# Is there a star over my head?

Say I'm standing up straight, and I draw a straight line from my core through the top of my head (perpendicular to the ground). What is the probability that that line intersects with a star?

EDIT: I'm not trying to exclude any stars. This should include stars which we've observed and stars which we haven't yet observed but can predict due to other things we've determined (like the overall star density of the universe). Also it should include all stars regardless of naked eye magnitude limit.

• Presumably you mean a naked eye magnitude star ? As the magnitude limit increases towards fainter stars, the probability is going to get very close to 1... Sep 8 '19 at 14:50
• @astrosnapper that's not obvious because of the finite age of the universe. Sep 8 '19 at 15:32
• Related: Olbers' paradox Sep 8 '19 at 16:45
• – user8021
Sep 9 '19 at 8:07
• @TaW: Not sure how that's relevant? But if our heads are about 6 inches apart, there's an angular difference of about 1 microdegree. The Sun is about half a degree across from Earth, so two people could both be under the Sun. (In fact, over 40000 people are under the Sun on average.) Sep 9 '19 at 10:24

### Summary

There's a 1 in 500 billion chance you're standing under a star outside the Milky Way, a 1 in 3.3 billion chance you're standing under a Milky Way star, and a 1 in 184 thousand chance you're standing under the Sun right now.

Big, fat, stinking, Warning! I did my best to keep my math straight, but this is all stuff I just came up with. I make no guarantees it's completely accurate, but the numbers seem to pass the sanity check so I think we're good.

Caveat the First: The numbers for stars other than the Sun are based on data with a great deal of uncertainty, such as the number of stars in the universe and the average size of a star. The numbers above could easily be off by a factor of 10 in either direction, and are merely intended to give a rough idea of how empty space is.

Caveat the Second: The numbers for the Sun and the Milky Way are based on the assumption that you are standing (or floating) at a random point on Earth. Anyone outside the tropics will never have the Sun over their head. People in the northern hemisphere are more likely to have Milky Way stars over their head, with the best odds being people near 36.8° N, because at that latitude straight up passes through the galactic center once a day.26

Note: You can mostly ignore everything in this answer and just look up the solid angle of the Sun to get the same result. All the other stars are really far away and very spread out. The difference in solid angle subtended is five thousandths of a percent more when we add the rest of the universe to the Sun.

### Background

Let's try to get a somewhat realistic, hard number. To do that, we'll need some assumptions.

As pointed out in Michael Walsby's answer1, if the universe is infinite (and homogeneous2), there is only an infinitesimal chance of there not being a star overhead, which normal math treats as exactly zero chance. So let's presume the universe is finite.

### Presumptions

• Specifically, let's presume the universe only consists of the observable universe. (Look up the expansion of the universe3 for further information.)
• Further, let's presume the contents of the observable universe are measured at their current (presumed) positions, not the position they appear to be. (If we see light from a star from 400 million years after the universe began, we would measure it as being about 13.5 billion light years away, but we calculate that it's likely closer to 45 billion light years away due to expansion.)
• We'll take the number of stars in the observable universe to be $$10^{24}$$. A 2013 estimate4 was $$10^{21}$$, a 2014 estimate5 was $$10^{23}$$, and a 2017 estimate6 was $$10^{24}$$, with each article expecting the estimate to increase as we get better telescopes over time. So we'll take the highest value and use it.
• We'll take the size of the observable universe7 to be $$8.8\cdot10^{26} \text{m (diameter)}$$, giving a surface area8 of $$2.433\cdot10^{54} \text{m}^2$$ 9, and a volume10 of $$3.568\cdot10^{80} \text{m}^3$$ 11.
• We'll take the average size of a star to be the size of the Sun, $$1.4\cdot10^{9}\text{m (diameter)}$$ 12. (I can't find any sources for average star size, just that the Sun is an average star.)

### Model

From here, we're going to cheat a bit. Realistically, we should model each galaxy separately. But we're just going to pretend the entire universe is perfectly uniform (this is true enough as we get farther away from Earth in the grand scheme of the cosmos). Further, we're going to start counting far enough out to ignore the Milky Way and Sun entirely, then add them back in later with different calculations.

