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We are throwing tetrahedron dice (like in board games) $n>0$ times. Let X and Y denote minimum and maximum from the results respectively. For what $n,k$ variables $(X \mod{k})$ and $(Y\mod{k})$ are independent?

Let $A=(X\mod k), \ B=(Y\mod k)$. These are discrete random variables, so I need to check when for all $a,b\in\left\{ 0,...,k-1\right\}$ we have: $P(A=a \wedge B=b)=P(A=a)\cdot P(B=b)$,right? But completely don't know how.

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up vote 3 down vote accepted

Let $X^\ast=X\bmod k$ and let $Y^\ast=Y\bmod k$. (For the sake of simplicity of notation, we do not explicitly mention $k$.)

Before computing, we can make an observation. If $k \ge 5$, then $X^\ast=X\bmod k=X$ and $Y^\ast=Y\bmod k=Y$. Now $\Pr(X=2)\ne 0$, and $\Pr(Y=1)\ne 0$, but $\Pr((X=2)\land (Y=1))=0$. Thus $\Pr(X^\ast=2)\ne 0$, and $\Pr(Y^\ast=1)\ne 0$, but $\Pr((X^\ast=2)\land (Y^\ast=1))=0$. So $X^\ast$ and $Y^\ast$ are not independent if $k \ge 5$.

The same argument works if $k=4$.

At the other extreme, if $k=1$, then $X^\ast$ and $Y^\ast$ are each $0$ with probability $1$, so $X^\ast$ and $Y^\ast$ are trivially independent.

So the only unsettled cases are $k=2$ and $k=3$. It is perhaps worthwhile to find the distributions of $X$ and $Y$. And maybe even the joint distribution. Even though finding the distributions is not really necessary, they give a "natural" approach to the problem, so we might as well take that path.

We have $X=4$ iff we get all $4$'s. Thus $\Pr(X=4)=(1/4)^n$. We have $X=3$ if we get all $3$'s and/or $4$'s, but not all $4$'s. Thus $\Pr(X=3)=(2/4)^n-(1/4)^n$. We have $X=2$ if we get all $2$'s, $3$'s, or $4$'s, but not all $3$'s and/or $4$'s. Thus $\Pr(X=2)=(3/4)^n-(2/4)^n$. Finally, $\Pr(X=1)=1-(3/4)^n$.

Similar reasoning gives the distribution of $Y$. We have $Y=1$ if we get all $1$'s. Thus $\Pr(Y=1)=(1/4)^n$. We have $Y=2$ if we get all $1$'s and/or $2$'s, but not all $1$'s. Thus $\Pr(Y=2)=(2/4)^n-(1/4)^n$. Similarly, $\Pr(Y=3)=(3/4)^n-(2/4)^n$, and $\Pr(Y=4)=1-(3/4)^n$.

Now we deal with the case $k=2$. We have $X^\ast=0$ if $X=2$ or $X=4$. Thus $\Pr(X^\ast=0)=(3/4)^n-(2/4)^n+(1/4)^n$. And of course $\Pr(X^\ast=1)=1-\Pr(X^\ast=0)$.

Similarly, $\Pr(Y^\ast=1)=(3/4)^n-(2/4)^n+(1/4)^n$ and $\Pr(Y^\ast=0)=1-\Pr(Y^\ast=1)$.

To study independence, we unfortunately need more information. Let us calculate, for example, the probability that $X^\ast=0$ and $Y^\ast=1$. This happens if the minimum is $2$ or $4$, and the maximum is $1$ or $3$. That can only happen if the minimum is $2$ and the maximum is $3$. So we need all $2$'s and/or $3$'s, but not all $2$'s and not all $3$'s. That has probability $(2/4)^n-2(1/4)^n$.

Simplify. We saw that the probability that $X^\ast$ and $Y^\ast=1$ is $(2^n-2)/4^n$. But $\Pr(X^\ast=0)=(3^n-2^n+1)/4^n$, and $\Pr(Y^\ast=1)=(3^n-2^n+1)/4^n$. So we ask, is it possible that $(2^n-2)/4^n=(3^n-2^n+1)^2/4^{2n}$?

The answer is no. For the above equation simplifies to $4^n(2^n-2)=(3^n-2^n+1)^2$. This does not hold at $n=1$. And if $n\ge 2$, then we can rewrite the equation as $(4^n)(2)(2^{n-1}-1)=(3^n-2^n+1)^2$. The highest power of $2$ that divides the left-hand side is odd, while the right-hand side is a perfect square. This is not possible. Applied nNumber theory!

We conclude that $X^\ast$ and $Y^\ast$ are not independent if $k=2$.

Finally, the case $k=3$ is easy to deal with. For now $X^\ast$ takes on the values $0$, $1$, or $2$, and $X^\ast=0$ precisely if the minimum is $3$. And $Y^\ast=2$ precisely if the maximum is $2$. These each have non-zero probability, but it is impossible for the minimum to be $3$ and the maximum to be $2$. It follows that $\Pr((X^\ast=0)\land (Y^\ast=2))\ne \Pr(X^\ast=0)\Pr(Y^\ast=2)$, so we do not have independence.

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thank you, finally understood :-) – xan Nov 6 '12 at 19:16

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