## Probabilty of your opposite getting their feet wet.

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Red Hal
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### Probabilty of your opposite getting their feet wet.

This question came out of a discussion elsewhere on the fora about creating an "Earth Sandwich". In theory, align two slices of bread precisely opposite each other on the earth's surface and, voila, you have an "Earth Sandwich".

In practice, assuming that bread placed on a body of water quickly goes soggy, there are surprisingly few parts of the Earth's surface where this can be achieved. To wit, there are very few areas of dry land that have another piece of dry land opposite them, making the tasty morsel a relatively difficult feat.

I got to thinking about this, and why it should be so. So my question is this:

The surface of a sphere has been completely shaded in two colours, amber and blue. Given any ratio of amber to blue, what is the probability that two points on the sphere that are directly opposite each other will be shaded the same colour?

Also, given the ratio, and the knowledge that one point is shaded a particular colour, what is the probability that it's opposite will also be that colour?

I would imagine that the probability distribution for the first case would approximate a shallow parabola, but what would the shape for the second case be?

Any ideas on where to start?
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M.qrius
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### Re: Probabilty of your opposite getting their feet wet.

Isn't the second one easy? Since a point has no size, you can define the point to be your color, and the chance that the other side is the same color is just the part of the earth that's shaded in that color. As in, if you have 0.3 chance of blue, and you pick your point as blue, there's 0.3 chance of having the other point blue as well. If you pick it as amber, there's 0.7 chance of the other side being amber.

Perhaps something similar can be done for the other case?

Edit:
Liiiike.... let's say we have a distribution which gives us 0.3 chance of blue, at any point. Then let's add the chance of 2 blues to the chance of 2 ambers, as in, 0.3*0.3 + 0.7*0.7. Or, for a more general case, x2+y2 = P, if your ratio is x/y.

These solutions assume your point has no size. If you want 2 patches of land to be the same color, say, patches which are slice-of-bread sized, you have to take the reduced surface area into account, and change the ratio accordingly.

I leave it to someone else to turn this relation into a formula dependent on the ratio, take reduces surface into account, and to make graphs. I'm supposed to be studying... (I have a test in 2 hours)
Last edited by M.qrius on Tue Nov 04, 2008 10:12 am UTC, edited 1 time in total.

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### Re: Probabilty of your opposite getting their feet wet.

doesn't this assume there is no correlation which clearly isn't true for land/sea
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M.qrius
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### Re: Probabilty of your opposite getting their feet wet.

Aye, I assumed no correlation. It's hard to grasp the correlation in formulas though.

(Also note the edited previous post)

jestingrabbit
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### Re: Probabilty of your opposite getting their feet wet.

There's not enough information to answer this at present. For instance, say you have it half one colour and half the other. There are colourings where every point is coloured differently to its opposite, and there are colourings where every point is coloured the same as its opposite.

So it comes down to what you want to assume...
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Red Hal
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### Re: Probabilty of your opposite getting their feet wet.

@Evilbeanfiend, I agree, but one must start somewhere.

@M.qrius. A fair point, and one which I hadn't considered (a case of wood and trees! ).

How would we then utilise this in the Earth Sandwich case? What proportion of the Earth's surface would we reasonably expect to be suitable for this?

The proportion of the Earth's surface that is land is 29.07% [Science Desk Reference American Scientific. New York: Wiley, 1999: 180.], so working with your assumption, we should expect that the proportion of the Earth's surface (in the absence of other factors) that is opposed by land should be in the region of 9% (30% of 30%).

@jestingrabbit, I concur that for any given ratio, one can colour the sphere such that there are no two antipodal points that are the same colour, hence the use of probabilities.

What I was trying to tease out was what the statistical likelihood was of our current configuration of land, compared to the random case.
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### Re: Probabilty of your opposite getting their feet wet.

Yeah, a point of land is more likely to be next to another point of land rather then sea, in the real world. Which makes it's opposite less likely to be land. If you really wanted to know the chances in the real world you could take a "two hemispheres style" map of the earth, and take the flipped mirror image on one and overlay it on the other hemisphere and see where it overlaps.
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jestingrabbit
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### Re: Probabilty of your opposite getting their feet wet.

The 9% figure is good enough for a "what you might expect" guess.

But, it misses some of the point, and now part of me is thinking about sigma algebras on sigma algebras. Congratulations on your nerd snipe.
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Red Hal
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### Re: Probabilty of your opposite getting their feet wet.