Given the above presumptions, we can easily calculate the stellar density of the observable universe to be $$\delta = \frac{10^{24}\text{stars}}{3.568\cdot10^{80} m^3} = 2.803\cdot10^{-57} \frac{\text{stars}}{\text{m}^3}$$ 13.

Next, we need to calculate the solid angle14 subtended by a star. The solid angle of a sphere is given by $$\Omega=2\pi\left(1-\frac{\sqrt{d^2-r^2}}{d}\right)\text{ sr}$$ 15, where $$\Omega$$ is the solid angle in steradians16 (sr), $$d$$ is the distance to the sphere and $$r$$ is the radius of the sphere. Using $$D$$ as the diameter, that converts to $$\Omega=2\pi\left(1-\frac{\sqrt{d^2-\left(\frac{D}{2}\right)^2}}{d}\right)\text{ sr}$$. Given the average diameter presumed above ($$1.4\cdot10^{9}\text{m}$$), this gives an average solid angle of $$\Omega=2\pi\left(1-\frac{\sqrt{d^2-4.9\cdot10^{17}\text{m}^2}}{d}\right)\text{ sr}$$ 17.

At this point, we could set up a proper integral, but my calculus is rather rusty, and not very sharp to begin with. So I'm going to approximate the answer using a series of concentric shells, each having a thickness of $$10^{22}\text{m}$$ (about a million light years). We'll put our first shell $$10^{22}\text{m}$$ away, then work our way out from there.

We'll calculate the total solid angle of each shell, then add all the shells together to get the solid angle subtended by the entire observable universe.

The last problem to fix here is that of overlap. Some stars in the farther shells will overlap stars in the nearby shells, causing us to overestimate the total coverage. So we'll calculate the probability of any given star overlapping and modify the result from there.

We'll ignore any overlap within a given shell, modeling as if every star in a shell is at a fixed distance, evenly distributed throughout the shell.

### Probability of Overlap

For a given star to overlap closer stars, it needs to be at a position already covered by the closer stars. For our purposes, we'll treat overlaps as binary: either the star is totally overlapped, or not overlapped at all.

The probability will be given by the amount of solid angle already subtended by previous shells divided by the total solid angle in the sky ($$4\pi\text{ sr}$$).

Let's call the probability a given star, $$i$$, is overlapped $$P_i$$, the solid angle subtended by that star $$\Omega_i$$, and the number of stars $$n$$. The amount of non-overlapped solid angle subtended by a given shell, $$k$$, is then $$\Omega_{kT}=(1-P_1)\Omega_1+(1-P_2)\Omega_2+\ldots+(1-P_n)\Omega_n\frac{\text{ sr}}{star}$$. Since we've said stars in a shell don't overlap each other, $$P_i$$ is the same for all $$i$$ in a given shell, allowing us to simplify the above equation to $$\Omega_{kT}=(1-P_k)(\Omega_1+\Omega_2+\ldots+\Omega_n)\frac{\text{ sr}}{star}$$, where $$P_k$$ is the probability of overlap for shell $$k$$. Since we're treating all the stars as having the same, average size, this simplifies even further to $$\Omega_{kT}=(1-P_k)\Omega_k n\frac{\text{ sr}}{star}$$, where $$\Omega_k$$ is the solid angle of a star in shell $$k$$.

### Calculating Solid Angle

The number of stars in a shell is given by the volume of the shell times the stellar density of said shell. For far away shells, we can treat the volume of the shell as being its surface area times its thickness. $$V_\text{shell}=4\pi d^2 t$$, where $$d$$ is the distance to the shell and $$t$$ is its thickness. Using $$\delta$$ as the stellar density, the number of stars is simply $$n=\delta V_\text{shell}=\delta 4\pi d^2 t$$.

From here, we can use the calculation for solid angle of a shell (from Probability of Overlap, above) to get $$\Omega_{kT}=(1-P_k)\Omega_k \delta 4\pi d^2 t\frac{\text{ sr}}{star}$$.

Note that $$P_k$$ is given by the partial sum of solid angle for all previous shells divided by total solid angle. And $$\Omega_k$$ is given by $$\Omega_k=2\pi\left(1-\frac{\sqrt{d_k^2-4.9\cdot10^{17}\text{m}^2}}{d_k}\right)\frac{\text{ sr}}{star}$$ (from Model, above).

