[unreality]

You have a perfect square... at the exact center lies a point, P.

You then randomly select three points along the perimeter of the square, which form a triangle. What is the probability that P will lie inside of this triangle?

edit: I guess we can assume that the three points won't fall along the same side (which would happen 1/16 of the time), or perhaps we have to count that as a "triangle that doesn't contain P" and have 1/16 at least be NO, but it's not even a triangle. So do whatever is easier for your method of solution: counting that 1/16 chance as one of the triangles that don't surround P, or assume that a re-randomization will take place if that occurs. It's up to you

Edited November 4, 2008 by unreality

[Guest]

well it depends do you want an equal triangle? cuz then it won't end up on the inside i believe

EDIT: I hate geometry

Edited November 4, 2008 by ledger

[Guest]

I think it will be easier to start with a circle, and then work from there. From the center's perspective a triangle inscribed on a circle is just like a triangle inscribed on a square, only with the probability spread uniformly. I think this will make it easier to figure out "winning" conditions.

The center lies within the triangle if the maximum angle formed between the radial lines is less than 180 degrees.

For a circle, the probability is 1/2. I would expect it to be more for a square, but I'm not positive.

I'm also not sure how useful that circle comparison is, but there you go.

[unreality]

I agree, the first thing I thought of too was a circle:

which is 1/2, as you showed

[Guest]

Alright, I think I have a better way of looking at this: let's break it down by how the points are split among the 4 sides:

There are 4 ways to arrange the points: 3 on one side, 2 on one side with the other adjacent, 2 on one side with one opposite, or 3 on different sides.

First we can find the probability of each of those 4 arrangements:

3 on one side: P=1/16

2 on one side, one adjacent: 6/16

2 on one side, one opposite: 3/16

3 on adjacent sides: 6/16

Then we can multiply those by the probability of the center being in the triangle for each of those 4 cases. That parts a lot trickier, though. I'll try to get back to it later.

[unreality]

Wouldn't 2 on one side, 1 on opposite be:

2/16?

you have one fixated which is given. Then one on same side, one on opposite side is (1/4)(1/4) = 1/16. Or both could be on the opposite side, which is also 1/16. Thus 1/16 + 1/16 = 2/16. Or am I missing something?

I guess you could add another 1/16 by saying that the other two "could be in any order", but wouldn't you do that to the other one too? I'm probably wrong of course

[Guest]

Wouldn't 2 on one side, 1 on opposite be:

2/16?

you have one fixated which is given. Then one on same side, one on opposite side is (1/4)(1/4) = 1/16. Or both could be on the opposite side, which is also 1/16. Thus 1/16 + 1/16 = 2/16. Or am I missing something?

I guess you could add another 1/16 by saying that the other two "could be in any order", but wouldn't you do that to the other one too? I'm probably wrong of course

I get the same probabilities as Chuck. First, the probability of having at least two points on one side is 5/8. The probability of all three on the same side is 1/16. So, the probability of only two on the same side is 9/16. 2/3 of that time, the other will be on an adjacent side, for 3/8, and the remaining 3/16, it will be opposite.

[Guest]

Alright, I think I have a better way of looking at this: let's break it down by how the points are split among the 4 sides:

There are 4 ways to arrange the points: 3 on one side, 2 on one side with the other adjacent, 2 on one side with one opposite, or 3 on different sides.

First we can find the probability of each of those 4 arrangements:

3 on one side: P=1/16

2 on one side, one adjacent: 6/16

2 on one side, one opposite: 3/16

3 on adjacent sides: 6/16

Then we can multiply those by the probability of the center being in the triangle for each of those 4 cases. That parts a lot trickier, though. I'll try to get back to it later.

The probability of the triangle enclosing the center is obviously zero for the first two, and 1/3 for the fourth case. Still working on #3

[Guest]

When your done - check three random points IN the square!

[Guest]

umm how can you calculate the probability? isnt there an infinite number of combinations of points on the perimeter you can pick? (you know, like 1/2 way along a side, or a 1/3 or 1/4 or 1/5 or 1/6...and so forth ....)

edit: hmm wait that might not necessarily mean that you cant work the probability out but i've gotta go now...

Edited November 5, 2008 by lemonymelon

[Prime]

I don't think the type of figure matters.

Consider immediate distances between two points if you travel from one to another along the perimeter of the figure.

For a center point to fall inside the triangle, the sum of any two immediate distances between two points must be greater than the third.

I.e., say the points were named A, B, and C. Then AB + AC > BC and AB + BC > AC and AC + BC > AB.

Again, those are not straght distances, but perimeter traversal.

What's the probability for such distribution? I'd have to think about it. But it seems, it should not dependent on the type of figure -- circle, or what.

