If you break a stick at two points chosen uniformly, the probability the three resulting sticks form a triangle is 1/4. Is there a "nice" proof of this fact? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-24T22:45:23Z http://mathoverflow.net/feeds/question/2014 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/2014/if-you-break-a-stick-at-two-points-chosen-uniformly-the-probability-the-three-re If you break a stick at two points chosen uniformly, the probability the three resulting sticks form a triangle is 1/4. Is there a "nice" proof of this fact? Michael Lugo 2009-10-23T02:28:05Z 2013-03-22T18:21:16Z <p>There is a standard problem in elementary probability that goes as follows. Consider a stick of length 1. Pick two points uniformly at random on the stick, and break the stick at those points. What is the probability that the three segments obtained in this way form a triangle?</p> <p>Of course this is the probability that no one of the short sticks is longer than 1/2. This probability turns out to be 1/4. See, for example, problem 5 in <a href="http://www.isds.duke.edu/courses/Fall05/sta104/hw/hw08sol.pdf" rel="nofollow">these homework solutions.</a></p> <p>It feels like there should be a nice symmetry-based argument for this answer, but I can't figure it out. I remember seeing once a solution to this problem where the two endpoints of the interval were joined to form a circle, but I can't reconstruct it. Can anybody help?</p> http://mathoverflow.net/questions/2014/if-you-break-a-stick-at-two-points-chosen-uniformly-the-probability-the-three-re/2016#2016 Answer by Kevin P. Costello for If you break a stick at two points chosen uniformly, the probability the three resulting sticks form a triangle is 1/4. Is there a "nice" proof of this fact? Kevin P. Costello 2009-10-23T02:43:55Z 2009-10-23T02:50:55Z <p>Here's what seems like the sort of argument you're looking for (based off of a trick Wendel used to compute the probability the convex hull of a set of random points on a sphere contains the center of the sphere, which is really the same question in disguise):</p> <p>Connect the endpoints of the stick into a circle. We now imagine we're cutting at three points instead of two. We can form a triangle if none of the resulting pieces is at least 1/2, i.e. if no semicircle contains all three of our cut points. </p> <p>Now imagine our cut as being formed in two stages. In the first stage, we choose three pairs of antipodal points on the circle. In the second, we choose one point from each pair to cut at. The sets of three points lying in a semicircle (the nontriangles) correspond exactly to the sets of three consecutive points out of our six chosen points. This means that 6 out of the possible 8 selections in the second stage lead to a non-triangle, regardless of the pairs of points chosen in the first stage. </p> http://mathoverflow.net/questions/2014/if-you-break-a-stick-at-two-points-chosen-uniformly-the-probability-the-three-re/2020#2020 Answer by Jason Dyer for If you break a stick at two points chosen uniformly, the probability the three resulting sticks form a triangle is 1/4. Is there a "nice" proof of this fact? Jason Dyer 2009-10-23T02:59:03Z 2009-10-23T02:59:03Z <p>Is the argument you remember along the lines of: picking three points on a circle, what is the probability they lie in the same semicircle?</p> <p>The problem is discussed here:</p> <p><a href="http://godplaysdice.blogspot.com/2007/10/probabilities-on-circle.html" rel="nofollow">http://godplaysdice.blogspot.com/2007/10/probabilities-on-circle.html</a></p> http://mathoverflow.net/questions/2014/if-you-break-a-stick-at-two-points-chosen-uniformly-the-probability-the-three-re/2029#2029 Answer by Ilya Nikokoshev for If you break a stick at two points chosen uniformly, the probability the three resulting sticks form a triangle is 1/4. Is there a "nice" proof of this fact? Ilya Nikokoshev 2009-10-23T03:57:55Z 2009-10-23T15:59:55Z <p>Yes, here's a nice and beautiful argument!</p> <p>First you should draw a picture of axes <code>a</code> and <code>b</code>. You're asked to select uniformly a point in the square <code>[0,1]x[0,1]</code>. Now because of the symmetry (sic!) it's equivalent to choosing the points <code>a</code> and <code>b</code> uniformly in the triangle cut from the square by <code>b &gt; a</code>.</p> <p>So you're actually uniformly selecting a point inside triangle defined by lines <code>a&gt;=0</code>, <code>b&lt;=1</code>, 'b>=a'.</p> <p>Now let's find the conditions to be able to make a triangle of short sticks. We should have <code>a + (1-b) &gt; b-a</code>, <code>b-a + (1-b) &gt; a</code> and <code>b &gt; 1 - b</code> which indeed, as you say, boils down to </p> <pre><code>b &gt; 1/2, a &lt; 1/2, b-a &lt; 1/2 </code></pre> <p>It remains to note that those lines create inside the big triangle a small triangle which is similar to big but with all lengths <code>1/2</code> of the big, so this small triangle has area of exactly 1/4 of original!