Probability that two consecutively generated integers with normal distribution are the same Given a random number generator that generates numbers $x\in\mathbb{R}$ with a normal probability distribution, with mean $\mu$ and standard deviation $\sigma$, and then rounds these numbers to the closest integer $y=\lfloor x+\frac{1}{2}\rfloor$.
I'd like to know if there's a way (other than simulation) to calculate the probability that two consecutive values generated by the generator are the same?
For example, if $\mu = 100.0$, $\sigma = 50.0$ and $y = 123$, what's the probability of this occurrence?
That is to say, what's the combined probability of two two consecutively generated raw numbers of the generator to be in the range $[y-\frac{1}{2},y+\frac{1}{2})$?
I'm guessing something like $(\Phi(y-\frac{1}{2})-\Phi(y+\frac{1}{2}))^2$ or a scaled version thereof?
 A: As others have stated in the comments, since the Normal distribution is continuous it will almost surely not return an integer value.
I will answer your question in the context of a general discrete random variable $X$ taking values in $\mathbb Z$ with
$$\mathbf P[X = x] = p_x.$$
We can use the law of total probability to find the probability of two consecutive independent samples $X_1,\,X_2$ taking the same value:
\begin{align*}
\mathbf P[X_1 = X_2] & = \sum_{x} \mathbf P[X_1 = X_2 \, | \, X_1 = x] \mathbf P[X_1 = x] \\
& = \sum_{x} \mathbf P[X_2 = x \, | \, X_1 = x] \mathbf P[X_1 = x] \\
& = \sum_x \mathbf P[X_2 = x] \mathbf P[X_1 = x] \\
& = \sum_x p_x^2.
\end{align*}
In general one would not expect this to series to admit a particularly pleasant closed formula.

Example
For instance if we consider the case that $X$ is Poisson distributed
$$\mathbf P[X = x] = e^{-\lambda} \frac{\lambda^x}{x!},$$
then we have
$$\mathbf P[X_1 = X_2] = e^{-2\lambda} \sum_{x} \frac{\lambda^{2x}}{(\lambda!)^2} = e^{-2\lambda} I_0(2\lambda),$$
where $I_0$ is a modified Bessel function of the first kind.

Update
Since my original answer the question has been updated to clarify that the distribution of interest is
$$ Y = \textstyle{\lfloor X + \frac12 \rfloor},$$
where $X \sim N(\mu,\sigma^2)$. We have for $y \in \mathbb Z$
\begin{align*}
p_y &= \mathbf P[Y = y] \\
& = \int_{y}^{y+1} \frac{1}{\sqrt{2 \pi \sigma^2} } \exp \left( - \frac{(x - \mu)^2}{2 \sigma^2} \right) dx \\
& = \Phi \left( \frac{y+1 - \mu}{\sigma} \right) - \Phi \left( \frac{y - \mu}{\sigma} \right).
\end{align*}
Hence we have an expression for the probability that two independent samples are equal
$$ \mathbf P[Y_1 = Y_2] = \sum_{y} \bigg( \Phi \left( \frac{y+1 - \mu}{\sigma} \right) - \Phi \left( \frac{y - \mu}{\sigma} \right) \bigg)^2. $$
As remarked above, I would not expect this to admit a simple closed form.
