[Guest]

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

[Guest]

since when was there an infinite amount of stars in the universe? last i checked it was a really large number but by no means infinite.

[bonanova]

1/4 for covering the center of a circle when the triangle fits within the circle.

That's independent of radius, so just let the radius increase without bound.

Voila! The answer for the plane is 1/4.

I've come full circle.

In the finite circle case, the covering probability is 1/4 only when we consider the center of the circle.

The covering probability decreases continuously as the point moves from the center to the circumference, where it's zero.

For example, at half-radius from center, the probability is 1/8.

And for a random point within the circle, it's 0.0739.

Now, we can scale up to infinity for any of these points just as easily as we can for the center point.

So while the rest of this post uses 1/4, perhaps it should be some lower number, perhaps more reasonably .0739

Now the center of the circle need not be at the origin to allow its increasing radius to swallow up the plane.

So this result applies to any point - not just to (0, 0)

Thus a "random" triangle covers every point in the plane 1/4 of the time!

This means the average area of a random triangle is 1/4 the area of the plane.

An interesting result, and one that negates any notion of a random triangle having a finite area.

The OP was formulated naively I think, as it probably anticipated that random triangle had finite area.

But perhaps not. Only if it's finite is the covering probability zero.

So I now raise the question I kept silent on while the debate brewed:

Is it possible to select at random a real number from the interval [-inf, inf]?

This process is required, to specify the vertices of - and even to discuss - a random triangle in the plane.

If you feel that it is possible, can you describe the process?

And, especially if we conclude a random triangle in the plane cannot be constructed,

how surprising is it that we nevertheless know its average size!

[Guest]

1/4 for covering the center of a circle when the triangle fits within the circle.

That's independent of radius, so just let the radius increase without bound.

Voila! The answer for the plane is 1/4.

I've come full circle.

In the finite circle case, the covering probability is 1/4 only when we consider the center of the circle.

The covering probability decreases continuously as the point moves from the center to the circumference, where it's zero.

For example, at half-radius from center, the probability is 1/8.

And for a random point within the circle, it's 0.0739.

Now, we can scale up to infinity for any of these points just as easily as we can for the center point.

So while the rest of this post uses 1/4, perhaps it should be some lower number, perhaps more reasonably .0739

Now the center of the circle need not be at the origin to allow its increasing radius to swallow up the plane.

So this result applies to any point - not just to (0, 0)

Thus a "random" triangle covers every point in the plane 1/4 of the time!

This means the average area of a random triangle is 1/4 the area of the plane.

An interesting result, and one that negates any notion of a random triangle having a finite area.

The OP was formulated naively I think, as it probably anticipated that random triangle had finite area.

But perhaps not. Only if it's finite is the covering probability zero.

So I now raise the question I kept silent on while the debate brewed:

Is it possible to select at random a real number from the interval [-inf, inf]?

This process is required, to specify the vertices of - and even to discuss - a random triangle in the plane.

If you feel that it is possible, can you describe the process?

And, especially if we conclude a random triangle in the plane cannot be constructed,

how surprising is it that we nevertheless know its average size!

indeed it is a wierd result. as i said before you technically will always choose inf or -inf because the probablity of choosing any number between them is 0. pick a bounds lets say -10^1000! to 10^1000! even with such large bounds there are infinitely many more numbers outside those bounds that would make picking any number within those bounds have a probability of 0. of course as you have pointed out a probability of 0 doesnt mean impossiblity.

As for how to pick a random number between 0 and infinity? its not harder than picking any other random number tbh. just pick a random number between 0 and pi for example and take the limit from the right of secant of that number. giving that in such a small domain there is an infinite range you'd be basically using a finite range of numbers to choose from an infinte range of numbers. it must be noted though, that although the domain has finite bounds, it is still infinite because there are infinite fractions, decimals etc, that fit into that domain. and thus cantor sets are another wierd topic.

but now i pose this question to you ( though maybe this should go in its own forum altogether ill let bonanova be the judge of that) in the case that you decided to use this type of method to randomly pick a number, would the probability of any particular number being chosen be affected by what exact function you are using? for example would it be different if we used tangent instead of secant?

[Guest]

Random is an impossible concept just like infinity. Even a coin flip isnt truly random its just very hard to purposefully predict so for all intents and purposes its random. Its very possible that's there no such thing. You can say there are infinity numbers, but numbers are mostly just a made up concept that helps us define and understand the world. but this does not mean that these concepts do not prove incredibly useful even though they are hard/impossible to wrap your head around.

