[Math] Let’s play a dice game

diceprobability

Roll a die with 100 faces, labeled from 1 to 100.

  1. You get to roll once and receive the amount of dollars labeled on the face. How much would you like to pay for this roll?

  2. How much would you pay if you can re-roll the dice if you are unsatisfied with the first outcome?

  3. You can roll the dice infinitely many times, and each roll costs 1 dollar except the first one. What is your strategy?

My thoughts:

  1. The expectation of a 100 faces dice is 50.5, so this is the fair value.

  2. If I can re-roll once, I would re-roll if I rolled between 1 and 50. Now the expectation would be $0.5 * 50.5 + 0.5 * 75.5 = 63$.

  3. I guess my strategy would be keep re-rolling until the expectation improvement would be less than the cost of 1 dollar to re-roll? Can someone give me a closed form solution ?

Best Answer

Your answers for $1$ and $2$ are right.

For $3$, say you reroll if you roll at most $x$. Then your expected payoff is

$$ E=\frac x{100}(E-1)+\left(1-\frac x{100}\right)\frac{x+1+100}2\;, $$

and solving for $E$ yields

$$ E=\frac{10100-3x-x^2}{2(100-x)}\;. $$

Then setting the derivative with respect to $x$ to zero yields

$$ 10100-3x-x^2=(2x+3)(100-x) $$

with solution $x=10(10\pm\sqrt2)$. Only the smaller solution is feasible, $x=10(10-\sqrt2)\approx85.86$, and substituting the two adjacent integers, $85$ and $86$, into the expression for $E$ yields $\frac{262}3\approx87.33$ for $x=85$ and $\frac{1223}{14}\approx 87.36$ for $x=86$.

Thus a maximal profit of about $\$87.36$ is achieved if you reroll whenever you have at most $86$.

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