This gives us $$\Omega_{kT}=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)2\pi\left(1-\frac{\sqrt{d_k^2-4.9\cdot10^{17}\text{m}^2}}{d_k}\right) \delta 4\pi d^2 t\text{ sr}$$. Given that each shell is $$10^{22}\text{m}$$ away, we can substitute $$d_k$$ with $$k 10^{22}\text{m}$$. Likewise, $$t$$ can be substituted with $$10^{22}\text{m}$$. And we already calculated $$\delta=2.803\cdot10^{-57} \frac{\text{stars}}{\text{m}^3}$$ (from Model, above).

This gives us
$$\Omega_{kT}=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)2\pi\left(1-\frac{\sqrt{(k 10^{22}\text{m})^2-4.9\cdot10^{17}\text{m}^2}}{k 10^{22}\text{m}}\right) 2.803\cdot10^{-57}\frac{\text{stars}}{\text{m}^3} 4\pi (k 10^{22}\text{m})^2 10^{22}\text{m}\frac{\text{ sr}}{star}$$

$$=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)\left(1-\frac{\sqrt{k^2 10^{44}-4.9\cdot10^{17}}}{k 10^{22}}\right) 2.803\cdot10^{-57} 8\pi^2 k^2 10^{66}\text{ sr}$$

$$=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)\,2.213\cdot 10^{11}\,k^2\left(1-\frac{\sqrt{k^2 10^{44}-4.9\cdot10^{17}}}{k 10^{22}}\right)\text{ sr}$$

From here, we can just plug the numbers into a calculation program.

$$\Omega_{T}=\sum_{k=1}^{k_\text{max}} \Omega_{kT}$$

Where $$k_\text{max}$$ is just the radius of the observable universe divided by the thickness of a given shell. Thus $$k_\text{max}=\frac{4.4\cdot 10^{26} \text{m}}{10^{22} \text{m}}=4.4\cdot 10^4=44000$$

$$\Omega_{T}=\sum_{k=1}^{44000} \Omega_{kT}$$

### Results

Because of the large numbers involved, it's difficult to just run this in a program. I resorted to writing a custom C++ program using the ttmath library18 for large numbers. The result was $$2.386\cdot 10^{-11}\text{ sr}$$, or $$1.898\cdot 10^{-12}$$ of the entire sky. Conversely, there's about a 1 in 500 billion chance you're standing under a star right now.

Note that we ignored the Milky Way and the Sun for this.

The C++ program can be found at PasteBin25. You'll have to get ttmath working properly. I added some instructions to the top of the C++ code to get you started if you care to make it work. It's not elegant or anything, just enough to function.

### The Sun

WolframAlpha helpfully informed me the Sun has a solid angle of about $$6.8\cdot 10^{-5}\text{ sr}$$, or about 2.8 million times more than all the stars in the universe combined. The solid angle formula above gives the same answer18 if we provide the Sun's 150 gigameter distance and 0.7 gigameter radius.

### The Milky Way

We could get an approximation for the Milky Way by taking its size and density and doing the same calculations as above, except on a smaller scale. However, the galaxy is very flat, so the odds greatly depend on whether you happen to be standing in the galactic plane or not. Also, we're off to one side, so there are far more stars towards the galactic center than away.

If we approximate the galaxy as a cylinder with a radius of $$5\cdot 10^{20}\text{ m}$$ (about 52000 light years) and a height of $$2\cdot 10^{16}\text{ m}$$ (about 2 light years), we get a volume of $$1.571\cdot 10^{58}\text{ m}^3$$ 20.

Current estimates of the galaxy's radius are closer to 100000 light years21 22, but I'm presuming the vast majority of stars are a lot closer than that.

There are estimated to be 100 to 400 billion stars in the Milky Way21. Let's pick 200 billion for our purposes. This puts the density of the Milky Way at $$\delta = \frac{200\cdot10^{9}\text{stars}}{1.571\cdot 10^{58}\text{ m}^3} = 1.273\cdot10^{-47} \frac{\text{stars}}{\text{m}^3}$$ 22, or about 4.5 billion times denser than the universe at large.

