A Visit from our Neighbours

Probability Level 4

Some time in the near future (about $300$ million years from now), our neighbours from the planet $\text{Alpha Centauri Bb}$ decide to visit Earth. Since they have no idea about the places of interest on Earth and there are no humans alive (on Earth) to guide them either, they decide to pick a random longitude and latitude and visit that place. The aliens will be thrilled iff the chosen coordinate is underwater.

Let the probability that the aliens are thrilled be $P$ . Find $\lfloor 100 P \rfloor$ .

Details and assumptions :

Assume that Earth is a perfect sphere, and $70.8 \%$ of Earth's surface is underwater.
The longitude and latitude are chosen randomly and uniformly from $(-180 ^{\circ} , 180 ^{\circ} ]$ and $[- 90 ^{\circ}, 90 ^{\circ} ]$ respectively.
Check out a follow-up tougher problem in my solution once you have attempted this problem (or revealed solution).

Image credit: ESO/L. Calçada/Nick Risinger (skysurvey.org)

Insufficient Information 95 29 70 42 100 50 63

1 solution

Pranshu Gaba
May 8, 2015

tl;dr: We do not know the distribution of water on the surface of Earth. $P$ depends on the distribution, hence information is insufficient.

Follow-up Question: Find the complete range of $P$ in terms of $\Delta$ , and also find the distribution of water at the extreme values of $P$ .

The way in which the point is chosen makes a difference. If the point were to be chosen randomly on the planet, then each point is equally likely to be chosen and the answer would have been $\lfloor 70.8\rfloor = 70$

However, in this problem, latitude and longitude are chosen. This makes some points to be more likely to be chosen than others. The points near the poles are more likely to chosen than at the equator. This is because the circumference of the circle of latitude decreases as we move away from the equator.

If we prove that for some two distributions, the value of $P$ changes, then we have proven the $P$ depends on the distribution.

The surface area of a part of a sphere with radius $a$ is $2 \pi a (x_2 - x_1 )$ and is derived here . This result will be required in the next steps.

Suppose $\Delta$ fraction of the Earth's surface is covered with water. (The following figures are projections of the Earth with the equator as the $x$ axis, blue is water and green is land).

We will see two very different cases:

Case 1 : All the water is near the equator

case 1 case 1

Suppose the water lies in the latitudes $(- \alpha ^{\circ} , \alpha ^{\circ} )$ . Using trigonometry and formula of surface area of sphere, it can be shown that $\alpha^{\circ} = \sin^{-1}( \frac{\Delta}{2})$ . $~~P_1 = \frac{2 \alpha }{180}$ . When $\Delta = 0.708,$ $P_1$ turns out to be about $0.23$

Case 2 : All the water is near the poles

case 2 case 2

Suppose the water lies in the latitudes $[-90^{\circ}, (-90 + \beta)^{\circ} ] \cup [ ( 90 - \beta )^{\circ}, 90^{\circ} ]$ . In this case, $\beta^{\circ} = \cos^{-1} (1 - \frac{\Delta}{2})$ . $~~P_2 = \frac{2 \beta }{180}$ . When $\Delta = 0.708$ , $P_2$ is about $0.55$ .

Since $P_1 \neq P_2$ , we can conclude that $P$ depends on the distribution of land and water on the planet, and we can't tell the exact value of $P$ without knowing the distribution.

Please comment below if you find any error in the reasoning or if any step is not clear and I will explain it in detail.

Lovely problem!

Calvin Lin Staff - 6 years, 1 month ago

Thank you!!!! :)

Pranshu Gaba - 6 years, 1 month ago

However your starting statements are wrong:

"The way in which the point is chosen makes a difference. If the point were to be chosen randomly on the planet, then each point is equally likely to be chosen and the answer would have been

There are just as many points at latitude 0 as there are at latitude 89. I will agree that the circumference of these circles is different, but the number of unique/distinct points does not change (both infinite.) If you are choosing a random lat and long, then you are uniquely determining a point on the surface of the earth, and all points are equally likely (points have no area, so the claim that some points are more likely than others does not logically follow.)

