Conditional distribution of $Z = (X-Y) / 3$, given $Y = y$ I'm trying to solve this problem, but without much success. Here we have that $X \sim \operatorname{Geo}(\theta)$, with $p(x) = \theta (1 - \theta)^x$, and 
$Y = X \mod 3$, thereby having $p(y) = \theta\frac{ (1 - \theta)^y}{
1 - (1 - \theta)^3}$, with $y = 0, 1, 2.$
Now I have to find the conditional distribution of $ Z = (X-Y)/3$, given $Y = y.$ I haven't found anything like this here (sorry if I duplicated a question, I tried not to). Can anyone help? 
 A: Perhaps a solution based on understanding what geometric distributions mean would be of more interest than a purely algebraic one.
Preliminaries: notation; Geometric distributions
Recall that a Geometric distribution with parameter $\theta$ describes the chances of observing a sequence of $x\in\{0,1,2,\ldots\}$ failures before the first success in a series of independent Bernoulli$(\theta)$ trials, whose values I will write $U_1,U_2, \ldots, U_n,\ldots,$ encoding (as usual) $U_i=1$ to represent success. Writing $p_\theta(x)$ for those quantities, independence implies
$$p_\theta(x+1) = p_\theta(x)\Pr(U_{n+1}=1) = p_\theta(x)\theta.$$
Conversely, this relation completely determines the distribution from the facts that (1) all probabilities must sum to unity and (2) the nonzero probabilities are those for $x\in\{0,1,2,\ldots\}$.
Solution
Let's interpret the $Y$ and $Z$ of the problem.  It is an obvious number-theoretic fact that any possible value $x$ of $X$ can be written in the form $$x=3z+y$$ where $y\in\{0,1,2\}$ and $z\in\{0,1,2,\ldots\}.$   When we condition on $Y=y$, we're saying there initially are $y$ failures and then there are $z$ groups of three failures each before success is observed.  The independence of the Bernoulli trials in each group of three implies any sequence of three failures has a chance
$$\rho = (1-\theta)^3.$$
Consequently, the independence of each (nonoverlapping) group of three failures implies
$$\Pr(Z=z+1) = \Pr(Z=z) (1-\theta)^3= \Pr(Z=z) \rho$$
for any $z=0,1,2,\ldots.$  Thus, conditional on $Y=y$, $Z$ has a Geometric distribution with parameter $\rho$.
Among the salient implications of this observation--ones that immediately solve the problem--are

*

*The distribution of $Z$ is independent of $Y$.


*The distribution of $Z$ is Geometric with parameter $\rho=(1-\theta)^3.$


*You can write the probabilities down immediately using the usual formulas for the geometric distribution, using the parameter $\rho$.

For those who prefer pure algebra, an amusing (and perhaps surprising) solution method is provided by the technique of decimation described at https://stats.stackexchange.com/a/35138/919.  This directly gives the distributions of $(Y,Z)$, from which the conditional distribution of $Z$ is found by dividing by the chance that $Y=y$.