This time, we'll take shells $$10^{17}\text{ m}$$ thick (about 10 light years) and go out from there. But we need to re-organize the math into a spherical form, so we'll presume the galaxy has the same volume, but is a sphere. This gives it a radius of $$1.554\cdot 10^{19}\text{ m}$$ 24, or 155.4 shells. We'll round to 155 shells.

$$\Omega_{T}=\sum_{k=1}^{155} \Omega_{kT}$$

Using our formula from above (Calculating Solid Angle), we can start substituting numbers.

$$\Omega_{kT}=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)2\pi\left(1-\frac{\sqrt{d_k^2-4.9\cdot10^{17}\text{m}^2}}{d_k}\right) \delta 4\pi d^2 t\frac{\text{sr}}{\text{star}}$$

$$=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)2\pi\left(1-\frac{\sqrt{(k\cdot 10^{17}\text{ m})^2-4.9\cdot10^{17}\text{ m}^2}}{k\cdot 10^{17}\text{ m}}\right) 1.273\cdot10^{-47} \frac{\text{stars}}{\text{m}^3} 4\pi (k\cdot 10^{17}\text{ m})^2 10^{17}\text{ m}\frac{\text{sr}}{\text{star}}$$

$$=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)\left(1-\frac{\sqrt{k^2\cdot 10^{34}\text{ m}^2-4.9\cdot10^{17}\text{ m}^2}}{k 10^{17}\text{ m}}\right) 1.273\cdot 10^{-47} \frac{\text{stars}}{\text{m}^3} 8\pi^2 k^2 10^{51}\text{ m}^3\frac{\text{sr}}{\text{star}}$$

$$=\left(1-\frac{\Omega_{(k-1)T}}{4\pi}\right)\cdot\,1.005\cdot 10^6\,k^2\,\left(1-\frac{\sqrt{k^2\cdot 10^{34}-4.9\cdot10^{17}}}{k 10^{17}}\right)\text{ sr}$$

Plugging this into the program gives $$3.816\cdot 10^{-9}\text{ sr}$$, which is $$3.037\cdot 10^{-10}$$ of the total sky. The odds you're standing under a star in the Milky Way are about 1 in 3.3 billion.

### Solid Angle Totals

Solid angle is:

• Sun, $$6.8\cdot 10^{-5}\text{ sr}$$
• Milky Way, $$3.816\cdot 10^{-9}\text{ sr}$$
• Universe, $$2.386\cdot 10^{-11}\text{ sr}$$
• Total, $$6.800384\cdot 10^{-5}\text{ sr}$$ (the extra digits are basically meaningless, adding about five thousandths of a percent to the Sun's solid angle)
• Milky Way plus Universe, $$3.840\cdot 10^{-9}\text{ sr}$$ (about 0.6% more than just the Milky Way)

