[Guest]

You draw a random triangle on the XY plane. What is the probability that origin (0,0) is contained in the triangle?

[Guest]

lets start with 2 points and then ask the question whats the chance of a randomly placed 3rd point making a triangle that will contatin the origin.

If the 2 points are in the same quadrant then for the 3rd point to make a triangle containing the origin, it must be in the opposite quadrant (reverse the signs of x and y)

If the 2 points are in adjacent quadrants and we call them A nd B, then the range of angles of ABC that can be made from adding C in one of the other quadrants in a way such that (0,0) is included plus the range of anlges of ABC from adding C in the other of the 2 other quadrants ( sorry for the confusing wording) should = 90.

In the case that the first 2 points are in opposite quadrants, and they dont lie on either the lines x=y or x=-y, one there is a quadrant in which the third point must be placed to make the triangle contain (0,0)

Therefore the answer is 25% until you consider the possibilities of one of the points being the origin, or the lines x=y, x=-y, x=0, or y=0, being formed. the probability of any of those given its an infinite place is the limit of 1/t as t approaches infinity for each one which is basically 0 so 25% is my answer. Just try to imagine any of these cases and you should understand.

[bonanova]

Assume wolog the first point is in Q1.

There are 10 combinations of quadrants for the other two, and 5 of them make a triangle impossible.

In the 5 favorable combinations, a triangle is not guaranteed, and I will quickly estimate a 50% success rate.

[bonanova]

After further review ...

Create your random points as follows:

Specify two positive lengths - large as you like but finite - and an angle different from 180^o.

By SAS, you've created a triangle and, by so doing, three points.

Now give your points coordinates by throwing your triangle, at random, onto the plane.

Stand at a random point on the plane, throw it in a random direction for a random distance, and spin it so it lands with a random orientation.

The OP then translates to the following question.

What is the probability that when your triangle lands it will cover the origin?

The answer is simply: p = A_triangle/A_plane = 0.

[Guest]

After further review ...

Create your random points as follows:

Specify two positive lengths - large as you like but finite - and an angle different from 180^o.

By SAS, you've created a triangle and, by so doing, three points.

Now give your points coordinates by throwing your triangle, at random, onto the plane.

Stand at a random point on the plane, throw it in a random direction for a random distance, and spin it so it lands with a random orientation.

The OP then translates to the following question.

What is the probability that when your triangle lands it will cover the origin?

The answer is simply: p = A_triangle/A_plane = 0.

this answer makes no sense.

[Guest]

After further review ...

Create your random points as follows:

Specify two positive lengths - large as you like but finite - and an angle different from 180^o.

By SAS, you've created a triangle and, by so doing, three points.

Now give your points coordinates by throwing your triangle, at random, onto the plane.

Stand at a random point on the plane, throw it in a random direction for a random distance, and spin it so it lands with a random orientation.

The OP then translates to the following question.

What is the probability that when your triangle lands it will cover the origin?

The answer is simply: p = A_triangle/A_plane = 0.

I think your error is in the first line when you assume the triangle has to be finite.

[bonanova]

this answer makes no sense.

Sure it does.

The odds of a random point being near [whatever near means on the infinite plane] the origin are vanishingly small.

If we think of three points each individually appearing in quadrants 1-4 with equal likelihood, we're making a non-random

assumption about the location of the origin.

Imagine displacing the origin by 10²¹ units in some arbitrary direction.

The equal likelihood for the 4 quadrants just went out the window.

Nothing in the process of creating three random points on the infinite plane puts us "near" the origin as a starting point.

The answer has to be the same if you consider covering any particular point in the plane - e.g. the point [10²¹, 0]

On the other hand, limit the three points to a finite area, for example a circle centered on - or covering - the origin,

and a non-zero probability is easily calculated.

It will again involve the ratio of areas, but this time both areas will be finite.

[bonanova]

I think your error is in the first line when you assume the triangle has to be finite.

The OP asks to draw a triangle on the plane.

If one or more of the vertices may be placed at infinity then yes, you get a different [undetermined]

probability. But by the time you drew that triangle, none of us would be here to discuss it.

\

[Guest]

Place any two random points (a,b), (c,d) on the plane. The chace thay the third point forms a triangle that will cover (0,0) is

arctan((b/a - d/c)/(1 + bd/ac)) / 2pi

... I think .