The fact is tho in this problem we don't really need to use randomness, or infinity. First it is important to mention that we do have to consider infinity in one quick way. The fact that infinity does not have a center doesn't matter. Pick a point and say "why is your center any more viable then this point" well the point you just imagined is a finite point. Zoom out far enough and for all intents and purposes both points are now the center.

But anyway we can sum all possible triangles (even though it is impossible to define there number or position) and find the chance that it happens. Now we can find a formula in the finite dimension that answers the OP question without even considering the radius or the finite nature of the circle. You have to admit that a random point has to be in some exact direction from every point, but what you call this direction can be invented as long as it is constant for the rest of the problem. So put the point on the x axis. Now theoretically there are equal number of points on any radian from the center. So theoretically there is an equal chance that any point is on any radian or angle. So put another random point down which turns out to be at angle B. The third point can be at any angle from pi to pi+B and the new triangle covers the center. B can take any angle from 0 to pi (any larger is redundant as all that matter is the difference in the angle to A). Now even though this is infinite and random once again we can very well average this as a number, pi/2. Thus on average the point can be anywhere on a quarter of the plane (there is no quarter but there is a function that describes it and the points).

So voila. Now if this formula had anything to do with r, we would be screwed or at least hard pressed to do much. but the point is yes the triangles cover 1/4 of the plane on average. why not. any point that you look at its true for and why is a triangle covering an infinite any less likely then a circular plane. Now its true that any triangle you put down is finite, but so is anything you do to try to define the plane.

[bushindo]

Is it possible to select at random a real number from the interval [-inf, inf]?

As sparx1 observed, this is not much harder than usual. In fact, there's already a famous distribution that already satisfy this- the gaussian distribution. Any number has a chance to show up in the gaussian distribution, although larger numbers are less likely to do so. In fact, any unbounded density function will satisfy this criteria. I think the harder question, and one intended by bonanova, is

Is it possible to uniformly sample a real number from the interval [-inf, inf]?

[bonanova]

As sparx1 observed, this is not much harder than usual. In fact, there's already a famous distribution that already satisfy this- the gaussian distribution. Any number has a chance to show up in the gaussian distribution, although larger numbers are less likely to do so. In fact, any unbounded density function will satisfy this criteria. I think the harder question, and one intended by bonanova, is

Is it possible to uniformly sample a real number from the interval [-inf, inf]?

Exactly.

Random generally implies uniform distribution.

One authority offers this guidance:

.

1. A random number is a number chosen as if by chance from some specified
   distribution such that selection of a large set of these numbers reproduces
   the underlying distribution. .
2. It is impossible to produce an arbitrarily long string of random digits and
   prove it is random.

.

If [2] is true, then random triangles and numbers only exist in bounded regions.

A clue, perhaps, to the difficulty of this discussion.

[Guest]

I just wanted to add (before you guys start talking about something metaphysical and the world explodes from being hit with some knowledge) that with my solution not only is a quarter of every point covered by each triangle, but a each triangle is in a quarter of all triangles. this is quite simple to prove. But i found it cool.

[Guest]

Final,

The world may not explode, but my head is about to.

Most of the math I deal with daily involves more of the finite realm stuff (error analysis, linear programming, etc). I haven't dealt with (voluntarily or involuntarily) this type of math for more than 10 years (and i managed to never, ever, ever, have to take ANY statistical type studies...weak excuse, I know, but an anomaly in how the math coursework was laid out when i went to school).

So needless to say, my head has been spinning the past couple days with some of these interesting discussions.

[Guest]

Yah i have trouble with this too because each triangle holds 0% of the points but a point is held by 1/4 of the triangles. its self contradictory but such is infinity. I find when talking about this stuff just act like your emphatically right. Eventually people will just say your right so you'll go away. I find it works better then actually thinking.

Ill have to admit that one night i was thinking while not sleeping and was convinced that you were right, but then i managed to convince myself out of it.

I took statistics but my teacher was obsessed with calculator functions and programs, so all we memorized was what buttons to push. but luckily enough i had to take the course over again and actually learned stuff.

hopefully the next problem will involve addition, I could use a break from conceptual math.

[Guest]

Final,

The world may not explode, but my head is about to.

Most of the math I deal with daily involves more of the finite realm stuff (error analysis, linear programming, etc). I haven't dealt with (voluntarily or involuntarily) this type of math for more than 10 years (and i managed to never, ever, ever, have to take ANY statistical type studies...weak excuse, I know, but an anomaly in how the math coursework was laid out when i went to school).