This problem seems similar to the one where stick was cut into three parts randomly to form or not to form a triangle.

[unreality]

This problem seems similar to the one where stick was cut into three parts randomly to form or not to form a triangle.

I think I posted that one But I didn't make this one up - in fact, I don't even know the answer ;D

[Prime]

My estimate of probability is 1/4 for the center to fall inside such inscribed triangle. No matter what regular poligon, or circle has 3 random points on its perimeter.

The illustration shows two random points on the perimeter of the square. Straight lines are drawn from those points through the center until intersecton with the perimeter on the other side. The portion of the perimeter marked with red shows possible positionning for the third point so that the center would fall inside inscribed triangle.

Note that the section of the perimeter enclosed between two points is equal in length to the portion of the perimeter where to place the third point in order to catch the center inside the triangle. Inscribed circle illustrates same idea.

The shorter distance between two points on the perimeter ranges uniformly between 0 and 1/2 of the perimeter, for an average 1/4 of the perimeter. So the probability to catch the center is 1/4.

[bonanova]

25.0194% contained the center.

[bonanova]

Call C the center, P1 P2 and P3 the three points.

For C to lie within triangle P1 P2 P3, the points must not all lie on the same side of any line drawn through C.

For this to obtain, it is necessary and sufficient that:

1. P2 and P3 lie on opposite sides of line P1-C, and
   
2. P1 and P3 lie on opposite sides of line P2-C.
   

Since any line through C bisects the square, both probabilities are 1/2.

Edit:

From the same two lines, Prime argues average perimeter length.

This argument multiplies two coin tosses.

Same result.

Edited November 6, 2008 by bonanova
Add comment on Prime's approach

[unreality]

It seems that the answer is,

1/4

[Prime]

25.0194% contained the center.

For a million samples, the deviation of +0.02% from the theoretical seems a bit large. I have a theory ...

There is a special case, where two random points P₁ and P₂ are diametrally opposite. So that if C is the center, then P₁C and P₂C lie on the same straight line. In this case, I am guessing, the software simulation counted the center point as "captured".

If we deal with mathematical zero-dimension poits, then there are infinitely many diametrally opposite points. However, there are infinitely many times more random points, which are not diametrally opposite. (It's an infinity times infinity type of deal, where one infinity infinitely greater than the other.) So those special cases should not affect the theoretical probability of 1/4.

Whereas software generates random numbers at certain minimum finate interval. Here the ratio of diametrally opposite points to total possible points has a definite finite ratio. Hence such large deviation.

It would be interesting to run the same experiment not counting "special cases" center capturing.

[Guest]

For a million samples, the deviation of +0.02% from the theoretical seems a bit large

It's actually well within what you'd expect to see. That binomial distribution will have a mean of 250,000 and a std. deviation of sqrt(187500), and n is so large that a normal approximation should be very very good.

The probability of an experimental value being as good or better than 25.0194 is .3459 (this counts from 24.9806 up to 25.0194).

That being said, I don't know how his simulation dealt with cases where the center fell on the line.

[bonanova]

I don't know how his simulation dealt with cases where the center fell on the line.

Chuck,

That's a basically zero-probability case; but if it happened I included it.

I included the cases where the triangle areas having the center as one vertex equaled* the area of the triangle that didn't.

*To within a comparison tolerance of 10^-13.

Whereas software generates random numbers at certain minimum finate interval. Here the ratio of diametrally opposite points to total possible points has a definite finite ratio. Hence such large deviation.

It would be interesting to run the same experiment not counting "special cases" center capturing.

Prime,

1. Simulations are finite samples, like all measurements.
   If you see someone's data points lying exactly on the theory curve, take it to the bank the data were doctored.
   My purpose was to establish 1/4 as the answer, as opposed say to 1/2 or some other simple fraction.
   I was satisfied that it did that.
   
   
   
2. The deviation is not systematic. That would indicate a poorly designed experiment.
   Equal areas computed to a finite number of decimal places have an unbiased even chance of meeting the criterion.
   
   
   
3. Saying "on the line" points are exterior is no more valid than saying they are interior.
   
   
   
4. The number of "on the line" cases has Lesbegue integral measure of zero; the point is moot.
   

[Prime]

I wasn't so much criticizing the experiment as the means for veryfying theoretical results, as pointing out an interesting difference between theoretical probability and the experiment design. Even though I was wrong attributing deviation to that difference, still...

Suppose random number generator produces numbers with 10 significant (decimal) digits. For a total number of points 10¹⁰. The ratio of number of pairs of diametrally opposite points to total number of pairs is 10^-10, thus affecting the experiment results ever so slightly. In case of mathematical zero-dimension points that ratio is zero. Therefore, it does not matter, in theory whether to include center capture into diameter cases, or not.

