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This question was asked in a test and I got it right. The answer key gives $\frac12$.

Problem: If 3 distinct points are chosen on a plane, find the probability that they form a triangle.

Attempt 1: The 3rd point will either be collinear or non-collinear with the other 2 points. Hence the probability is $\frac12$, assuming that collinearity and non-collinearity of the 3 points are equally likely events.

Attempt 2: Now suppose we take the midpoint (say $M$) of 2 of the points (say $A$ and $B$). We can draw an infinite number of lines passing through $M$, out of which only 1 line will pass through $A$ and $B$. Keeping this in mind, we can choose the 3rd point $C$ on any of those infinite lines, excluding the one passing through $A$ and $B$. Now it seems as if the probability will be tending to 1.

What is wrong with attempt 2? Or is the answer actually 1 and not $\frac12$?

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  • $\begingroup$ Comments are not for extended discussion; this conversation has been moved to chat. $\endgroup$
    – Daniel Fischer
    Oct 28, 2016 at 17:28
  • $\begingroup$ Shreyas, you quoted "Assuming probability of collinearity equally likely". That's not exactly true. There are more non collinear points on the plane than there are collinear points(ideally they're both infinite, but sometimes one infinity is greater than another. Like in limits to infinity for f(x)=x and f(x)=x^2) $\endgroup$
    – Pritt Balagopal
    Apr 11, 2017 at 2:22

11 Answers 11

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There is no such thing as a uniform distribution on the plane. Without specifying how the points are chosen, the question is not properly stated. However, if the points are chosen independently from some continuous distribution (absolutely continuous with respect to Lebesgue measure), the probability of the third point lying exactly on the line through the first two is $0$.

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    $\begingroup$ +1, for the only answer that mentions explicitly independence ;-) $\endgroup$
    – Jean-Claude Arbaut
    Oct 11, 2016 at 11:24
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    $\begingroup$ ...and I gave you the 50th upvote. By the way, what's the probability of three points picked in an $m×n$ lattice not forming a triangle? This is different, but has it been asked before? $\endgroup$
    – Parcly Taxel
    Oct 11, 2016 at 17:41
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    $\begingroup$ Not downvoting, yet not sure why this was accepted. Yes, this answer is correct on some level, and nicely short. This is a school test though, so a few assumptions can be made (i.e., if something is not specified, like dependency, then it is not meant to apply). You do not need any kind of special features (Lebesgue etc.) to get a correct proof here; you only need to have infinitely many picks in both dimensions (the answer would be the same for N^2 or Q^2). No need for knowledge about measures, integrals, countability, continuity etc.. No special statistic/stochastic prerequisites either. $\endgroup$
    – AnoE
    Oct 11, 2016 at 22:38
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    $\begingroup$ @AnoE The answer is absolutely correct saying "Without specifying how the points are chosen, the question is not properly stated." There is nothing more you can say to the question as it was posed. The rest is a good account of what the answer might be to some similar but correctly posed question. $\endgroup$
    – Colin McLarty
    Oct 11, 2016 at 23:29
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    $\begingroup$ Yes, @SlippD.Thompson $\endgroup$
    – AnoE
    Oct 12, 2016 at 6:33
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Nothing can be said about this as long as nothing has been said about the distribution (justifying the comment of angryavian).

Expressions "at random" or "are chosen" do not speak for themselves because there is no natural uniform distribution on $\mathbb R^2$.

If the distribution is absolutely continuous wrt the Lebesgue measure (i.e. if the distribution has a PDF) then automatically the answer is $1$ because every line in the plane $\mathbb R^2$ has Lebesgue measure $0$ (which is probably what Kaj means to say).

So in that case for any fixed line the probability that the third point is chosen on it equals $0$.

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    $\begingroup$ Thanks, I understand your explanation...this question was asked at school level, hence no details were given regarding the distribution and stuff...(seems complex to me) $\endgroup$
    – Serenity
    Oct 11, 2016 at 7:37
  • $\begingroup$ Glad to hear that. You are welcome. $\endgroup$
    – drhab
    Oct 11, 2016 at 7:38
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    $\begingroup$ why is the probability of a 3rd point being on the fixed line equal to 0 ? :( $\endgroup$
    – Ciprian Tomoiagă
    Oct 11, 2016 at 8:45
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    $\begingroup$ @ShreyasS what does school level mean? High school? Middle school? College? Have you asked your teacher to explain?\ $\endgroup$
    – DRF
    Oct 11, 2016 at 9:41
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    $\begingroup$ @Shreyas The explanation in "attempt 1" is plainly wrong. You can tell your teacher that he/she needs to brush up his/her probability ;) $\endgroup$
    – tomsmeding
    Oct 11, 2016 at 11:24
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This is similar to this probability "joke":

Given a bowl with 9 black balls and 1 white ball, what's the chance that you pick a white ball? $\frac{1}{2}$, either you pick it or you don't.