Thanks! Although it was a genuine question. Flipping the globe has already been done, I suppose I just found the figure counter-intuitively low, and wondered if it was my intuition at fault (yep! ) or whether there were other forces at work.
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phlip
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### Re: Probabilty of your opposite getting their feet wet.

BoomFrog wrote:If you really wanted to know the chances in the real world you could take a "two hemispheres style" map of the earth, and take the flipped mirror image on one and overlay it on the other hemisphere and see where it overlaps.

I have that image handy... from that other thread. I did notice when I first made that picture, how North America seems to fit quite cleanly opposite the Indian Ocean, and Australia the Atlantic... pretty much the only intersections of major landmasses are South America/Southeast Asia, and assorted Arctic Circle places with Antarctica.

I also happen to have the ETOPO5 data handy... please don't ask me why. From that data, I get about 28.6% of the Earth being above sea level (that is, a value of >= 0 in ETOPO5). Of that landmass, about 12.8% has antipodal land (that is, about 3.67% of the Earth). These figures were calculated assuming the Earth is a perfect sphere (ie the area covered by a paticular datum is proportional to its latitude)... calculations with an ellipsoidal Earth would be slightly different (though I'd expect them to be within a reasonable range). I'd download the newer ETOPO1, but I really don't think the increased resolution is that important here.
Last edited by phlip on Tue Nov 04, 2008 11:15 am UTC, edited 1 time in total.

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Red Hal
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### Re: Probabilty of your opposite getting their feet wet.

So, the actual value differs from the expected value by a factor of approximately 2.5. That's quite a difference! I wonder how likely that is statistically?
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phlip
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### Re: Probabilty of your opposite getting their feet wet.

I just want to point out (mainly to convince myself) that this is different to the "at least one of the two kids is male, what's the probability that the other one is too?" trap... as when a point is land, and its antipodal point is water, then it counts as one point without antipodal land... but when a point is land and its antipodal point is land, it counts as two points with antipodal land. So while the answer to the question "I picked two points at random, and at least one is land, what's the chance that the other one is?" is less than 30%, the answer to the question "how many points of land are antipodal to another point of land" would be 30% if the land-ness of all the points on the Earth were independent (which they're not, of course, but that's not important, at least as far as this post's concerned). The equivalent question in the boy/girl setup would be "What percentage of boys born in two-child families have brothers?" which is, indeed, 50%.

I'm sure all of you figured that out already, but my sleep-deprived brain was naggling at me, worried that we'd fallen into a bad-statistics trap, by claiming that the expected figure is 30%² = 9%. I'm now mostly sure we haven't.

Also: a point to further drag in the nerd-snipees... the land-ness of a point is not a binary proposition... points have a height above, or a depth below, sea level... and a point, say, 1000m above sea level is much less likely to be near water as a point, say, 10m above sea level. Some modelling is clearly necessary on what is a "likely" heightmap for a planet, which will most likely put the large majority of the land in discrete continents, and be less likely to describe a planet which is basically just one big random-scatter archipelago. This, in turn, will almost certainly affect the expected amount of antipodal land on the planet.

And, on that note, I leave you for bed.

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M.qrius
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### Re: Probabilty of your opposite getting their feet wet.

Going on speculation here:

The fact that land tends to come in large masses, reduces the chance1 when the ratio is low.

1chance of earth-earth antipodal, when there's more water than earth on the globe.

*Tries to think of arguments for hypothesis

Intuitively, I'd say something like, if you start with a point of earth, you don't influence the ratio of land/water in the area outside your point, since a point has no size. But if you take the landmass into account, if you start with a point of earth, it is also very likely that you have pinpointed an entire mass of earth. This would reduce the amount of earth outside your landmass. I think outside your landmass the chance for earth is still the (new2) ratio, no matter if it's massed together or not.

2new as in, the ratio when you take away your pinpointed landmass

Also, I feel offended3 that people assume only land above sea level is land. For the record: I'm from the Netherlands.

3not really

phlip
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### Re: Probabilty of your opposite getting their feet wet.

Yeah, well, given the topological data was the only thing I had to hand, I used that. If you know of some data I could use that would let me include the Netherlands (and exclude lakes, and similar above-sea-level bodies of water), then I'd be happy to recalculate it all...

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Seraph
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### Re: Probabilty of your opposite getting their feet wet.