</p> http://mathoverflow.net/questions/2014/if-you-break-a-stick-at-two-points-chosen-uniformly-the-probability-the-three-re/29539#29539 Answer by Michael Lugo for If you break a stick at two points chosen uniformly, the probability the three resulting sticks form a triangle is 1/4. Is there a "nice" proof of this fact? Michael Lugo 2010-06-25T19:34:21Z 2010-06-26T00:47:49Z <p>One reference for a solution to this problem is Carlos d'Andrea and Emiliano Gomez, "The broken spaghetti noodle", American Mathematical Monthly 113 (2006), p. 555. More generally the probability that an interval broken at n-1 points chosen uniformly at random is broken into pieces which can be rearranged to form an $n$-gon is $1 - n/2^{n-1}$.</p> http://mathoverflow.net/questions/2014/if-you-break-a-stick-at-two-points-chosen-uniformly-the-probability-the-three-re/66784#66784 Answer by Mort Schwartz for If you break a stick at two points chosen uniformly, the probability the three resulting sticks form a triangle is 1/4. Is there a "nice" proof of this fact? Mort Schwartz 2011-06-03T00:12:42Z 2011-06-03T00:12:42Z <p>SOLUTION USING ONLY ORIGINAL LINE: Call the end points A and B and the midpoint M. Let N be the midpoint of MB. Call P the first random point and Q the second. In full generality, we can consider P as lying between M and B. On AM, let H be the point for which HP is half the length of the original stick. For the 3 sticks to make a triangle, it is necessary and sufficient that the total length of any 2 sticks be greater than the third. Therefore no piece can exceed half the length of the original stick. That tells us immediately that Q cannot lie on AH or PB, immediately eliminating half the possible points for Q. Likewise all points on MP are also eliminated. Only points on <em>HM</em> qualify as possible Q points. If we let AB=1, the probability that the 3 sticks can form a triangle is then the average length of HM for all Ps. Note that if P=N, the midpoint of MB, then HM=MP, so the qualifying HM points comprise 1/4 the length of the stick and p=1/4 that a triangle can be formed after randomly choosing a Q for this P. Similarly, for every P to the right of N there is a matching P an equal distance to the left of N, so that the average probability (HM length) for these two Ps is 1/4. That yields an overall average probability for <em>all</em> Ps of 1/4.</p> http://mathoverflow.net/questions/2014/if-you-break-a-stick-at-two-points-chosen-uniformly-the-probability-the-three-re/66797#66797 Answer by Peter Shor for If you break a stick at two points chosen uniformly, the probability the three resulting sticks form a triangle is 1/4. Is there a "nice" proof of this fact? Peter Shor 2011-06-03T05:18:48Z 2011-06-04T11:26:10Z <p>Consider an equilateral triangle with altitude 1. It is not hard to show that if you choose a point randomly in this triangle, the distances to the three sides gives the same distribution of lengths that you obtain by breaking a stick at two random points. Now, the locus of points for which no distance is longer than 1/2 is the smaller equilateral triangle formed by joining the midpoints of the edges, which has area 1/4 that of the original triangle.</p> <p><img src="http://math.mit.edu/~shor/MO/triangle.jpg" alt="triangle figure"></p> http://mathoverflow.net/questions/2014/if-you-break-a-stick-at-two-points-chosen-uniformly-the-probability-the-three-re/66888#66888 Answer by Christian Blatter for If you break a stick at two points chosen uniformly, the probability the three resulting sticks form a triangle is 1/4. Is there a "nice" proof of this fact? Christian Blatter 2011-06-04T13:00:40Z 2011-06-04T14:06:46Z <p>A triangle is possible iff no part is $>{1\over2}$. With probability ${1\over2}$ both cuts are on the same side of the midpoint $M$, in which case no triangle is possible. If the cuts $x$ and $y$, $\ x &lt; y$, are on different sides of $M$ then with probability ${1\over 2}$ the point $x$ is further left in its half than $y$ is in the right half. In this case there is no triangle possible either. It follows that only ${1\over 4}$ of all cuts admit the forming of a triangle.</p> http://mathoverflow.net/questions/2014/if-you-break-a-stick-at-two-points-chosen-uniformly-the-probability-the-three-re/66898#66898 Answer by Allen Hatcher for If you break a stick at two points chosen uniformly, the probability the three resulting sticks form a triangle is 1/4. Is there a "nice" proof of this fact? Allen Hatcher 2011-06-04T14:26:01Z 2011-06-04T14:26:01Z <p>It seems natural to rephrase the question in terms of barycentric coordinates in a triangle. These coordinates are numbers $x$, $y$, $z$ in the interval $[0,1]$ satisfying the equation $x+y+z=1$. We are looking for triples $(x,y,z)$ of such numbers satisfying the three triangle inequalities $x \le y+z$, $y\le x+z$, and $z\le x+y$. Replacing the relations "$\le$" by "$=$", we get line segments joining the midpoints of the edges of the triangle. These line segments cut the triangle into four congruent subtriangles. The central one of these four subtriangles is the region where all three triangle inequalities hold, and this region has area equal to one quarter of the area of the big triangle.