[Guest]

Sure it does.

The odds of a random point being near [whatever near means on the infinite plane] the origin are vanishingly small.

If we think of three points each individually appearing in quadrants 1-4 with equal likelihood, we're making a non-random

assumption about the location of the origin.

Imagine displacing the origin by 10²¹ units in some arbitrary direction.

The equal likelihood for the 4 quadrants just went out the window.

Nothing in the process of creating three random points on the infinite plane puts us "near" the origin as a starting point.

The answer has to be the same if you consider covering any particular point in the plane - e.g. the point [10²¹, 0]

On the other hand, limit the three points to a finite area, for example a circle centered on - or covering - the origin,

and a non-zero probability is easily calculated.

It will again involve the ratio of areas, but this time both areas will be finite.

this is just plain wrong. if you were to displace the origin by whatever. then the same principle would take effec, just displace the quadrants appropriately. nevertheless that is irrelevant because the origin isn't something we are "randomly assuming" the origin is very much defined as (0,0). if you could why not try this. use a random number generator to choose coordinates randomly. then see if they form a triangle that contains the origin. Do you really think you would never get one?

and again using ur logic that can be applied to any single point in R². therefore by your reasoning any randomly placed triangle would have a 0 % chance of containing any point in the XY plane, which clearly isn't true.

[Guest]

almost

If your only going to read one read the second

put a point down... you have determinged L1(not going to use this but just saying youve determined one part).

put point two down. you have determined the base B

now re ordinate (rotate) the coordinate plane so the center of the base is on the positive x axis just because its easier to visualize

at significant distances the difference in distance between the two is negligible (or also the second point will equally increase and decrease the viable angle as they move closer and farther

now draw a triangle to the center accept draw the sides through the center into infinity

the probability that the third point will make a triangle to cover the center is the area on the other side of the line

or since both your angles at the base are equal. (pi-2(theta))/2pi

distance along x axis =R

lets consider the base to be the distance from one each point to the x axis or half the actual base

now theta equals inverse tan (R/B)

I would put forward as the direction or R or B do not matter and they are completely random, independent, and of the same order

R/B=1

so inverseTan(1)=pi/4

so the whole thing (pi-pi/4)/2pi=3/8

put the first point right on the axis

the second point makes the triangle with the center

draw a line from the x axis to the second point making a right triangle inside of the other triangle with height H

and distance along the axis X

invTan(H/X)=the angle in the unit (and world) circle of points for covering the point.

now the average length of H and X are = just as the average X coordinate would be the same as the average Y. Thus the math in the previous spoiler stands and 3/8

it also occured to me but i think this answer is iffy(er)

take a circle area pir^2

take a triangle inscribed in the circle area 1/2bh

biggest b or h is D. average b or h is 1/2D=r so (1/2r^2)/(pir^2)=1/(2pi)

this one is alot of estimation work so Im going with my second answer which is supported by my first

thanx for reading

[Guest]

to make the problem simpler, lets reduce it to the number line.

you place two points at random on the number line. what are the chances that the line will cover the number zero?

i'd say its 1/2. though there are uncountably infinite number of points on on the number line, there are is equal chances of each individual point being on the left or the right.

now to apply the logic to the original post, you have a 2d grid, and you are now selecting 3 points such that they are forming a triangle, and want to know what the chances are of the triangle covering the origin.

so first we'll need the chances that the 3 points lie along the same line.

the first two points are guaranteed to lie along the same line, and therefore an infinite number of lines can be formed, each of infinite length.

however, there are infinitely more points not on the line.

basically, you have x^2 /x^x, and want to know the limit as x approaches infinity.

this gives probability of zero.

therefore the three points most definitely do not lie on the same line, if chosen randomly.

from this then we are guaranteed to have a triangle from 3 random points.

so now we need the chances of the triangle covering the origin.

since the points of the triangle can extend to infinity, the area inside the triangle extends to infinity. so now the question becomes, are there infinitely more points outside the triangle, are the number of points equal, or what?

you can make an argument for all three possibilities in my opinion.

[bonanova]

... by your reasoning any randomly placed triangle would have a 0 % chance of containing any point in the XY plane, which clearly isn't true ...