### References

1 Michael Walsby's answer to this question, is there a star over my head?. https://astronomy.stackexchange.com/a/33294/10678
2 A Wikipedia article, Cosmological principle. https://en.wikipedia.org/wiki/Cosmological_principle
3 A Wikipedia article, Expansion of the universe. https://en.wikipedia.org/wiki/Expansion_of_the_universe
4 A UCSB ScienceLine quest, About how many stars are in space?, from 2013. https://scienceline.ucsb.edu/getkey.php?key=3775
5 A Sky and Telescope article, How Many Stars are There in the Universe?, from 2014. https://www.skyandtelescope.com/astronomy-resources/how-many-stars-are-there/
6 A Space.com article, How Many Stars Are In The Universe?, from 2017. https://www.space.com/26078-how-many-stars-are-there.html
7 A Wikipedia article, Observable universe. https://en.wikipedia.org/wiki/Observable_universe
8 A Wikipedia article, Sphere, section Enclosed volume. https://en.wikipedia.org/wiki/Sphere#Enclosed_volume
9 A WolframAlpha calculation, surface area of a sphere, diameter 8.8*10^26 m. https://www.wolframalpha.com/input/?i=surface+area+of+a+sphere%2C+diameter+8.8*10%5E26+m
10 A Wikipedia article, Sphere, section Surface area. https://en.wikipedia.org/wiki/Sphere#Surface_area
11 A WolframAlpha calculation, volume of a sphere, diameter 8.8*10^26 m. https://www.wolframalpha.com/input/?i=volume+of+a+sphere%2C+diameter+8.8*10%5E26+m
12 A nineplanets.org article, The Sun. https://nineplanets.org/sol.html
13 A WolframAlpha calculation, (10^24 stars) / (3.568⋅10^80 m^3). https://www.wolframalpha.com/input/?i=%2810%5E24+stars%29+%2F+%283.568%E2%8B%8510%5E80+m%5E3%29
14 A Wikipedia article, Solid angle. https://en.wikipedia.org/wiki/Solid_angle
15 Harish Chandra Rajpoot's answer to a geometry.se question, Calculating Solid angle for a sphere in space. https://math.stackexchange.com/a/1264753/265963
17 A WolframAlpha calculation, 2*pi*(1-sqrt(d^2-(1.4*10^9 m/2)^2)/d). https://www.wolframalpha.com/input/?i=2*pi*%281-sqrt%28d%5E2-%281.4*10%5E9+m%2F2%29%5E2%29%2Fd%29
18 Website for ttmath. https://www.ttmath.org/
19 A WolframAlpha calculation, 2*pi*(1 - sqrt(d^2 - r^2)/d), where d = 150 billion, r=0.7 billion. https://www.wolframalpha.com/input/?i=2*pi*%281+-+sqrt%28d%5E2+-+r%5E2%29%2Fd%29%2C+where+d+%3D+150+billion%2C+r%3D0.7+billion
20 A WolframAlpha calculation, pi * (5*10^20 m)^2 * (2*10^16 m). https://www.wolframalpha.com/input/?i=pi+*+%285*10%5E20+m%29%5E2+*+%282*10%5E16+m%29
21 A Wikipedia article, Milky Way. https://en.wikipedia.org/wiki/Milky_Way
22 A Space.com article from 2018, It Would Take 200,000 Years at Light Speed to Cross the Milky Way. https://www.space.com/41047-milky-way-galaxy-size-bigger-than-thought.html
23 A WolframAlpha calculation, (200*10^9 stars) / (1.571*10^58 m^3). https://www.wolframalpha.com/input/?i=(200*10^9+stars)+%2F+(1.571*10^58+m^3)
24 A WolframAlpha calculation, solve for r: (4/3)*pi*r^3 = 1.571*10^58 m^3. https://www.wolframalpha.com/input/?i=solve+for+r%3A++%284%2F3%29*pi*r%5E3+%3D+1.571*10%5E58+m%5E3
25 My C++ program code on PasteBin. https://pastebin.com/XZTzeRpG
26 A Physics Forums post, Orientation of the Earth, Sun and Solar System in the Milky Way. Specifically, Figure 1, showing angles of 60.2° for the Sun, and 23.4° less than that for Earth. https://www.physicsforums.com/threads/orientation-of-the-earth-sun-and-solar-system-in-the-milky-way.888643/

• Comments are not for extended discussion; this conversation has been moved to chat. Sep 11 '19 at 16:02

In short: no one knows for sure, but currently it looks that the probability is 1.

Longer: On our current understanding, the Universe is probably infinite in space. This depends on the recent WMAP satellite results, which have shown a zero curvature of the Universe below measurement precision. The other two options were a positive curvature (thus, we would live a 4D sphere), or a negative: If the curvature is exactly zero (the last option on the picture), or it is negative, and the Universe doesn't have some exotic topology, then it is infinite.

And an infinite Universe has infinite many stars, thus it doesn't matter, where do you see, somewhere you will find a star.

However, most likely you have no option to actually see it - it is nearly surely over the cosmological horizon, thus there is no way to get any information from it, or interact with it in any sense, due to the expansion of the Universe. Note, the currently accelerating expansion continuously reduces even the count of the stars inside the cosmological horizon.

Without an universal expansion, the whole sky would be filled with stars and it would be so light than the Sun (Olbers paradoxon).

If you count only the stars beside the cosmological horizon, then the probability is very small. The typical size of the stars is in the order of $$\approx$$ 1million km, and they are some light years away from each other ($$\approx 10^{13}$$ km). They are $$\approx 10^7$$ times more away from each other than their diameter. And even this calculation doesn't count that most space of the Universe isn't filled with any galaxy - the galaxies are disc-like objects around 20 times more far from each other than their diameter. You can find a more exact calculation in MichaelJ's pretty answer.

• Comments are not for extended discussion; this conversation has been moved to chat. Sep 11 '19 at 17:14

I don't want to recapitulate all MichaelS's lovely work above, so I'll do a quick back-of-the-envelope calculation borrowing from his numbers.

