$100$-sided die probability follow-up: Why doesn't the maximizing EV strategy change?

Question

This is a follow up to the question here: 100-sided die probability

The question is as follows: You are given a 100-sided die. After you roll once, you can choose to either get paid the dollar amount of that roll OR pay one dollar for one more roll. What is the expected value of the game? There is no limit on number of rolls.

It is said at the linked question that the strategy to maximize expected value is to reroll when you get $86$ or lower, and the stop when you get $87$ or higher.

Question. Why doesn't the strategy change if you've rolled the dice once versus if you're now rolling the dice for the $10$th time?

Thoughts. The expected value of the game doesn't change. Each of the rolls are independent, hence not factoring in any losses from previous rolls. It's a closed-form game. The expectation doesn't depend on any previous rolls. So we consider each roll individually and don't consider the history.

Answer 1

Let $V$ denote the expected value.

Assume that you adopt the initial strategy that you will reroll if and only if you roll $A$ or lower.

Then

$$V = \frac{100 - A}{100} \times \left[ ~\frac{100 + (A + 1)}{2} ~\right] + \frac{A}{100} \left[ ~V - 1 ~\right]. \tag1 $$

(1) above needs some explaining.

The probability is $\dfrac{100-A}{100}$ that you will roll higher than $A$. When that happens, your average roll will be $\dfrac{100 + (A + 1)}{2}.$

The complementary event, not rolling above $A$, has probability of $~\dfrac{A}{100}~$ of occurring. When that happens, your expectation on the next roll will be $V - 1$, since you will have to pay $1\$$ to re-roll.

Now, suppose that you have rolled $n$ times, without exceeding $A$. What is your expectation on the $(n+1)$-st roll?

The die has no memory. The $(n-1)\$$ that you have paid so far is gone. So, the equation in (1) above still applies on the $(n+1)$-th roll.

Thus, the equation in (1) above always pertains, no matter how many re-rolls have been attempted. Therefore, the equation in (1) is the sole basis for determining the optimal value of $A$, regardless of how many re-rolls you take.

The idea is that any strategy that you adopt will be based on the idea that you re-roll if and only if you roll $A$ or lower.

Inherent in (1) above is the constraint that infinite re-rolls are available at $1\$$ each. This is built in to the

$$ \frac{A}{100} \left[ ~V - 1 ~\right] ~~\text{term}.$$

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Addendum

Since the linked answer does not use the following approach to compute the optimal value of $A$, I will include it:

$$V = \frac{100 - A}{100} \times \left[ ~\frac{100 + (A + 1)}{2} ~\right] + \frac{A}{100} \left[ ~V - 1 ~\right] \implies $$

$$V ~\left[ ~1 - \frac{A}{100} ~\right] = \frac{100 - A}{100} \times \left[ ~\frac{100 + (A + 1)}{2} ~\right] + \frac{- A}{100}\implies $$

$$V ~\left[ ~\frac{100 - A}{100} ~\right] = \frac{100 - A}{100} \times \left[ ~\frac{100 + (A + 1)}{2} ~\right] + \frac{100 - A}{100} - 1 \implies $$

$$V = \left[ ~\frac{100 + (A + 1)}{2} ~\right] + 1 - \frac{100}{100 - A} \implies $$

$$V = \frac{103}{2} + \frac{A}{2} - \frac{100}{100 - A} \implies \tag2 $$

$$\frac{dV}{dA} = \frac{1}{2} - \frac{100}{\left( ~ 100 - A ~\right)^2}.$$

So

$$\frac{dV}{dA} = 0 \iff (100 - A)^2 = 200 \iff 100 - A = 10\sqrt{2} \approx 14.14.$$

This indicates that the optimal value of $A$ is either $85$ or $86$.

Using (2) above:

• For $A = 85$:
  $\displaystyle V = \frac{188}{2} - \frac{100}{15} = 87.\overline{3}.$

• For $A = 86$:
  $\displaystyle V = \frac{189}{2} - \frac{100}{14} = 87.3\overline{571428}.$

So, $A = 86$ is superior to $A = 85.$