Let's compare zero probability of covering the origin with some related statements.

A finite triangle is placed at random on the infinite plane.

Consider the following statements:

1. There are no points inside the triangle
2. The probability of any point in the plane falling inside the triangle is zero.
3. The probability of any point in the plane, that was chosen before the triangle was placed, falling inside the triangle is zero.

.

Now consider:

1. If a finite, randomly placed triangle covers the origin 25% of the time, with what probability does such a triangle cover the point (0, 1)?
2. Does a finite, randomly placed triangle cover some points in the plane more often than it covers other points?
3. Can a finite, randomly placed triangle cover every point in the plane 25% of the time?

Finally consider:

An event with zero probability can nonetheless happen.

A deck has an uncountably infinite number of cards, all different, that includes the Ace of Spades.

1. Is it possible to cut the Ace of Spades?
2. Does the probability of cutting the Ace of Spades differ from that of cutting any other card in the deck?
   
3. What is the probability of cutting the Ace of Spades?

[bonanova]

... place two points at random on the number line. what are the chances that the line will cover the number zero?

i'd say its 1/2. though there are uncountably infinite number of points on on the number line, there are is equal chances of each individual point being on the left or the right.

Choose two real numbers at random.

If the probability of their falling on opposite sides of the number zero is 1/2,

with what probability do they fall on either side of the number

.

1. one
2. two
3. pi
4. one million
5. 10¹⁰⁰
6. ten raised to the power of the previous number?

.

It is tempting to treat the continuum as if it were bounded, with elements that fall in equal measure about a single, unique point.

[Guest]

i decided to try writing a cpu program to determine the percent, in a fairly restricted way.

my first assumption was that the triangle is formed by placing 1 point at a time.

then i limited the size of the coordinate plane. then i assumed that the points of the triangle could be any points within that limit.

form here, i got a probability of 26.7%

here's my code in case you are interested.

import random

import math

total = 0

points =[0]*6

quadrant = [0]*3


for i in range(0,1000000):

   for j in range(0,6):

      points[j] = random.uniform(-10,10)    

   for j in range(0,6,2):

      if points[j] >=0 and points[j+1] >0:

         quadrant[j/2] = 1

      elif points[j] <0 and points[j+1] >=0:

         quadrant[j/2] = 2

      elif points[j]<= 0 and points[j+1] <0:

         quadrant[j/2] = 3

      else:

         quadrant[j/2] = 4

   if quadrant[0] == quadrant[1] and quadrant[0] == quadrant[2]:

      pass

   elif quadrant[0] == quadrant[1] and (quadrant[2] +2)%4 != quadrant[0]:

      pass

   elif quadrant[0] == quadrant[2] and (quadrant[1] +2)%4 != quadrant[0]:

      pass

   elif quadrant[1] == quadrant[2] and (quadrant[0] +2)%4 != quadrant[1]:

      pass

   else:

      if quadrant[0] + quadrant[1] != 5 and quadrant[0] != quadrant[1]:

         slope1 = (points[1] -points[3])/(points[0] -points[2])

         yint1 = points[1] -points[0]*slope1

         if quadrant[0] +quadrant[2] != 5 and quadrant[0] != quadrant[2]:

            slope2 = (points[1] -points[5])/(points[0] -points[4])

            yint2 = points[1] -points[0]*slope2

         else:

            slope2 = (points[3] -points[5])/(points[2] -points[4])

            yint2 = points[3] -points[2]*slope2

      else:

         slope1 = (points[1]-points[5])/(points[0] -points[4])

         yint1 = points[1] -points[0]*slope1

         slope2 = (points[3]-points[5])/(points[2] -points[4])

         yint2 = points[3] -points[2]*slope1

      if yint1 >0 and yint2 <= 0:

         total = total +1

      elif yint1 <=0 and yint2 >0:

         total = total +1


print total, float(total)/1000000

[bonanova]

i decided to try writing a cpu program to determine the percent, in a fairly restricted way.

my first assumption was that the triangle is formed by placing 1 point at a time.

then i limited the size of the coordinate plane. then i assumed that the points of the triangle could be any points within that limit.

form here, i got a probability of 26.7%

here's my code in case you are interested.

