In a dice game where a player's payoff is whatever the die is rolled, each player can roll how many times they want. Each payoff gets added cumulatively (e.g. roll 5 and 5, then payoff = 10). The catch is that if the user rolled a six at any point in time, they get 0 (e.g. rolled 5 and 6, then payoff = 0).

Intuitively, raising $n$ (the number of die rolls) could raise the expected payoff but also increases the probability of at least rolling one six and getting 0 as a payoff. For example if you picked a really large $n$, the likelihood of rolling at least 1 six and getting 0 payout is extremely likely. But going from $n=1$ to $n=2$ you get a little higher expected payoff (checked the math manually in lieu of a general formula).

Going with the weighted average approach to come up with a formula for expected value, one part of the formula must be the weighted average of getting a 0 payout, i.e. the probability of rolling at least one six out of $n$ dice rolls equals $1-\left(\frac{5}{6}\right)^n$. As such, we get:

$E(X) =$ weighted average for each payoff

$E(X) =$ weighted average of getting 0 + the rest of the weighted average of payoffs

$E(X) = \left(1-\left(\frac{5}{6}\right)^n\right) 0$ + the rest of the weighted average of payoffs

However, I'm having trouble wrapping my head around coming up with the rest of the equation. How can I solve this problem?

  • $\begingroup$ Does the player roll all $n$ dice at once, or does he roll them one at a time, making a decision after each roll on whether or not to roll again? $\endgroup$ – awkward Oct 18 '17 at 13:27
  • $\begingroup$ Player pick n dice to roll at once. Although now I'm interested in how the calculation would differ if the player decides after each roll. $\endgroup$ – rawr rang Oct 18 '17 at 16:17
  • $\begingroup$ According to my calculations, it makes no difference in the final answer if the player makes a decision after each roll, but it is a different approach. $\endgroup$ – awkward Oct 20 '17 at 12:44

If no die equals $6$, the expected value for each die equals $\frac{1+2+3+4+5}{5} = 3$. There are $n$ dice, so the expected value equals $3n$. We thus get:

$$E(X) = \left(1 - \left(\frac{5}{6}\right)^n\right) 0 + \left(\frac{5}{6}\right)^n 3 n = \left(\frac{5}{6}\right)^n 3 n$$

The highest expected payout is achieved for $n=5$ and $n=6$, with $E(X) \approx 6.028$.


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