The area of a human head as viewed from above (or below) is, umm, let's see, average head width 6 to 7 inches, convert to modern units, ignore that heads aren't round -- that's about $$17cm$$ across, which makes just under $$0.03m^2$$ per head.

The Earth's surface area appears to be about $$500\cdot 10^{12} m^2$$. That area corresponds to a full spherical surface at a distance of one Earth radius from Earth's center.

From this, we can determine that one head, seen from Earth's center, covers about $$6\cdot 10^{-17}$$ of the full sky.

If we assume those $$10^{24}$$ stars (there may be more or fewer) are evenly distributed (they aren't), there are... lots and lots and lots of stars over your head at any given moment! Over a million, in fact.

Probably, maybe.

There are at least two ways of answering the question. One is to ask what were your coordinates when you wrote the question and exactly what time it was. Then we'll need to draw a line in a model to see what you hit and whether any of those hits are stars. This assumes a complete map, which is a problem. The answer is different for everyone on Earth and changes constantly. It becomes the right question if we are in a starship. Given the vastness of space, it’s probably better to ask “How far until we hit something.”

The other answer is about probability. How often is a star directly overhead? I'll suggest one way to reason about it. There seem to be a lot of limiting factors. I'll point out a few of those, too.

First, a gut check. Our Sun is directly overhead for a good area of the Earth at all times. The sun is relatively near, so it’s coverage is special. That trillions of billions of other stars have the rest of the planet covered nonetheless seems likely.

An excellent detail of this question is whether the line you are imagining intersects with a star. I take this to mean whether the abstract line passes through any part of the star's mass, not just its center of mass or other centers.

The odds are that we are not at the center of the Universe, if "center of Universe" even has any meaning. It can be argued (it is argued) that we are at the center of the observable universe, essentially because we're looking in all directions with the same limited gear. So we can imagine a giant sphere of observability, just to give this problem some space. Imagine ourselves as a grain of sand floating at the center of a big balloon. In truth, the grain of sand is far too big in proportion to any real balloon, but imagine we are in the dead center of a balloon on an impossibly small grain.

For the balloon's dimensions, consider a sphere with a radius of 4, where the units are $$1.1\times 10^{26}$$ meters. The surface of that sphere is going to be $$4\pi r^2$$, or $$64\pi$$ square units. If we prefer not to talk in terms with a "$$\pi$$" mixed in, it's roughly 200 of these large square units.

Imagine that this is the area that we are gazing upon from inside the center of the balloon, sitting on our microscopic and impossibly concentric grain of sand. We can only see half of the area at once (even less, really), but we are spinning around. So we can canvas the entire inside surface of the balloon over the course of the day.

So there we are, on this spec of sand, looking at the part of the balloon we can see. One of us has a laser pointer that we can use point to different parts of the balloon and talk about them. In fact, it might be fun to imagine the laser pointer having a kind of "light pen" mode that we can use to draw inscriptions on the surface of the balloon. Plastering your name across the night sky would make for quite a show. For the sake of the illustration, you have to imagine these props having metaphysical properties. We aren't really concerned with the light pen. It's just to imagine that we're drawing lines.

Now imagine that we tried to place inside of the balloon, at scale, all of the stuff of the observable universe, or, for the sake of the question, just the stars. We would put everything inside the balloon precisely where it would be relative to our vantage point.

Now we can go through, one at a time, and consider each star individually. Each time we examine a star, we could draw the line from us to it with our laser pointer. We could use the light pen to trace the outline of the star with the laser pointer, inscribing a little circle on the surface of the balloon behind it. Every time we did this with a particular star, we would add a circle on the balloon to build up a flat map of the stars. We could process each star, one by one, and eliminate each star until the balloon is empty again. It's just us, looking back up at the map we made.

Now let's say the balloon was originally red and our light pen was drawing in green. Let's also say that the green circles we drew were colored in, filled with green. After we've processed all the stars, we've got green dots all over the inside of the balloon. The size of each green dot would first be a function of the size of the star. Bigger stars would tend to draw relatively bigger circles on the map.