import random
import math
total = 0
points =[0]*6
quadrant = [0]*3

for i in range(0,1000000):
for j in range(0,6):
points[j] = random.uniform(-10,10)
for j in range(0,6,2):
if points[j] >=0 and points[j+1] >0:
quadrant[j/2] = 1
elif points[j] <0 and points[j+1] >=0:
quadrant[j/2] = 2
elif points[j]<= 0 and points[j+1] <0:
quadrant[j/2] = 3
else:
quadrant[j/2] = 4
if quadrant[0] == quadrant[1] and quadrant[0] == quadrant[2]:
pass
elif quadrant[0] == quadrant[1] and (quadrant[2] +2)%4 != quadrant[0]:
pass
elif quadrant[0] == quadrant[2] and (quadrant[1] +2)%4 != quadrant[0]:
pass
elif quadrant[1] == quadrant[2] and (quadrant[0] +2)%4 != quadrant[1]:
pass
else:
if quadrant[0] + quadrant[1] != 5 and quadrant[0] != quadrant[1]:
slope1 = (points[1] -points[3])/(points[0] -points[2])
yint1 = points[1] -points[0]*slope1
if quadrant[0] +quadrant[2] != 5 and quadrant[0] != quadrant[2]:
slope2 = (points[1] -points[5])/(points[0] -points[4])
yint2 = points[1] -points[0]*slope2
else:
slope2 = (points[3] -points[5])/(points[2] -points[4])
yint2 = points[3] -points[2]*slope2
else:
slope1 = (points[1]-points[5])/(points[0] -points[4])
yint1 = points[1] -points[0]*slope1
slope2 = (points[3]-points[5])/(points[2] -points[4])
yint2 = points[3] -points[2]*slope1
if yint1 >0 and yint2 <= 0:
total = total +1
elif yint1 <=0 and yint2 >0:
total = total +1

print total, float(total)/1000000

Since finite realms do have midpoints, and thus are intuitive and programmable I wrote for the unit circle.

[Guest]

Let's compare zero probability of covering the origin with some related statements.

A finite triangle is placed at random on the infinite plane.

Consider the following statements:

1. There are no points inside the triangle
2. The probability of any point in the plane falling inside the triangle is zero.
3. The probability of any point in the plane, that was chosen before the triangle was placed, falling inside the triangle is zero.

.

Now consider:

1. If a finite, randomly placed triangle covers the origin 25% of the time, with what probability does such a triangle cover the point (0, 1)?
2. Does a finite, randomly placed triangle cover some points in the plane more often than it covers other points?
3. Can a finite, randomly placed triangle cover every point in the plane 25% of the time?

Finally consider:

An event with zero probability can nonetheless happen.

A deck has an uncountably infinite number of cards, all different, that includes the Ace of Spades.

1. Is it possible to cut the Ace of Spades?
2. Does the probability of cutting the Ace of Spades differ from that of cutting any other card in the deck?
   
3. What is the probability of cutting the Ace of Spades?

sorry i lst you at ace of spades but there is no reason to assume the triangle is finite. if anthing you should assume a random triangle is infinite because it has random coordinates. if any of those coordinates are infinty then so is the triangle's area.

consider this for the highest point in the triangle is it likely to land below or above 10^12332? technically there are infinitely many more points above 10^12332 than below and therefore it would be higher. apply that same reasoning to any finite number and you get a "0 percent chance" of the point being between that point and zero. therefore it must be above all finite numbers. now the lowest point same reasoning. most likely, using the your type of logic, this triangle is going to have infinite area anyway.

[Guest]

Randomly select three points in a plane by any means that has a uniform distribution. If you consider their polar coordinates, it is easy to see that only their angular coordinates are of relevance in determining whether the origin is inside the triangle or not. I see no reason to believe that the probability density of the angular coordinate is anything but uniform (not so for the distance from the origin, but that no longer matters). It gets a bit boring from here, but it's not too difficult to show that the probability that the origin is contained within a triangle constructed this way is 1/4.

[Guest]

After reading a few more posts, I'm starting to think that we've all been wrong about this one, and that the answer is actually undefined.

