<p>Well, depending on where you live, the probability is either 0 or 1, modulo cloudy weather.</p>
<p>From the picture below (from <a href="http://www.lunar-occultations.com/iota/planets/1207venus.htm" rel="nofollow noreferrer">this site</a>), 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).</p>
<p>But with binoculars (or without, if you have very sharp eyes), you should be able to see it over all of the US.</p>
<p><a href="https://i.sstatic.net/iS2wG.jpg" rel="nofollow noreferrer"><img src="https://i.sstatic.net/iS2wG.jpg" alt="occultation map" /></a></p>
<p><sup>(cyan=occultation at moonrise/moonset; red dotted=daytime occultation; blue=twilight occultation; white=nighttime occultation)</sup></p>
<hr />
<strong>Miscalculation</strong>
<p>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 &quot;true for all practical purposes, until disproved&quot;. Many, many factors go into a calculation like this (all the way down to mathematical axioms).</p>
<p>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.</p>
<strong>Uncertainties</strong>
<p>Of course there are always <em>uncertainties</em> associated with such calculations, which gives an uncertainty in the position of celestial bodies. But if you take a look at <a href="https://www.youtube.com/watch?v=yox5Isqwk4Q" rel="nofollow noreferrer">this video</a>, 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 <span class="math-container">$\sim15$</span> arcmin, the probability of which is virtually zero.</p>

CC BY-SA 3.0
2015-11-26T10:19:54.593Z
What is the probability of Venus not moving directly behind the moon at all?
user5434678
CC BY-SA 3.0
2015-11-26T10:22:45.330Z
Well, Venus is _always_ behind the Moon, as seen from some position. Usually, this position is somewhere in space, but on Dec 7, the position happens to be on Earth for a short while. The probability of this happening is 1.
pela
CC BY-SA 3.0
2015-11-26T10:26:55.037Z
Or do you mean, "What is the probability of astronomers making a wrong calculation"? The probability of this is of the order of zero. :)
pela
CC BY-SA 3.0
2015-11-26T10:31:08.793Z
@pele-The Odds of a Wrong calculation is what im asking yes,and can you provide a link to why there can be no human error in calculation envolved in this kind of event.
user5434678
CC BY-SA 3.0
2015-11-26T11:03:19.680Z
@TheVoid: Okay, I see. 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). [cont'd below]
pela
CC BY-SA 3.0
2015-11-26T11:05:26.743Z
But a calculation like this has to do with celestial mechanics, which is very well understood, and which continuously makes accurate prediction 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.
pela
CC BY-SA 3.0
2015-11-26T12:00:02.447Z
@TheVoid: I incorporated these comments in an edit and also wrote about uncertainties.
pela
CC BY-SA 3.0
2015-11-30T22:53:49.013Z
side comment: kind of like asking what are the odds that the sun will rise tomorrow morning
Jack R. Woods
CC BY-SA 3.0
2015-11-26T10:06:04.530Z
2015-11-26T11:59:20.557Z
2020-06-17T09:47:40.653Z
pela
<p>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?</p>

<the-moon><venus><planetary-transits>
Venus transiting behind the Earth’s moon December 7, 2015
user5434678
# Venus transiting behind the Earth’s moon December 7, 2015

(Asked by: username_0 on 2015-11-26)

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?

**Tags:** the-moon, venus, planetary-transits

---

## 1 answers

### Answer by username_1 on 2015-11-26. Score: 5 | Accepted answer

Well, depending on where you live, the probability is either 0 or 1, modulo cloudy weather.

From the picture below (from [this site](http://www.lunar-occultations.com/iota/planets/1207venus.htm)), 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.

[![occultation map](https://i.sstatic.net/iS2wG.jpg)](https://i.sstatic.net/iS2wG.jpg)

(cyan=occultation at moonrise/moonset; red dotted=daytime occultation; blue=twilight occultation; white=nighttime occultation)

---

**Miscalculation**

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](https://www.youtube.com/watch?v=yox5Isqwk4Q), 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.

> **Comments:**
> **username_0**: What is the probability of Venus not moving directly behind the moon at all?  
> **username_1**: Well, Venus is _always_ behind the Moon, as seen from some position. Usually, this position is somewhere in space, but on Dec 7, the position happens to be on Earth for a short while. The probability of this happening is 1.  
> **username_1**: Or do you mean, "What is the probability of astronomers making a wrong calculation"? The probability of this is of the order of zero. :)  
> **username_0**: @pele-The Odds of a Wrong calculation is what im asking yes,and can you provide a link to why there can be no human error in calculation envolved in this kind of event.  
> **username_1**: @TheVoid: Okay, I see. 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). [cont'd below]  
> **username_1**: But a calculation like this has to do with celestial mechanics, which is very well understood, and which continuously makes accurate prediction 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.  
> **username_1**: @TheVoid: I incorporated these comments in an edit and also wrote about uncertainties.  
> **username_2**: side comment: kind of like asking what are the odds that the sun will rise tomorrow morning  

---