So needless to say, my head has been spinning the past couple days with some of these interesting discussions.

The anomaly here lies in the fact that you use math and didn't need statistics not that you study it. at my school people like me who major in pure math can avoid ever taking a stats course (which i plan to do) but people who actually have a reason to learn math aka aplied math majors must take it. it's either head blows up on a daily basis or deals with statistics.

[Guest]

I've enjoyed watching the math wizardry unfold, peppered though it was by my own novel, yet well intentioned misapplications. It is precisely challenges such as these that not a few years past I wisely abandoned a goal of engineering greatness in favor of an academic pursuit leading to a particularly satisfying career elsewhere. In short, math was a snap till stats and calculus did me like Roundup does dandelions.

Question concerning zero and 'approaching zero'. The probably of cutting an ace of spades out of an infinite deck containing no ace of spades I see as unquestionably zero. So if there is an ace of spades in the deck, the probably can't also be zero, can it?

One other related question. Rather than a triangle, lets draw a random circle on the plane x,y. What then would be the probably of the origin being contained within the circle?

[bushindo]

I've enjoyed watching the math wizardry unfold, peppered though it was by my own novel, yet well intentioned misapplications. It is precisely challenges such as these that not a few years past I wisely abandoned a goal of engineering greatness in favor of an academic pursuit leading to a particularly satisfying career elsewhere. In short, math was a snap till stats and calculus did me like Roundup does dandelions.

Question concerning zero and 'approaching zero'. The probably of cutting an ace of spades out of an infinite deck containing no ace of spades I see as unquestionably zero. So if there is an ace of spades in the deck, the probably can't also be zero, can it?

One other related question. Rather than a triangle, lets draw a random circle on the plane x,y. What then would be the probably of the origin being contained within the circle?

Interesting question. Originally, I thought that we can specify a random circle on the plane by picking a random coordinate for the center, and then uniformly sample a radius from the interval (0, inf). That didn't work out so well in computing the probability of containing the origin.

Here's another way to construct the random circle. Uniformly sample 3 points on the plane, and use it to construct the circle. All circles can be uniquely determined by 3 points. Since the constructed circle will also include the triangle specified by the 3 points, the chance that the circle include the origin is greater than that of the triangle, which we know is 1/4.

But here's another way to construct the random circle. Uniformly sample 3 points on the plane, and use it to construct a triangle. Now, we construct the circle as the biggest circle that can fit inside the triangle we just made. Since the constructed triangle is smaller than the triangle, the chance that the circle contains the origin is less than 1/4. In this example and the one immediately above, we could estimate the chance of the circle containing the origin by finding the ratio of the area between the circle and the triangle.

The 3 methods above equates to different density distribution for the circle on a plane. There are many other ways of specifying a random circle on the plane, though.

Edited August 3, 2009 by bushindo

[Guest]

Interesting question. Originally, I thought that we can specify a random circle on the plane by picking a random coordinate for the center, and then uniformly sample a radius from the interval (0, inf). That didn't work out so well in computing the probability of containing the origin.

Here's another way to construct the random circle. Uniformly sample 3 points on the plane, and use it to construct the circle. All circles can be uniquely determined by 3 points. Since the constructed circle will also include the triangle specified by the 3 points, the chance that the circle include the origin is greater than that of the triangle, which we know is 1/4.

But here's another way to construct the random circle. Uniformly sample 3 points on the plane, and use it to construct a triangle. Now, we construct the circle as the biggest circle that can fit inside the triangle we just made. Since the constructed triangle is smaller than the triangle, the chance that the circle contains the origin is less than 1/4. In this example and the one immediately above, we could estimate the chance of the circle containing the origin by finding the ratio of the area between the circle and the triangle.

The 3 methods above equates to different density distribution for the circle on a plane. There are many other ways of specifying a random circle on the plane, though.

Very interesting indeed... this is the same as the length of chord problem where definition of how you choose "random" cricle in this case changes the probability!

Here's another one: this one was far easier - rectangle containing origin! Choose two random points in the plane. Consider them the opposite vertices of a rectangle and draw the rectangle. The probability that it contains the origin is 1/4 given by whether the two points that are chosen lie in different quadrants and the quadrants are across!

Now consider the example for circle using two points. Draw line joining them and take mid-point of this line as centre of the cricle and radius is ofcourse the distance from this point to the two random points chosen earlier.