Let's go back to Phillip1882's number line example, and simplify further by considering an arbitrary (finite) interval P1-P2, and select a point P0 such that P1 <= P0 <= P2 (randomly or otherwise). Now randomly select another point, say P3. I think everybody will agree that the probability that P3 lies on the interval P1-P0 is (P0-P1)/(P2-P1). Great. No problem so far. But, what happens to that probability when we let P1 and/or P2 -> -/+ \infty.

O.o??

[Guest]

ok first off i was really tired when i did my last post and theres a few flaws in some of it, but its close

in the first one it should have said 1/4 if you look at it i had 2*theta and then only used theta

in the second what I got was 1/8. But i think this was off by a factor of 2 because it only works for two quadrants at a time

the third spoiler I think is the probability of a random point in a circle falling in a randomly drawn triangle. plus the assumptions about b and h are probably off

To look at this clearer

you can think of the first line being the x axis. The second being on any radian. the average angle between them would be pi/2 so the answer is like 1/4

or you can think of it as two lines starting at the origin with area between them of

Ax+C-(Bx+d) c=0 d=0

so Ax-Bx (in the first quadrant) and percentage (Ax-Bx)/2pi

it can be quickly surmised that as one line goes around the area goes from 0to1/2to0 the average area being 1/4

now i attached a quick picture to explain what im talking about

If you think of the problem as the placing of that last point it becomes clearer.

probablyunnecessary.bmp

[Guest]

thinking about it now, your post about a unit circle kind of proves this . They should have the same answer

Any random triangle will land in some circle with center at the origin. therefore any random triangle within that particular circle has a 1/4 chance of including the origin according to Deegee's explanation in the other thread. The chance of it landing in any particular circle is incredibly small but it doesnt matter. You know it must land in one and no matter which it lands in, it's chances of containing the origin will be the same: 1/4.

[Guest]

thinking about it now, your post about a unit circle kind of proves this . They should have the same answer

Any random triangle will land in some circle with center at the origin. therefore any random triangle within that particular circle has a 1/4 chance of including the origin according to Deegee's explanation in the other thread. The chance of it landing in any particular circle is incredibly small but it doesnt matter. You know it must land in one and no matter which it lands in, it's chances of containing the origin will be the same: 1/4.

I was just about to post a long argument against your previous post, but thankfully I read your most recent post. I'm thinking through your explanation, and it completely makes sense. The infinite nature of the plane vs the finite nature of the triangle still makes me want to disagree, but there seems to be no rational argument against your explanation. I'll surely keep pondering it though!

[Guest]

Given that an infinite set of triangles (or rectangles, or circles or any other shape or a combination thereof) can be randomly placed on the plane xy, then there must be a subset of that set which includes the origin. As a percentage of the original set, this subset, though infinte itself, is infintesimal, or 0%. Thus, the probability is zero, even though the possibility exists.

[Guest]

to Brighterthan1000

while when this subset is a constant value your right but when the subsets size is directly proportional to the whole set you actually have to do work (to prove it either way). Ill see if i can write a program that somehow does this problem.

[Guest]

to Brighterthan1000

while when this subset is a constant value your right but when the subsets size is directly proportional to the whole set you actually have to do work (to prove it either way). Ill see if i can write a program that somehow does this problem.

Another way to look at it. Given (repeat) given that there are an infinite number of stars in the universe.

Some are red and some are white. Pick one at random. The probability of it being red is 0. Red is an infinate subset of an infinite set, so it is an infinitely small subset, thus a probability of 0. The same is true for white.

Now say that 25% of the stars are red and 75% are blue. Then the probabilities become those percentages.

[Guest]

right but if you can prove that the red stars are N for every N^2 other then as N goes to infin you get 0

if you prove it is of the same order tho it has a value.

So since we are proving that red are directly dependent and proportional to the other ones there is a value

take those stars and say pick three stars what the chance one of them is red (back to your original example of red is finite white is finite stars are infinite)

0

pick four (any number really doesnt matter) stars whats the chance that one of the white stars are in the polygon.

well lets say a quarter of the stars are in each direction or quadrant compared to the white stars when you zoom out far enough (or distributed uniformly so that the density of stars is the same in each direction). straight away whats the chance that one of each of the four stars is from each quadrant (a portion of the successes) 6/(4^3)

this is compareable to our question

Edited July 31, 2009 by final