"Exact" probability from Normal distribution I have a small problem. Lets say I have a system that emits random distances from a certain point in space, its a real number from a normal distribution with mean = 0 and a set variance I know (lets say its 1 for this example). Note that distance of -x and x are the same, so practical distribution is like a "folded" bell starting at 0 with densities being double at every x>0.
My problem is that I am now analyzing a stream of those distances and I need to determine the probability that the system would emit that value. Say, I need to know what the probability is that system emits 1.23. 
I know how to calculate P(X < dist), its CDF(dist) - CDF(-dist), but I have no idea how to work with a single point. Do I have to use some fixed ε environment e.g. 2*(CDF(x+ε)-CDF(x-ε))?
 A: The probability of any single value from a continuous distribution is zero. This is what makes a continuous distribution different from a discrete distribution. The normal distribution is a continuous distribution and hence the probability of getting exactly 1.23 is zero. Further, the probability of getting either 1.23 or -1.23 is also zeo. Said differently, the set $\{-1.23;1.23 \}$ has measure zero (the probability that any of its values are produced by a normal is zero). The set of rational numbers has also measure zero. Depending on what you are interested in, a way might be to work with intervals which seems to be what you have suggested. You might also consider the probabilities of distances greater than a certain value.
If however you are interested in estimating the parameters of your normal distribution, then the likelihoodfunction might be what you are interested in. Assuming that the system creates distances independently (each distance is drawn independently from other distances) and according to the same folded normal distribution, then this likelihoodfunction is given by the product of the densities of each distance. This however isn't equal to the probability of the observed distances.