In this case, the probability would be given by chances that the distance of the nearer point to origin is LESS than half the distance between the two points and that the two points are in different quadrants and the slopes of lines connecting the two points are different depending on which quadrants they are in and .... "all hell breaks loose"

Edited August 3, 2009 by DeeGee

[Guest]

lets start with the fact that thier are infinte circles with center at the origin. any random circle's center would land on one of those circles. now lets pick 2 random numbers: 1 would be the radius of the random circle in question the 2nd would be the radius of the circle on which it's center lands. if the circle in question has a larger radius than the circle on which its center lands then it will contain the origin. if it has a smaller radius than the circle on which its center lands, it wont contain the origin. on the off chance the 2 are the same then it will contain the origin.

now lets simplify this. we pick 2 random numbers. 1 of them will be the distance of the circles center to the origin. and the second will be the radius of the circle.

the probability of these 2 numbers being the same is 0 due to the fact that thier is an infinite range for both of them the chance of 1 being larger than the other is equal to the chance of vice versa. the 3 probabilities add up to 100 %

por eso,50%is my answer

[unreality]

I know I came into this kind of late but it reminds me of a (semi-joke) problem my uncle's friend told me:

Suppose you have a triangle in an infinite plane. Now you place a point anywhere on this plane. What is the probability it is inside the triangle?

1/8

Extend the sides of the triangles to make three infinite unbounded lines. The point must be on *this* side of *that* line (1/2) for all three lines... 1/2 * 1/2 * 1/2 = 1/8

Not sure how accurate that it is, I guess it depends on your random scheme for drawing the triangle and placing the point...

edit ~ anyway it's possible that the questions are asking the same thing (with the 'motion' just relative to each other) but I'd have to think about it more

Edited August 5, 2009 by unreality

[Guest]

Ok just a quick response here. I'll try to show all the work for it soon but here it is.

Start with one random point A in a random quadrant (I, II, III, IV). Then add a second random point B in a random quadrant. The probability of A and B being in the same quadrant is 1/4 or 25%. The probability of those points being in different quadrants is 3/4 or 75%.

Now for the work. If A and B are in the same quadrant, use the origin as point C. The shadow, or the angle opposite ACB will be the area where any other C can be placed to have the triangle include the origin. This area can be maximized to the whole quadrant opposite the first containing points A and B by moving A and B towards seperate axis'. Thus having a 25% chance of including the origin. This might be confusing but I can draw it out later.

I figured when A and B and in different quadrants, the chance of a third point C randomly place to have the triangle include the origin went up to 1/2 or 50%. This is shown by moving A and B to the same axis yet still remaining in their respective quadrants.

So long story short, 1/4 * 1/4 for A and B in same quadrants and then 3/4 * 1/2 for A and B in different quadrants. Add those together to get 7/16 or 43.75%.

[Guest]

Here is what I must write as solution for this problem:

Every random triangle in the plane defines also a circumscribed circle. First let's look at the probability that this circle will contain the origin.

Let's look at the line from point O(0,0) (the origin) and the center of the circle O'. The circle intersects with that line at points A1 and A2. The circle will contain the origin of the distance |OO'| is greater than |OA1|. We have four variants for the two random points O and A1:

-------------O----A1---O'------

-------A1----O---------O'------

-------O'----O----A1-----------

---O'----A1--O-----------------

Two of the solutions contain the origin and two don't. So the probability that a random circle will contain the origin is 1/2.

Now let's assume that the circumscribed circle contains the origin. What is the probability that a random triangle with vertexes on the circle edge contains the origin? I already gave answer to - it is 3/8

So the final answer is P = (1/2)*(3/8) = 3/16

[Guest]

Let's draw a line from each vertex to the origin. This gives us three angles. Now, if the largest angle, viewed from the third vertex, is more than 180 degrees, then the origin is contained within the triangle. Thus, probability is 0.5

[Guest]

Let's draw a line from each vertex to the origin. This gives us three angles. Now, if the largest angle, viewed from the third vertex, is more than 180 degrees, then the origin is contained within the triangle. Thus, probability is 0.5

If we mark the angles as "alpha", "beta" and "gamma", where alpha<beta<gamma, then the triangle will contain the origin if 180+alpha<gamma<180+beta... How did you came to answer 180?