While there are indeed both $\infty$ points which are collinear and $\infty$ points which are non-collinear, they're not quite the same $\infty$, so $\frac{\infty}{\infty+\infty}\neq \frac{1}{2}$. (See also Hilbert's Hotel on different levels of infinity)

As a matter of fact, since for collinear points, the choice of $x$ fixes the choice of $y$, there is only one level of infinity. For the non-collinear points, however, there are two levels of infinity: Both $x$ and $y$ can take infinite values. Thus, $\frac{P(\textrm{collinear})}{P(\textrm{non-collinear})}=\frac{1}{\infty}$, which tends to zero. In other words, $P(\textrm{non-collinear})\approx 1$.

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    $\begingroup$ No, the number of points in the real plane and on the real line are exactly the same infinity ($\beth_1$). It's also not clear that these "levels of infinity" arguments work, given that we're talking about uncountable sets. Hilbert's hotel deals with countable infinities. $\endgroup$
    – David Richerby
    Oct 11, 2016 at 14:48
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    $\begingroup$ You could use the same argument to "prove" that $\mathbb{Z}^2$ is bigger than $\mathbb{Z}$, but Hilbert's hotel shows us that those sets have the same cardinality. For example, the existence of space-filling curves proves that there are bijections between $\mathbb{R}$ and $\mathbb{R}^2$, which proves that they have the same cardinality. $\endgroup$
    – David Richerby
    Oct 11, 2016 at 16:11
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    $\begingroup$ I wanted to +1 for the joke, which I find illuminating and think will help the OP, but the rest of the answer (starting at the first mention of $\infty$) is too vague and imprecise to help rather than confuse, I think. $\endgroup$
    – ShreevatsaR
    Oct 11, 2016 at 16:16
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    $\begingroup$ The last sentence is the only part of this answer that should remain $\endgroup$
    – mhodges
    Oct 11, 2016 at 16:30
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    $\begingroup$ This argument is completely incorrect. $\endgroup$
    – djechlin
    Oct 12, 2016 at 23:49
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I see nothing wrong with the reasoning in Attempt 2, but Attempt 1 is all kinds of wrong.

Just because there are two possible outcomes, it does not follow that the probability of one of them is 0.5. This is only the case when each outcome is as likely as the other, such as with a coin toss.

To randomly pick a third point, out of all the infinite number of points on the plain, that happens to lie exactly on the line AB is hugely unlikely. Infinitely unlikely, in fact. Probability of a triangle = 1.

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  • $\begingroup$ I get it, but I don't. I understand that the odds are Infinitesimally small and are approaching zero, but if I can list infinitely many trios of points that are on the same line, it's hard to accept that the probability of a triangle equals 1. $\endgroup$
    – J.R.
    Oct 13, 2016 at 20:31
  • $\begingroup$ There are an infinite number of points between 0 and 1. But choosing any particular one (if we are choosing randomly) is zero. You also have to be convinced that 0.99999... (recurring) equals 1 :-) $\endgroup$
    – Gazzer
    Oct 14, 2016 at 10:06
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There is no obvious, 'natural' probability distribution of 'choosing points from a plane'. Hence a question starting like

If 3 distinct points are chosen on a plane, what is the probability...

without indicating a specific method of choosing points makes no sense, and the only two answers to it I can think of are 'the probability is any you can think of' or just 'get off'.

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Attempt 2 is flawed in the assumption that a point C on a line will be between A and B on a line segment. Since a line stretches on forever in both directions, there are an infinite number of points that could be C that are not between A and B. But this is still not on a plane, just a line.

The actual answer is that the probability of a point being on the line segment that connects any two points "approaches" zero, and because division by infinity is required, IS zero. In other words, infinitesimally small, and effectively zero. There are infinite possible points on the line segment, but there are also infinite possible line segments with infinite points in the plane:

$(AB)\infty / (AB)\infty*\infty$

Essentially, you are calculating the odds of a point falling on a specific line segment out of an infinite number of line segments:

$1 / \infty$

But division by infinity is effectively zero. So, the probability of a third point being on that specific line segment is actually zero. (This actually makes more sense, when you consider that by definition, a point actually has no size, and a line actually has no width...)

