between these two approaches or if the solutions should indeed be considered differently. Any insights or further explanations would be greatly appreciated.
