Observing from Earth what are the odds that Venus does not line up in conjunction with Earth’s moon and does not transit behind the moon but slightly below or above with a complete visual of Venus during the entire conjunction of the two? What are the chances of a miscalculation in this event?
Well, depending on where you live, the probability is either 0 or 1, modulo cloudy weather.
From the picture below (from this site), you see that you'll need to be in the US to see the occultation. However, only in the regions outlined in cyan (West Alaska, East Siberia, East Canada, and in Caribbean islands east of the Dominican Republic) will it be during darkness (just before sunrise / after sunset).
But with binoculars (or without, if you have very sharp eyes), you should be able to see it over all of the US.
(cyan=occultation at moonrise/moonset; red dotted=daytime occultation; blue=twilight occultation; white=nighttime occultation)
After your edit, I see that you are not referring to the probability of seeing the occultation, but the probability of a mis-calculation. I'd say that those odds cannot really be calculated, but it can be said that they are extremely small. No physical theory can ever be proved, but after a sufficient amount of verifications, we usually accept a theory as "true for all practical purposes, until disproved". Many, many factors go into a calculation like this (all the way down to mathematical axioms).
But a calculation like this has to do with celestial mechanics, which is very well understood, and which continuously makes accurate predictions and hence is continuously empirically verified. So the odds you request are definitely much, much smaller than, say, the odds of calculating the weather tomorrow, or the mass of a galaxy cluster.Uncertainties
Of course there are always uncertainties associated with such calculations, which gives an uncertainty in the position of celestial bodies. But if you take a look at this video, you'll see that the predicted path of Venus is more or less through the middle of the Moon. If the actual path should not cross the disk of the Moon, the calculation would be off by $\sim15$ arcmin, the probability of which is virtually zero.