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  • $\begingroup$ Your solution is correct, but Attempt 2 never said anything about C being between A and B. It said C being on a line that passes through A and B $\endgroup$
    – mhodges
    Oct 11, 2016 at 16:34
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    $\begingroup$ Any approach that relies on multiplying or dividing by "infinity" is doomed. $\endgroup$
    – David Richerby
    Oct 11, 2016 at 17:28
  • $\begingroup$ If the three points are chosen independently, Craig's answer above can be visualized by noting that the area of any line is zero. For a non-triangle to result, the third point would have to land in a region of area (measure to be technical) zero. $\endgroup$
    – ttw
    Oct 11, 2016 at 19:48
  • $\begingroup$ FYI - infinity divided by infinity is indeterminant. Infinity divided by infinity squared in indeterminant. Infinity divided by infinity to the one millionth power is indeterminant. IOW, Infinity divided by infinity squared is not 1 divided by infinity. At least that's how I am interpreting the reasoning for your answer. Anyways, I do agree that the answer is zero, I just don't agree with the formula you used. $\endgroup$
    – Dunk
    Oct 11, 2016 at 21:38
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Having placed the first 2 points, there is a single straight line that goes through these two points. There is also an infinite number of parallel lines that don't go through those 2 points. (The plane itself can be finite).

For a non-triangle to occur, the third point must go on that single line out of the infinite possibilities. Assuming independence the probability of this is zero. Hence the probability of a triangle is 1.

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  • $\begingroup$ "Assuming independence the probability of this is zero." That depends on the distribution. Suppose the line containing the first two points is selected with probability $\tfrac12$ and the other lines with probability summing to $\tfrac12$. The probability of hitting the first line is not zero. $\endgroup$
    – David Richerby
    Oct 14, 2016 at 9:07
  • $\begingroup$ Well, obviously. If we don't assume the points are placed randomly anywhere then the question is unanswerable anyway. If, for example, you have to place the points on a grid, but there's nothing in the question that makes this constraint. $\endgroup$
    – Gazzer
    Oct 14, 2016 at 9:54
  • $\begingroup$ In my example, the points are placed "randomly anywhere". "Randomly" does not mean "uniformly at random" and, in this case, it cannot mean "uniformly at random", since there is no uniform distribution on the real plane. $\endgroup$
    – David Richerby
    Oct 14, 2016 at 10:26
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What has happened to good old probability without measure theory ? You can't say that the question is dumb just because a less-than-100-year-old theory can't solve it. If a theory doesn't answer a question, then don't try to use it. The question is actually very good, if you ask me. No need to go into debate about different kinds of infinity. If this is a school-level problem, then it's assumed that each point on the plane is equally likely. If highly skilled mathematicians can't solve the problem without telling this young fellow about Lebesgue measure or absolute continuity and whatnot then just use conditional probability. We don't care where the first two points $A$ and $B$ end up. So assuming there are two points on the plane, what is the probablity that the third one $C$ form a triangle ? All that matters is, of all lines parallel to $(AB)$, which one is $C$ on ? If all points are equally likely, then so are all those lines. Looking at all those parallel lines is like looking at $\mathbb{R}$. And $(AB)$ is one point of the real-line. So the problem is like saying "Given $x$ in $\mathbb{R}$ what is the probability for a random $y$ that $y=x$ ?" If you think it's $0$ and you need a measure to prove it, just say : For any bounded interval $I$ containing $x$ probability that $y=x$ given that $y\in I$ is zero with respect to normalised Lebesgue measure on that interval. So without even assuming that $y\in I$, not a chance.

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    $\begingroup$ Did you just admonish the other answers for invoking the Lebesgue measure, and then use it yourself to answer the same question? $\endgroup$
    – ilkkachu
    Oct 13, 2016 at 15:05
  • $\begingroup$ No, I read a few statements higher up according to which the question wasn't soluble because of the fact that there is no uniform measure on the plane. $\endgroup$
    – James Well
    Oct 14, 2016 at 0:49
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Let's make it simple. No need to bother with any calculation here. Probability is equal to 1 by definition.

Indeed, whenever you have 3 distinct points in a plane, you have a triangle. And even in the case where the 3 points are aligned on the same line, we are just facing with a degenerate triangle.

P.S: this also applies in a 3D space.

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A possibly silly way to finesse the ambiguity in the initial distribution: put the points on the projective plane. Then uniform distribution is well-defined and clearly, in the presence of uniform distribution, it is legitimate to use it to define a set of three points chosen randomly. This, of course, again leads to the probability of three points forming a triangle being 1.

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Eventhough, the actual answer is 1, because the measure of any line is 0, I am wondering whether the actual formulation of the problem was just slightly different since it looks silly the provided textbook answers to be 1/2.

One suggestion for the true formulation is the following:

Given 3 points A,B and C, taken at random in the plane(here, to make the things proper, one may substitute that with 'uniformly at random in a sqaure $C_{n}$ with side n, $n\to \infty$'), what is the probability that ABC is a positive oriented triangle?

The variants where just one word as 'obtuse' or 'acute' in front of 'triangle' is missed doesn't make sense since in such cases the answer differs from 1/2.

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