This analogy is imperfect in many ways. It is imperfect here in an important respect. If you imagine that we are tracing the stars with a circular motion in the hand, which is natural, then we'll be distorting the map. The angle of the light pen in hand as we made a circular motion would be projected across a great distance. That map would be interesting for other reasons, but we're trying to identify just the areas that are on line with us, stars we are "under." We want the real size of the star to be on the map, not a size relative to the distance between us and it.

To stay true, we'll have to imagine that our map simply has a circle on it whose center is on line with us and the star it represents. The size of the star's circle is its actual size. Our sun is roughly 1.39 million kilometers across, so the circle it draws would have this diameter on our map. This is the area of points that would, regardless of distance, carry a line all the way between them and us to make a candidate for a star being "overhead".

The answer to whether at least one star is probably overhead at a given time is, in one way of thinking, the proportion of red and green on the map. How much of the entire map is green? That's roughly how likely we are to be on line with a star at any time.

If we want to keep going on this line of probability, this would be the time to get the average size of every observable star, calculate an average diameter, multiply it by the number of stars, and have an estimated area. This will be wildly off because we've flattened three or four dimensions into two and didn't account for overlap. Unfortunately, the overlap of what's overhead does not appear to be consistent. Note that when looking up at the night sky we can see the Milky Way, of which we are a part.

Also, to get those averages, you'd have to have really thoroughly indexed the observable Universe. A lot of people have been working on that for a long time, but it's very large. So if we had enough data to have reasonably good averages for things like the size of a star, we may as well forget the averages and make the actual map. We'd take care of overlapping circles that way, too. While we're at it, forget the map entirely. Just have the GPS in your phone feed your position on the globe into a model that will draw the line and check everything above you. It's the real problem we started with, just taking into appreciation that the vastness of the cosmos is so overwhelmingly large that the computation required to check what's overhead may have a shorter radius than the radius of the observable universe. Even a database to hold records of every star and its position is impossibly large.

I also read lately that the universe might be (these are guesses and arguments) at least 250 times larger than what we can observe. I have also read that the earth is flat. Maybe the universe goes on infinitely. Reasoning about that will have similar boundary conditions.

Your best bet is to actually feed your location into a model and limit the model so you can get a reasonably fast calculation. Change the question to: “What is the nearest star on this line, given a spacial and computational boundary?” You'll have to accept that somewhere beyond what can be calculated, even beyond what can be seen, there may still be a star.

• Welcome on the Astronomy SE! Look how nice-looking formulas I inserted into your post. It is because we have some Latex support. Type $4\pi r^2$ and you will get $4\pi r^2$. Sep 8 '19 at 18:15
• Very nice. Thank you! Sep 8 '19 at 18:19
• The angular size of the star is what we want to project onto our balloon, not the linear size. If the star were 0 distance away, it would take up half the sky (presuming the ground counts as "sky" here), but if it's infinity distance away it takes up zero sky. Your solution is greatly underestimating the amount of green if the balloon is always outside the farthest star being considered. Sep 8 '19 at 23:07

According to Olbers, of paradox fame, if the universe is infinite a line of sight in any direction should eventually reach a star. Why then was the night sky so dark when in theory it should be bright as day? Leaving that particular question aside, we have no proof that the universe is infinite, but it is sufficiently large that a line in any direction should sooner or later reach the surface of a star. Whether the line in question would have to travel only tens of light years to reach the star or many billions depends on where you are standing and at what particular moment you choose to draw the line. If you happened to be on the equator at the right time of year and the right time of day, the line might only have to travel a little over eight light minutes to reach a star. In the universe, as opposed to on paper, you cannot have really straight lines, only an approximation, because the universe is curved and any line defining device, a laser beam perhaps, would be acted on by a force and bent.

• This reasoning is wholly incorrect. Even if you have an infinite universe, there may not be infinitely many stars. Furthermore, even if you have an infinite universe with infinitely many stars, there are still distributions such that the probability of any given line hitting a star eventually is 0. Sep 9 '19 at 6:43
• @Carl-FredrikNybergBrodda: Note that Olber's Paradox is based on notions of homogeneity and isotropy, which together are known as the Cosmological Principle. It's generally presumed true, even if it's not logically guaranteed. This paper from 2016 suggests odds of 120000:1 in favor of isotropy. Sep 9 '19 at 9:36
• There's zero evidence or rationale here to support that the universe is "sufficiently large" for this to work out. How big is big enough? Sep 9 '19 at 13:22