P.S. If we rotate the coordinate system by angle alpha, then the equation will be 180<gamma<180+beta. So the problem is defined by two random variables... Still no clue how KlueMaster came to answer 180

Edited December 24, 2010 by Philip Petrov

[Guest]

If we mark the angles as "alpha", "beta" and "gamma", where alpha<beta<gamma, then the triangle will contain the origin if 180+alpha<gamma<180+beta... How did you came to answer 180?

P.S. If we rotate the coordinate system by angle alpha, then the equation will be 180<gamma<180+beta. So the problem is defined by two random variables... Still no clue how KlueMaster came to answer 180

I am not talking about the angles of the triangle, but angle between vertices with origin as the joining point.

So here gamma = alpha + beta. I hope that answers

[Guest]

Okay, my thoughts.

First of all, it is easy to find a probability when there is a one to something correspondence. It’s like when there is a one to one correspondence in and out of a set, the probability of choosing something in a set is one half, EVEN IF THE SET IS INFINITE. For example, there is the same number of perfect squares as the number of integers, as one can easily match every integer to its perfect square. Also there is the same number of even numbers as odd numbers, as one can easily match an even number n to an odd number n+1.

So now look at the question. For the purpose of my proof, I have established that rotations of a triangle are not the same.

(See attachment for photo)

It is clear that when translated, the triangle on the top contains the origin if and only if the bottom vertex is in the triangle on the bottom, which is a 180 rotation of the triangle on the top. This works also for other vertices, and rotations of the figure, but the rotations are not considered to be the same. Now the area within the lower triangle is finite, yet the entire plane is infinite. There for the possibility of a random triangle to contain the origin is very close to zero and will explode all the computers in the world if someone tried to compute the numbers. There are an infinite number of points in the triangle, yet there are an infinite² number of points in the entire plane (infinite number of x values and infinite number of y values multiplied together). We have set an infinite-to-infinte² correspondence, and so the probability is: infinity/infinity squared = 1/infinity (think of the number of parallel lines in a plane) = 0.000000000000000000000000000000000000000000000…(infinite amount of zeros)...000001. Therefore we say that the probability is 0%.

[Guest]

I am not talking about the angles of the triangle, but angle between vertices with origin as the joining point.

So here gamma = alpha + beta. I hope that answers

I am talking about the same. And all three angles are variables (they are random). If you don't write everything in details with graphics... it's not clear.

[Guest]

I will detail my solution above because I feel it's not clear for all:

Problem 1. What is the probability that a random circle in the plane contains the origin?

Solution: Let the radius-vector (the distance between the origin and the center of the circle) is "r". Obviously |r| is a random number. When the circle will contain the origin? Obviously when the radius of the circle R is larger than |r|. Well R is also a random number. So this problem is rephrased as: "We have two positive random numbers R and r. What is the probability that R>|r|?". I am not going to provide solution here (it's simple but long) - it's 1/2.

Problem 2. What is the probability that random triangle in the plane contains the origin?

Solution: Every triangle have a circumscribed circle. If the origin is in the triangle, then it will be in the circumscribed circle. So let's separate the question in two parts - first we will see when the origin is in the circle and then we will see if it is in the circle, then when it will be inside the triangle.

Well we know when the origin is in the circle - the probability is 1/2 (proven in Problem 1). (1)

Now we have the second part of the problem - we have a fixed circle which contains the origin somewhere inside (it is actually at random place). So we can look at the origin as random point. Also we have three random points on the circle (the vertexes of the triangle). So what is the possibility that the random point (origin) is inside the random triangle? It's the "Area of the triangle"/"Area of the circle".

We do know the area of the circle - it's PiR^2. What is the area of the triangle?

We can parametrize it by two angles and the radius of the circle. The area is:

S = 2R^2sin(alpha).sin(beta).sin(180-alpha-beta)

Where alpha is between 0 and Pi and beta is between 0 and Pi-alpha.

Now we simplify and integrate using the help of mathematica:

In[1] = (Integrate[(Sin[x] Sin[y] Sin[x + y]), {x, 0, Pi}, {y, 0, Pi – x}]2R^2)/(Pi R^2)

The total probability is the combination between (1) and (2):

P = (1/2).(3/8) = 3/16

It's not 3/16 as I wrote above because I miscalculated Shame on me

Edited December 24, 2010 by Philip Petrov

[Guest]

The answer is (3/4)*(1/2) = 3/8

[Guest]

The answer is (3/4Pi)*(1/2) = 3/8Pi

As per the correction in the other thread - P=(3/4Pi)*(1/2) = 3/8Pi