# Will the remainder of multiple dice rolls be fair if at least one roll is performed fairly?

Suppose Alice and Bob are playing a dice game. They each hold a six sided die and a cup. They shake their die in the cup, flip the cup on the table and reveal the roll at the same time. The result is the modular arithmetic of the sum of each die, so 2 + 3 = 5, 3 + 4 = 1, 5 + 3 = 2. The result is a value between 1 and 6 (the values of a single die), so any sum above 6 is "wrapped around" like a clock.

Now, Bob wants to cheat. He thinks he can manipulate the outcome of the roll by setting his die. If Alice continues to roll her dice fairly, can Bob change the outcome of the game?

So far, I don't think so. If Bob always rolled a 5, for example, Alice can roll 1 to get 6, roll 2 to get 1, 3 to get 2 and so on. Alice can effectively hit all the numbers. If Bob loaded the die to only roll 1, 3, 5, Alice can still hit 1-6 by rolling fairly.

Second question if the above is true: If the outcome is fair when at least one player acts fairly, can this fairness be extended to n-players? Example: if 25 people roll, and 24 of them set their die, can the 25th player ensure fairness by rolling fairly?

There's probably a simple proof or property out there that easily answers this, but I've had a bear of a time trying to find it.

• By 'fair', I assume you mean 'all results have equal probability'? – shardulc says Reinstate Monica May 25 '16 at 15:24
• Yes. The game would be unfair if Bob could manipulate the outcome of the end result. – John Nelson May 25 '16 at 15:35

Yes, one person rolling fairly is enough to ensure fairness. Let $X$ be the outcome of a fair die roll and let $Y$ be any random variable independent from $X$ taking values in $\{1,2,3,4,5,6\}$, with $P(Y=i)=p_i$ for $i=1,2,3,4,5,6$. Note that $\sum_{i=1}^6 p_1=1$. Then for $k \in \{1,2,3,4,5,6\}$,

\begin{align*}P(X+Y \equiv k \mod 6)&=\sum_{i=1}^6 P(X+Y \equiv k \mod 6 \text{ and } X \equiv i \mod 6)\\ &=\sum_{i=1}^6P(Y \equiv (k-i) \mod 6 \text{ and }X \equiv i \mod 6)\\ &=\sum_{i=1}^6P(Y \equiv (k-i) \mod 6 )P(X \equiv i \mod 6)\\ &=\sum_{i=1}^6p_{k-i}\cdot \frac{1}{6}\\ &=\frac{1}{6}\sum_{i=1}^6p_{k-i}\\ &=\frac{1}{6}\sum_{j=1}^6p_j\\ &=\frac{1}{6}. \end{align*}

Here the subscript $k-i$ in $p_{k-i}$ is taken to be the value $\ell \in\{1,2,3,4,5,6\}$ which is congruent to $k-i$ modulo 6. As $i$ ranges from $1$ to $6$, so does $\ell$.

Independence of $X$ and $Y$ is necessary to factor the probability in the third line. If the unfair player gets to see the fair die roll and then choose the value on their die, the argument breaks down.

This argument also shows that the result holds for the general case of $n$ players with at least one playing fairly. Let $X$ be the outcome of a fair player's roll, and $Y$ be the value in $\{1,2,3,4,5,6\}$ which is congruent to the sum of the other players' rolls modulo 6.

Yes, it is fair as long as one person rolls fairly and the unfair ones do not know the result of the fair roll in advance. Just imagine all the unfair rollers rolling first. Add up their rolls $\bmod 6$. Now when the fair roller rolls and the result is added in, the result is fair. This is an extension of your argument for two people.

Sum up and apply the modulus to all unfair rollers. You get a set of probabilities $p_i$ with $i=\{1..6\}$. Consider the additional fair roll which will have probabilities $p'_i=1/6$. Now consider the probability that the final result is, e.g. 3. This will happen in the following cases: $\{(1,2),(2,1),(3,6),(6,3),(4,5),(5,4)\}$. So its probability will be: $p''_3=p_1\frac16+p_2\frac16+p_3\frac16+p_6\frac16+p_4\frac16+p_5\frac16=1\cdot\frac16=\frac16$. You can follow the same procedure for the outcome and you'll find that all of them have probability $\frac16$.