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Infinite spaces; infinitesimal probabilities

March 12, 2008

No one commented on my brief post about a simple but interesting geometric probability problem.

There are a lot of interesting things to discuss about probabilities when the sample space is infinite. Such as:

  1. What is the probability that if a student guesses randomly on a math problem whose answer could be any rational number, he will get the problem correct? (See my post on cheating on the Math Olympiad to see why that problem comes to mind.) 😦
  2. What is the probability that if an integer is chosen at random, it will be even?
  3. Divide a rectangle into two parts, one with twice the area of the other. Pick a point randomly within the rectangle. What is the probability that the point is inside the smaller part?

Problem 1. exemplifies is something that bothers “beginners” about probability — that something that is “possible” can have probability zero.

Problem 2. raises many interesting questions. We know that the cardinality of the set of even integers is equal to the cardinality of all the integers. (For more on this, check the links here.) Yet, intuitively we know that the answer must be 1/2. There’s a fascinating discussion of this problem and more complicated versions at The Math Forum‘s Ask Dr. Math archive. In particular, I was surprised to read the following: “If you’re talking about the entire infinite set of integers, there is no way to do this [choose an integer at random] without some sort of a distribution function over the integers, and there is no such function that gives an “equal probability” for all integers.” (emphasis mine) The problem is solvable, of course, by using limits. And the answer, as we expect it must be, is indeed 1/2.

Problem 3 is a “basic” geometric probability problem. It was raised by another Ask Dr. Math reader (follow the link in the question) who argued that since the cardinality of each of the spaces is the same, the probability of landing in each must be the same. Doctor Tom replied, in part that “the cardinality of the set does not determine its measure” and that it is “measure” and not “cardinality” that is important in determining probabilities.

Then we come back to my original question:

  • ABCD is a square with side length 10. P is a point chosen at random inside ABCD. What is the probability that P lies on one of the diagonals of ABCD?

What I think is cool about this question is that the number of points on the diagonals is infinite, but it it’s size/area/measure (corrected in response to Pseudonym’s comment) is infinitesimal in comparison to the size of the area of the square. I suppose in that way it’s related to the problem below about choosing a rational number at random out of the set of all real numbers.

Some other interesting and related questions from Ask Dr. Math:

And my favorite:

Please post your comments on any of these problems that pique your interest or raise issues that you’d like to discuss further!

9 Comments leave one →
  1. Pseudonym permalink
    March 12, 2008 12:53 am

    What makes some of these questions hard to answer is that they bring in some quite advanced maths.

    Take the original question:

    ABCD is a square with side length 10. P is a point chosen at random inside ABCD. What is the probability that P lies on one of the diagonals of ABCD?

    You said “the number of points on the diagonals is infinite, but it is infinitesimal in comparison to the size of the area of the square”. Actually, that’s not true; the cardinality of both sets is the same (it’s C, the continuum).

    But still, there are intuitively “far fewer” points on the diagonals than in the square in total because the area (Lebesgue measure) of the diagonals is zero.

  2. March 12, 2008 12:59 am

    Indeed, you have the vocabulary down better than I do. I wasn’t sure if the cardinality of the 2 sets was the same, though I suspected it was. I knew intuitively that the answer to the question must be zero, and that it was simple to see why, given the area model of geometric probability, but I had never seen a question that asked about the probability of a point being on a particular line (or pair of lines) before. The typical geometric probability questions are all about the probability of a point chosen being in some smaller area than the whole area, and finding the answer comes down to finding the (ratio of) the areas.

  3. Alane Tentoni permalink
    March 12, 2008 8:02 am

    In your problem (with square ABCD), could you say that the diagonals have no area, since (being line segments) they have no width?

    You run into the same type of problem when you go from a binomial distribution to a continuous distribution.

    And now I’m going to go look up binomial & continuous distributions again! 🙂

  4. March 12, 2008 11:16 pm

    Yes, Alane, exactly, the diagonals have no area, so the probability of landing on them is zero. Because the “measure” that matters in this case is area. If we were talking about the probability of a randomly chosen point being, say, the midpoint of a line segment, the “measure” would be length (and the answer would still be zero).