That said, I disagree with multiplying probilities of points being on opposite side of center lines: (1/2)*(1/2) as a way of solving this problem.

Consider two deliberately selected (not random points) on the perimeter. Now select the third point at random.

Same reasoning leads to the same calculation of probability (1/2)*(1/2)=1/4. But actually the probability to catch the center varies from 0 to 1/2, depending on the positionning of the two initial points.

[bonanova]

The ratio of number of pairs of diametrally opposite points to total number of pairs is 10^-10, thus affecting the experiment results ever so slightly.

You're drawing a trend line through a single datum, which randomly happened to exceed 1/4 - by an amount commensurate with finite sampling.

I don't simply assert the deviation is random. Repeated simulations included results less than 25%.

Sampling noise swamps comparison tolerance noise, [of the order of 10^-13],

which itself swamps the effect of including or excluding the Lesbegue-nonexistent cases of the center-of-square landing on the triangle's perimeter.

The decision to include or exclude points on the line has zero effect. OK. That's the discussion of the simulation. On to the OP problem analysis...

That said, I disagree with multiplying probilities of points being on opposite side of center lines: (1/2)*(1/2) as a way of solving this problem.

Consider two deliberately selected (not random points) on the perimeter. Now select the third point at random.

Same reasoning leads to the same calculation of probability (1/2)*(1/2)=1/4. But actually the probability to catch the center varies from 0 to 1/2, depending on the positionning of the two initial points.

I'm thinking that you do disagree with the probability method. No discussion points there.

But the inconvenient problem with your objection is that you also disagree - every bit as much - with the perimeter method.

Consider two deliberately selected (not random points) on the perimeter. Now select the third point at random.

Actually the probability to catch the center varies from 0 to 1/2, depending on the positionning of the two initial points.

As clearly shown by the length of the red portion of the perimeter in your drawing.

Therefore the method of perimeters does not lead to the answer of 1/4.

All you've said is that the correct answer for a specific case is not the correct answer for the OP.

My preference, evident also in my roll-of-the-die analysis, is to find the shortest path to final answer.

Given the OP, .5 x .5 = 1/4 has a certain charm for me.

Your preference, there and here, is to analyze particular cases and sum, or average, afterward.

That's how you think.

The world needs us both.

[Prime]

You're drawing a trend line through a single datum, which randomly happened to exceed 1/4 - by an amount commensurate with finite sampling.

I don't simply assert the deviation is random. Repeated simulations included results less than 25%.

Sampling noise swamps comparison tolerance noise, [of the order of 10^-13],

which itself swamps the effect of including or excluding the Lesbegue-nonexistent cases of the center-of-square landing on the triangle's perimeter.

The decision to include or exclude points on the line has zero effect. OK. That's the discussion of the simulation. On to the OP problem analysis...

I think you misinterpret the point that I was making. And, I believe, including or excluding cases with two diametrically opposite points from center-capturing does bear on the oucome of the experiment. Even though it gets burried under normal deviation.

Change your randomizer to give only 8 possible points (from 0 to 7). And run the simulation again. Or, to convince yourself without experiment, imagine your randomizer gave only two possible values: 0 and 1/2, which represent diametrically opposite points on the perimeter.

I'm thinking that you do disagree with the probability method. No discussion points there.

But the inconvenient problem with your objection is that you also disagree - every bit as much - with the perimeter method.

Consider two deliberately selected (not random points) on the perimeter. Now select the third point at random.

Actually the probability to catch the center varies from 0 to 1/2, depending on the positionning of the two initial points.

As clearly shown by the length of the red portion of the perimeter in your drawing.

Therefore the method of perimeters does not lead to the answer of 1/4.

All you've said is that the correct answer for a specific case is not the correct answer for the OP.

You have omitted my assertion that all those individual cases are uniformly distributed. (Each is as likely as another.) And the overall probability is the average of their probabilities.

The only way I can construct a reasoning for (1/2)*(1/2) solution is as following:

1). Draw the line from P1 through the center. The probability for P3 to be, say, on the left of that line is 1/2.

2). Draw the line from P2 through the center. The probability of the point to be on the required (right) side of that second line ranges from 0 to 1 uniformly depending on where P3 or the line through P2 fell. The average of those probabilities is 1/2.

3). This sequence includes all possible variations for center-capturing point placement for a total probability of (1/2)*(1/2)=1/4.

My preference, evident also in my roll-of-the-die analysis, is to find the shortest path to final answer.

Given the OP, .5 x .5 = 1/4 has a certain charm for me.

Your preference, there and here, is to analyze particular cases and sum, or average, afterward.

That's how you think.

The world needs us both.

It is interesting -- the difference in our approaches to solving problems. My way tends toward sequential processing. Yours is more like parallel.