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> The math is the same as flipping a coin 7 times [...] I'm not saying that a particular hemisphere has a high change of having most of the land; I'm saying th

> The math is the same as flipping a coin 7 times [...] 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.

s1artibartfast 4y ago
I see, so if you have the freedom to define hemispheres however you want, what are the chances you can fit a definition that captures the majority. 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.
- mFixman 4y ago
  Your example is not the same for two reasons: 1. You are assuming there always are 7 continents and each of them contains 1% of all space on the Earth's surface. I assume there are 14 continents with 2% of the Earth's surface each, which is still incorrect but it's closer to reality. 2. I'm not defining hemisphere however I want: I'm using the canonical definition of hemisphere, but choosing any hemisphere for me dice. For example, two continents that appear on numbers 1 and 48 can never be in the same hemisphere. 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.
  - s1artibartfast 4y ago
    Thats interesting. A few questions: 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.
    - mFixman 4y ago
      The Northern, Southern, Eastern, and Western Hemispheres are just four of the infinite hemispheres of a sphere. 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.
      - s1artibartfast 4y ago
        I see, that joke went over my head. 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.
treeman79 4y ago
Take it account dice rolling in a small fixed area where they tend to get stuck together?

s1artibartfast 4y ago

I see, so if you have the freedom to define hemispheres however you want, what are the chances you can fit a definition that captures the majority. 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.

mFixman 4y ago
Your example is not the same for two reasons: 1. You are assuming there always are 7 continents and each of them contains 1% of all space on the Earth's surface. I assume there are 14 continents with 2% of the Earth's surface each, which is still incorrect but it's closer to reality. 2. I'm not defining hemisphere however I want: I'm using the canonical definition of hemisphere, but choosing any hemisphere for me dice. For example, two continents that appear on numbers 1 and 48 can never be in the same hemisphere. 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.
- s1artibartfast 4y ago
  Thats interesting. A few questions: 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.
  - mFixman 4y ago
    The Northern, Southern, Eastern, and Western Hemispheres are just four of the infinite hemispheres of a sphere. 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.
    - s1artibartfast 4y ago
      I see, that joke went over my head. 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.

treeman79 4y ago

Take it account dice rolling in a small fixed area where they tend to get stuck together?