Best strategy for 6 sided dice game: Roll as many dice as you want, you lose if at least one $1$ appears.

Question

I am new here because we got stuck with a question during the weekend's discussion.

Let's imagine we are playing a dice game. You can roll as many 6-sided dice as you want, and so can your opponents, simultaneously. The highest sum wins, but if there is at least one "$1$" in the roll, you score $0$ points in total.

Is there "the" best strategy with how many dice to roll in order to statistically (unlimited repeats) win the game? If there is, how can we prove it?

Our only starting point was the counterprobability for each throw. Each dice has a $5/6$ chance not to be a $1$. So $1-(5/6 + 5/6 + \ldots)$ should be at least a "$1$" in the throw. That's all we were able to manage, but we are missing the link to further calculations.

Thanks a lot for your help.

Edit: Thanks for the hint with 1−(5/6)^n

We are playing one round with only 2 players.

If it is a draw --> reroll with the selected number of dice

The other player does not know how many dice the other player selects.

The first win ends the game.

If you roll "1" the whole round scores "0"

To the question: Are you trying to win or trying to maximise the expected number of points? We are trying to find the best number of dice to win the game, but it never has occurred to me, the the number of dice of the other player is of relevance! Thanks for the input

Also thanks a lot for all of your comments. I really appreciate it.

Answer 1

The probability of not getting a "$1$" in $n$ rolls for a six-sided dice is $\left(\frac{5}{6}\right)^{n}$. With that in mind we can go onto calculating the expected value of each dice roll. To calculate value of one roll, $V_1$, we shall take the probability of each outcome with the value it has, so we have: $$V_1=6\cdot\frac{1}{6}+5\cdot\frac{1}{6}+4\cdot\frac{1}{6}+3\cdot\frac{1}{6}+2\cdot\frac{1}{6}+0\cdot\frac{1}{6}=\frac{20}{6}=\frac{10}{3}$$ From here we can see that the value of $n$ rolls will be $$V_n=n\cdot\frac{10}{3}$$ To get the expected value for $n$ throws, let's denote it with $E_n$, we just multiply it with probabilty of not getting a "$1$" in $n$ dice rolls: $$E_n=n\cdot\frac{10}{3}\cdot\left(\frac{5}{6}\right)^{n-1}$$

There are numerous ways to find maximum of $E_n$, but to keep real analysis out of this we can use simple old brute force and just start plugging in $n\in\mathbb{N}$ until we see that $E_n$ is dropping. I used python for this, and here is the graph of the calculation: $$ $$" />

From here you can see that $E_n$ peaks at $5$ rolls, where the expected value is $6.697959533607684$, while for $6$ rolls it is $6.697959533607683$. So in practice you'll have the same chance of winning with either of $5$ or $6$ rolls.

Answer 2

You could formulate this as a simultaneous game in game theory.

Both players simultaneously choose a positive integer number of dice. Then you need to define a utility function for each player, so you might choose:

• Rolling a 1 gives a utility of 0, regardless of the other player's roll
• Rolling a higher sum than the opponent gives a utility of 1
• Drawing with the opponent gives a utility of 1/2

Note that in this version of the game, both players get 0 if they both roll a 1.

You can then compute the expected utility of choosing $n$ dice when your opponent chooses $m$ dice. "Solving" this game would mean finding a Nash equilibrium (a strategy profile where no player has an incentive to deviate). I was able to compute the expected payoff for up to 5 dice each (left number refers to the payoff for the row player).

This seems to suggest that:

• If the opponent chooses 1 die, then the best response is to choose 2 dice.
• If the opponent chooses 2 dice, then the best response is to choose 3 dice.
• If the opponent chooses 3 dice, then the best response is to choose 4 dice.
• If the opponent chooses 4 or more dice, then the best response is to choose 1 die.

So there may be no pure strategy equilibrium because it looks like there's always an incentive to deviate. (This isn't a complete proof since I haven't computed beyond 5 dice each).

At least you can know how many dice to choose if you can somehow figure out how many dice your opponent is using!

Answer 3

This answer continues the line of reasoning in Gary Liang's good answer (with some modifications), wherein we realize the setup as a simultaneous choice game and the problem as finding Nash equilibria.

First, for any choice $(n_A, n_B)$ of number of dice for players $A, B$, respectively, assign to player $A$ utility equal to the probability that $A$ wins less the probability that $B$ wins, and assign player $B$ the negative of that quantity, so that the game is zero-sum. Using the specification that ties are resolved by rerolling again with the same die counts, exact computations give the following payouts to player $A$, rounded to the nearest $10^{-3}$: \begin{array}{c|ccccc} A \backslash B & 1 & 2 & 3 & 4 & 5\\ \hline 1 & 0 & \color{red}{-0.407} & \color{red}{-0.244} & \color{red}{-0.056} & \color{green}{\boxed{0.107}}\\ 2 & \color{green}{\boxed{0.407}} & 0 & \color{red}{-0.235} & \color{red}{-0.139} & \color{green}{0.017} \\ 3 & \color{green}{0.244} & \color{\green}{\boxed{0.235}} & 0 & \color{red}{-0.131} & \color{red}{-0.061} \\ 4 & \color{green}{0.056} & \color{green}{0.139} & \color{green}{\boxed{0.131}} & 0 & \color{red}{-0.066} \\ 5 & \color{red}{-0.107} & \color{red}{-0.017} & \color{green}{0.061} & \color{green}{\boxed{0.066}} & 0 \\ \end{array} The best responses to rolling $1,2,3,4$ dice are, respectively, rolling $2,3,4,5$ dice, and the best responses to rolling $5$ or more dice is rolling $1$ die (best responses to strategies of $B$ are boxed). (Since no die count $6$ or greater is a best response, they will never be part of a Nash equilibrium, so we prune those die counts from the game.) In particular, we see that advantage of strategies is nontransitive, that is, no matter how many dice $n_A$ you roll, there is some count $n_B$ of dice that would favors your opponent. Thus, the best we can do is to find mixed-strategy Nash equilibria, a straightforward calculation (made tolerable by employing a c.a.s.).

There are at least $4$ Nash equilibriua; computing directly gives the probabilities for each strategy (again they are rounded here to the nearest $10^{-3}$): \begin{array}{c|llll} \textrm{strategy} & \textrm{equilibrium a} & \textrm{equilibrium b} & \textrm{equilibrium c} & \textrm{equilibrium d} \\ \hline 1 & 0.147 & 0.289 & 0 & 0 \\ 2 & 0 & 0 & 0.195 & 0.298 \\ 3 & 0.260 & 0 & 0.054 & 0 \\ 4 & 0 & 0.469 & 0 & 0.075 \\ 5 & 0.592 & 0.243 & 0.752 & 0.627 \\ \end{array}