    Now you’ve got me running to look up binomial versus continuous distributions. 🙂

  5. Austin permalink
    March 13, 2008 2:58 pm

    Couldn’t you also say that it is impossible to pick a single integer from the countable set randomly rather than say probability 0 is not impossible?

    Also, Can one say that the probability of picking a random rational 1 is 50% also?

    What would be the probability of picking a random rational < 1/2 ?

  6. March 13, 2008 3:34 pm

    Austin, when there is an infinite set of possibilities, the probability of each is zero, but one will still be picked, so none is impossible.

    I’m not sure what you’re asking about with your 50% question. I think maybe some symbols got lost in the translation. Did you mean < 1?

    For the random rational < 1/2, do you mean out of the entire set of rationals? Or the reals?

    If you’re talking about choosing a random rational number, and determining the probability that it is either < 1/2 or < 1 that’s an interesting question, and I’m not sure of the answer. My intuition is that no matter which number you choose as your “dividing point” you’re equally likely to choose a number either greater or less than it, for a probability of 50%. But it does seem paradoxical that whatever number you choose you get the same probability. If the probability that the number chosen is less than one is 1/2 and the probability that the number chosen is greater than 1/2 is also 1/2, then the probability that the number chosen is both greater than 1/2 and less than 1 should be 1/4, but I think that the probability of the chosen number falling into any finite interval must be zero.

    I think that is where we get into needing to determine the distribution function over the rationals that we mean when we say “choosing a random rational number”.

  7. Austin permalink
    March 13, 2008 4:04 pm

    Yeah, i think some of the text i typed vanished.

    The distribution i would use would be:

    2 3 5 7 …
    0 1/2 2/3 4/5 6/7 …
    1 1/4 2/9 4/25 6/49 …
    2 1/8 2/27 4/125 6/343 …

    In this case, the probability of picking a random integer that is say 2 would be 1/4 * 2/3 * 4/5 * 6/7 … and the probability of picking 3 would be:
    1/2 * 2/9 * 4/5 * 6/7…

    To get the rationals, let each probability in the above table = 1/2 of the probability listed for those exponents >=1 and let 1/2 of the original probability be for negative exponents and 1/2 of it be for positive exponents.

  8. Ragtime permalink
    March 14, 2008 9:09 am

    One of your links referred to a different problem:

    Three Points are taken at random on an infinite Plane. Find the chance
    of their being the vertices of an obtuse-angled Triangle.

    My initial thought is that the answer is 100%. If we make line segment AB, and orient ourselves on the infinite plane such that AB is on the x-axis, then the triangle is obtuse any time point C is “left of A” or “right of B”, and only acute in the “smaller infinity” when C is between A and B.

    The fact that Lewis Carroll and Israeli mathematicians came up with different answers, though, makes me think that the problem couldn’t be as easy as I thought. Unfortunately, the link to the Israeli paper is broken.

  9. March 14, 2008 11:36 pm

    Austin, I’m not completely following your probability distribution, probably at least partially because your table doesn’t come out nicely in this format, but also because I’m not following what pattern you’re using in your numerators that begins (1, 2, 4, 6, …). (Also, I think you might have transposed a 1/2 and a 1/4 in your examples?)

    Ragtime, take a look at this article: A Statistical Paradox: (1) by M. J. O’Carroll for further discussion of the obtuse triangle problem. It begins by fixing the longest side, and shows that different answers can be found as to the probability of the obtuse triangle depending on exactly where you fix it. And it comes down to distribution functions again, I believe. The article states: It is not possible to take one point at random in an infinite plane, if you want “at random” to mean that it is no more likely to be in any one region than any other of the same area.

    There’s another article on the same problem called A Lewis Carroll Pillow Problem: Probability of an Obtuse Triangle by Stephen Portnoy

    I can only access the first page of each of those articles, but that much is interesting enough. 🙂

    I also just realized that this is somewhat related to a puzzle I read before about the probability of “a random triangle” being acute. Check out the discussions at JD2718 and SquareCircleZ

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