Red Hal wrote:I concur that for any given ratio, one can colour the sphere such that there are no two antipodal points that are the same colour, hence the use of probabilities.

I'm not sure I agree with you.

I would say that for any ratio other then exactly 50%, there will always be a set of antipodal points that are the same color.

If every point of Color A was antipodal to a point of color B, and vice-versa), then doesn't that imply you have a 50/50 mix of the two? Therefor if you don't have a 50/50 mix, then there exists some point who's antipode is the same color.

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### Re: Probabilty of your opposite getting their feet wet.

So we need some kind of clumping parameter to describe how land is, well, clumpy.

Part of what makes the percentage so low on the Earth is the pacific ocean, which covers almost an entire hemisphere.

In a sense, antipodal is a special case of "two points X distance apart, what is the probability they correlate". That also happens to describe a certain degree of clumping -- if two points near each other have a high correlation, that is a kind of clumping.

In the case of the spherical space, the set of points that are at X distance from a given point form a circle until it collapses to a point at the antipode (that a word?), or at X=0. The graph of "given that the center point is land, what percent of the circle at distance X is also land" and "given that the center point is water, what percent of the circle at distance X is water" might be pretty to look at.
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Red Hal
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### Re: Probabilty of your opposite getting their feet wet.

Seraph wrote:
Red Hal wrote:I concur that for any given ratio, one can colour the sphere such that there are no two antipodal points that are the same colour, hence the use of probabilities.

I'm not sure I agree with you.

I would say that for any ratio other then exactly 50%, there will always be a set of antipodal points that are the same color.

If every point of Color A was antipodal to a point of color B, and vice-versa), then doesn't that imply you have a 50/50 mix of the two? Therefor if you don't have a 50/50 mix, then there exists some point who's antipode is the same color.

Absolutely right! I was confusing the two cases and not thinking clearly. /slaps own wrist.
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### Re: Probabilty of your opposite getting their feet wet.

Red Hal wrote:
Seraph wrote:
Red Hal wrote:I concur that for any given ratio, one can colour the sphere such that there are no two antipodal points that are the same colour, hence the use of probabilities.

I'm not sure I agree with you.

I would say that for any ratio other then exactly 50%, there will always be a set of antipodal points that are the same color.

If every point of Color A was antipodal to a point of color B, and vice-versa), then doesn't that imply you have a 50/50 mix of the two? Therefor if you don't have a 50/50 mix, then there exists some point who's antipode is the same color.

Absolutely right! I was confusing the two cases and not thinking clearly. /slaps own wrist.

Are you sure? Just because there's a bijection between the points of two shapes, doesn't mean they have the same area. On second thoughts this looks right, but I can't see how to prove it one way or the other.
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### Re: Probabilty of your opposite getting their feet wet.

Even if there are some areas of measure zero which are of the dominant color, you still need measurable area covering over half the sphere, so that sounds right. Check this out

We put a measure on the projective plane by the induced measure of the upper half sphere (since the equator has measure zero, we don't lose anything as far as I can tell) Note that rotating the sphere is simply a translation of the sets on the sphere, and when projecting to the projective plane maybe splitting a set into two (but never more subdivisions than that) so all rotations of the sphere give the same measure. To be honest I'm a little fuzzy doing this, but it sounds like it gives us a measure, and we get that each point in the projective plane is measured either amber, blue or both (depending on whether it passes only through blue points, only through amber points, or both) We call the sets A and B for the amber points and the blue points, and we're interested in M(A-B) the measure of the set of points that are amber and not blue in P

Assume M(P) = 1 where P is the projective plane. Then A union B = P and M(P) = M(A) + M(B) - M(A intersect B). We get that the chance of making the sandwich is M(A-B) = M(A) - M(A intersect B) = M(P) - M(B)

IMPORTANT: These are all measures in the projective space, NOT the sphere (these aren't the same)

So, e.g. if we color one half of the sphere blue and one half amber, M(B) = M(P) so the chance of making a sandwich is zero (M(B) = M(P) as every projective point passes through a blue point on the sphere). Hence I conclude this is a pretty decent model. In particular, if the measure of the blue space is strictly less than half of the sphere, then M(B)<M(P) and there exists a non-zero probability you can make the sandwich. If the measure of the blue space is greater than half of the sphere in Euclidean space, then.... well, then you can't be sure actually. You could have half the sphere covered by blue, in which case you can't make the sandwich and on the other hand, you could have just a thin band of amber going around the equator, and you could make the sandwich with non-zero probability.

