Why is there land at all at the south pole? Can't just be coincidence.
One pattern that is curiously consistent is that broadly land tends to be antipodal to ocean (and not just because there’s more ocean in general - there’s a genuine statistical skew) - so having land at one pole and none at the other matches that overall pattern. (https://www.geodatos.net/en/antipodes)
Basically a surprisingly large proportion of antipodal pairs have land at one end and water at the other (although water-water is also common).
In other words it would certainly be more surprising to have land at both poles; and we should be unsurprised if we find land at neither or land at one and not the other.
So by that measure the poles are fairly typical of other antipodes.
Regardless, it's just an interesting snapshot in time considering continental drift.
It’s easy to fall into some counterintuitive fallacies reasoning about this though because we’re considering a spherical surface pairwise and you might be tempted to say ‘ah, but we’re counting the antipodal land twice so that probably accounts for the 15%/30% thing’ - but I don’t think that follows; that 15% with land antipodal to it can be thought of as two sets of 7.5% of the land, which are antipodal to one another - the ‘double counting’ doesn’t work that way to eliminate the anomaly.
We have a 30% chance of picking a random point on Earth and getting land. Because earth is so big and these theoretical points are so small I'm going to assume that one point of land existing doesn't effect the odds of another piece of land existing. So that's two antipodal points, each with a 30% chance of being land.
Putting that into a probability calculator (again, long time since statistics) that gives us odds of both point A & B being land as 9%. If we flip that around and go with the 30% land and 70% water odds we end up with a chance of 21%. It seems like there's actually MORE land that's antipodal to other land than we'd expect statistically? I must have some bad math here.
Not sure where 21% comes from - if 3/10 of points are on land, 9/100 antipode-pairs should be land-land, and that means 9/100 / 30/100 = 9/30 = 30% of land should have land opposite it, and conversely 70% should have water?
But the true numbers are 15%/85% - on the actual earth, land is more likely to have ocean opposite it than random chance would suggest.
Analyzing whether items sufficiently far from the ‘expected’ random distribution as to be statistically significant is a rather different exercise though.
This is the issue.
A large contributor of the "point antipodal to land is water" anomaly is the typical size of land masses. If, for example, the average size of land masses was 30% of earth's area, then we'd have a single large land mass (Pangaea style), and all antipodal points would be water.
The approximation of independence here doesn't reflect reality. Eurasia alone is more than 10% of the land area of Earth.
EDIT: also, do we know if there were no other continents that submerged?
If we distribute chunks of land randomly on a sphere we would expect roughly the same amount of land on any randomly-chosen hemisphere. But what can we say about the average amount of land in the hemisphere with the most land?
My intuition says that in an Earth-like sphere, with 29% land and 71% water, there's a high chance of there existing a hemisphere with almost all of the land.
I have no idea how to prove this empirically, but it seems like a really nice maths problem.
Suspect there’s something to do with the fractal dimension of the coastlines as well. Agreed - sounds like a rich vein of math.
The locations of the continents spend the vast majority of the time either approaching a clustered state, or leaving it.
Because the thick continents with land are a minority of the surface [2], They only briefly pass through a period of even dispersal when passing from one clustered state to another.
>My intuition says that in an Earth-like sphere, with 29% land and 71% water, there's a high chance of there existing a hemisphere with almost all of the land
Can you explain more? If you have 7 or so plates with land, why would you expect them to have a high chance of being in the same hemisphere? The math is the same as flipping a coin 7 times. Flip the first and see which side it lands on, then flip the remaining 6 and see how many match the same hemisphere. What am I missing?
https://www.youtube.com/watch?v=gQqQhZp4uG8
https://www.imperial.ac.uk/news/199334/scientists-find-link-...
I'm not saying that a particular hemisphere has a high change of having most of the land; I'm saying that there will be a hemisphere that has this chance.
29% of Earth is land and 80% of this land is in the land hemisphere, so this problem would be the equivalent of rolling a D48 14 times and placing the dice on a table so that any 11 of the rolled numbers are visible if I look at it from the top.
I think that is also fairly solvable. let me think on it. I cant understand your example, but is this the same?:
Roll a d100 once per continent (e.g. 7 times) and write the numbers down. What percent of the time can you pick a range 50 numbers that captures all 7 results. the range can wrap around from 100 to 0.
The problem would be easy to monte carlo, but I would have to think about how to solve it in closed form.
Will update if I have a solution.
My question is the probability of being able to choose 11 of the 14 continents so that there exists a hemisphere that contains all of them, like the land hemisphere contains 80% of the Earth's land.
1)When you say you are using the canonical definition of hemisphere, do you mean North and South? Or are you talking about East and West as well?
2) Out of curiosity, why 14 continents?
If you are using North and South as pre-defined hemispheres, ignoring East and West, I do think it can be reduced down to a coin flip if you want to prove that plates are unlikely to almost all be in the same hemisphere. This is because allowing them to overlap allows more clustering, not less.
12/14 continents is >80% and 11/14 is <80%.
If you drop a continent randomly, it has a 50/50 chance of being in the north or south.
The probability of 12 or more being in the northern hemisphere is 0.647%. The probability of 12 or more being in the southern hemisphere is 0.647%. Together, the probability of 12 or more being in the same N/S hemisphere is 1.294%
You can do the same for E and W, getting another 1.294%, for a total of 2.588% chance. This is allowing the continents to overlap, so the real chance would be lower.
By "canonical definition of hemisphere" I'm talking about the surface of any half-sphere that covers half of the Earth; the "joke" in my proof is that you can choose any one.
For any two continents there's always a hemisphere that contains both of them (assuming you don't worry about the semicircle in the border of the hemisphere). If you put three equidistant continents with the maximum possible distance between them (so that they form an equilateral triangle whose centre is the centre of the sphere) then you can't do anything about it.
Think it like shining a light from some point in space that lights 80% of the land in Earth.
That is what I initially meant by having "freedom to define hemispheres however you want", eg set arbitrary centers.
That certainly makes it a more interesting problem and increases the chances substantially.