So basically: Each point on the surface of the earth has a unique latitude and longitude. Each of these points is unique. And between any two such points there exists another point (properties of the real numbers.) Thus this is an even distribution of latitude, longitude pairs over the surface of the earth. Thus the probability can be determined because of the choice of the coordinate system used.

Jeff Ifland - 5 years, 7 months ago

Hmm.. I do not agree with your statements.

You said that "There are just as many points at latitude 0 as there are at latitude 89. I will agree that the circumference of these circles is different, but the number of unique/distinct points does not change (both infinite.) "

We cannot say that the number of points are equal just because they are both infinite. We know there are infinite irrational numbers and there are infinite real numbers, but there are more real numbers, aren't there?

Consider two different planets with the same area of land and for simplicity, say it is uniformly distributed across all longitudes. Planet A has all its land near the north pole, say from lat 90 to lat 90 - a. In this case, the probability of choosing land is $\frac{ a } { 180 }$ . Planet B has its land near the equator from lat 0 to lat b. The probability of choosing land is $\frac{ b } { 180 }$ .

Since the area of land is same in both cases, we can see that b ≤ a, with equality possible only when a = b = 90. This means $\frac{ b } { 180 } \leq \frac{a } { 180 }$ . See for the same area of land and different distribution of land, we get different probabilities. Therefore the probability does depend on the distribution of the land.

You mentioned that "Thus the probability can be determined because of the choice of the coordinate system used.". Can you determine the probability without knowing the distribution of the land ? I don't think that is possible.

As mentioned on this MathWorld page: Sphere Point Picking , Using spherical coordinates $\theta$ and $\phi$ where $\theta$ and $\phi$ are uniformly chosen from $\theta \in ( - 180 ^ {\circ } , 180 ^ {\circ} ]$ and $\phi \in [ -90 ^ {\circ } , 90 ^{\circ} ]$ is not a good idea if you want every point on the sphere to be chosen with equal probability. Using spherical coordinates results in points bunching up near the poles. This is basically what I meant when I said "The points near the poles are more likely to chosen than at the equator".

Pranshu Gaba - 5 years, 7 months ago

"We cannot say that the number of points are equal just because they are both infinite. We know there are infinite irrational numbers and there are infinite real numbers, but there are more real numbers, aren't there?"

You are right if we were dealing with different sets this might be an issue to consider and to prove one way or the other, however we are not. These sets are both the sets of real numbers between (-180,180] thus there are not just the same size, but in fact they are identical sets no matter the longitude. (There is a one to one correspondence between points at any latitude and any other latitude and each unique point at the first latitude maps to a unique point at the second.)

Your argument about the two planets is quite convincing however. While the number of points does not change from latitude to latitude, and all points have a unique coordinate pair, there is an issue where the density, while uniform around a latitude, is not uniform from latitude to latitude.

"Can you determine the probability without knowing the distribution of the land ? I don't think that is possible."

As you mention above, as long as the distribution of points covers the surface, and has uniform density across the surface.

"is not a good idea if you want every point on the sphere to be chosen with equal probability. Using spherical coordinates results in points bunching up near the poles. This is basically what I meant when I said "The points near the poles are more likely to chosen than at the equator"."

I also have some issue with this statement: every point on the sphere has equal probability of being picked, however the density does not remain constant. So what I take from that article is that while each point is equally likely, the likely hood of a point being in a small region on the sphere is not consistent, due to this change in density of the points.

So at this point I am agreeing with your answer (you have swayed me on the correct answer). However I am not agreeing with all of your arguments as to why.

Thank you for the problem, and for the discussion!

Jeff Ifland - 5 years, 7 months ago

A Visit from our Neighbours

Image credit: ESO/L. Calçada/Nick Risinger (skysurvey.org)

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