But going back to my original point, if the measure isn't split 50/50, WLOG we call the space with less measure B, and then there is a measurable region on the sphere that has amber on both sides
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### Re: Probabilty of your opposite getting their feet wet.

Has anyone considered the option that Earth is just not sticking to logical mathematics?
I think the Earth is broken.

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### Re: Probabilty of your opposite getting their feet wet.

jestingrabbit wrote:The 9% figure is good enough for a "what you might expect" guess.

But, it misses some of the point, and now part of me is thinking about sigma algebras on sigma algebras. Congratulations on your nerd snipe.

Right, we need a probability measure on the set of measure x subsets of the sphere, and I'm not sure if there are any. I've done a little work with random closed sets in a polish space, but those tend to have measure 0, so don't really help us here. The other thing that one would be inclined to do is try to get a measure on the set of subsets of the sphere by independently coloring each point amber with probability x, and blue with probability 1-x. My intuition tells me that you could indeed get a measure on a sigma-algebra of subsets of the sphere, but the probability of producing a set with measure x will not be 1, or even positive, but rather 0, and in fact you will almost certainly get a nonmeasurable set. So again, this doesn't give us a way of answering the question.

Macbi wrote:
Seraph wrote:I would say that for any ratio other then exactly 50%, there will always be a set of antipodal points that are the same color.

If every point of Color A was antipodal to a point of color B, and vice-versa), then doesn't that imply you have a 50/50 mix of the two? Therefor if you don't have a 50/50 mix, then there exists some point who's antipode is the same color.

Are you sure? Just because there's a bijection between the points of two shapes, doesn't mean they have the same area. On second thoughts this looks right, but I can't see how to prove it one way or the other.

Right, but in this case the bijection is the map [imath]x \mapsto -x[/imath], which is measure preserving. So if the image of A under that map is B, then A and B have the same measure. Office_Shredder has a more detailed (and possibly overcomplicated) explanation.
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jestingrabbit
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### Re: Probabilty of your opposite getting their feet wet.

skeptical scientist wrote:
jestingrabbit wrote:The 9% figure is good enough for a "what you might expect" guess.

But, it misses some of the point, and now part of me is thinking about sigma algebras on sigma algebras. Congratulations on your nerd snipe.

Right, we need a probability measure on the set of measure x subsets of the sphere, and I'm not sure if there are any. I've done a little work with random closed sets in a polish space, but those tend to have measure 0, so don't really help us here.

The sigma algebra I've been thinking about for all the measureable sets is generated by the topology associated with the metric [imath]d(A, B)= \mu(A \Delta B)[/imath] but its ugly: its not locally compact; there does seem to be a countable dense set so I think it is countably generated; its not immeadiately clear which sets (of sets) are measurable though; nor that the space is measurable in any nice way.

I suspect we can't put a probability measure on it, or at least not in a nice way.
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If we have a Lebesgue probability space [imath](X, \mathcal{B},\mu)[/imath] then there is a countable sequence [imath]\{A_i\}\subseteq \mathcal{B}[/imath] such that [imath]\mu(A_i)= 1/2[/imath] and [imath]d(A_i, A_j) = 1/2[/imath]. Now, we can put balls around the [imath]A_i[/imath] of radius 1/6 say, so that all the balls are disjoint and intuitively I want to give them all the same measure. So that kills the probability space right there.

I suspect you can modify that argument for the subsets of the measurable sets you were thinking about.

skeptical scientist wrote:The other thing that one would be inclined to do is try to get a measure on the set of subsets of the sphere by independently coloring each point amber with probability x, and blue with probability 1-x. My intuition tells me that you could indeed get a measure on a sigma-algebra of subsets of the sphere, but the probability of producing a set with measure x will not be 1, or even positive, but rather 0, and in fact you will almost certainly get a nonmeasurable set. So again, this doesn't give us a way of answering the question.

That would be a product of two point spaces, so Kolmogorov 0-1 springs to mind which would give you the "good" sets having measure 1 I believe, but its not a countable product of spaces, its cardinality c, and that isn't something I've seen before, and I don't think Kolmogorov even applies, if you could formalise what you mean by that product, though I could be wrong (you'd need some sort of transfinite limit... weird). You also wouldn't get measurable subsets of the sphere in general, you'd get weird sets.

Weird construction man. Weird.
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