</p> <p>This is essentially the same argument as in the answers by Peter Shor and Ilya Nikokoshev, particularly in the reformulation of the latter answer in Ori Gurel-Gurevich's comment</p> http://mathoverflow.net/questions/2014/if-you-break-a-stick-at-two-points-chosen-uniformly-the-probability-the-three-re/75869#75869 Answer by liberalcynic for If you break a stick at two points chosen uniformly, the probability the three resulting sticks form a triangle is 1/4. Is there a "nice" proof of this fact? liberalcynic 2011-09-19T16:24:56Z 2011-09-19T16:24:56Z <p>My answer is 1/4. Consider a stick of length L. We make a cut at distance x from one end. Second cut at distance y from the same end. No side of a triangle can be greater than the sum of the other two sides, so both x and (y-x) have to be less than L/2. Probability is the number of favorable outcomes divided by total number of outcomes, so we get ((L/2) * (L/2))/(L*L) = 1/4!</p> http://mathoverflow.net/questions/2014/if-you-break-a-stick-at-two-points-chosen-uniformly-the-probability-the-three-re/75876#75876 Answer by liberalcynic for If you break a stick at two points chosen uniformly, the probability the three resulting sticks form a triangle is 1/4. Is there a "nice" proof of this fact? liberalcynic 2011-09-19T17:28:02Z 2011-09-19T17:28:02Z <p>@Michael Kissner: Maybe I'm messing it up, but I kinda figured that the minimum value the product of x and (y-x) can take is 0 and the maximum value it can take is L times L, which is L^2. Actually, if we take the first cut at x and the second cut at a distance dx from x, we can integrate the product of x and dx between the limits of 0 and L/2 for the favorable outcomes, and between 0 and L for the total outcomes. This would also give us the answer as 1/4. OMG, if this logic is flawed, then I've stumbled on to this answer by accident, which is kinda cool in some weird way!</p> http://mathoverflow.net/questions/2014/if-you-break-a-stick-at-two-points-chosen-uniformly-the-probability-the-three-re/81692#81692 Answer by vivek for If you break a stick at two points chosen uniformly, the probability the three resulting sticks form a triangle is 1/4. Is there a "nice" proof of this fact? vivek 2011-11-23T07:52:43Z 2011-11-23T07:52:43Z <p>another way to get the answer is to write a computer program to generate 2 random numbers from unif(0,1), say x1,x2, whenever x1x2 swap their values), let l1=x1,l2= x2 - x1 , l3= 1 - x2 , make a function that returns 1 when l1 +l2 >l3 &amp;&amp; l1 +l3 >l2 &amp;&amp; l3 +l2 >l1 , 0 otherwise . now repeat this experiment n times , as ntends to infinity by the frequentists definition of probability we get a good estimate of the probable value. . . . . . it turns out to be 1/4 .</p> http://mathoverflow.net/questions/2014/if-you-break-a-stick-at-two-points-chosen-uniformly-the-probability-the-three-re/112517#112517 Answer by Neal for If you break a stick at two points chosen uniformly, the probability the three resulting sticks form a triangle is 1/4. Is there a "nice" proof of this fact? Neal 2012-11-15T20:31:30Z 2012-11-16T16:56:02Z <p>Practically, the likelihood is >25% because there would be a natural tendency to break the larger piece after the first break. This increases the likelihood of having no piece longer than 1/2 because the probability weighted range for the second break is decreased.</p> <p>If we are breaking sticks (typically considered small in length), it is also not practical to break into extra short sections because sticks are not infinitely thin. It is very difficult to break stick into smaller pieces than 2-3 times their thickness. This tilts the likelihood well into the >25% range.</p> http://mathoverflow.net/questions/2014/if-you-break-a-stick-at-two-points-chosen-uniformly-the-probability-the-three-re/125306#125306 Answer by Shaswata for If you break a stick at two points chosen uniformly, the probability the three resulting sticks form a triangle is 1/4. Is there a "nice" proof of this fact? Shaswata 2013-03-22T18:21:16Z 2013-03-22T18:21:16Z <p>Let AB be the stick. WLOG we may assume AB=1(Since the probability won't depend on the length of AB). Let the points at which the stick is broken be P and Q.</p> <p>AP=x,PQ=y and QB=z.</p> <p>Since $0\leq AP,PQ,QB \leq 1$ we need to consider all the points inside the $1\times 1\times 1$ cube. Futhermore the points lie on the x+y+z=1 plane.</p> <p><a href="http://postimage.org/image/cyzgqo0vd/" rel="nofollow">x+y+z=1 plane</a>(click on the link to see the image of the plane)</p> <p>On applying the triangle inequalities (i.e $x+y>z,y+z>x\text{ and }x+z>y$) we find that the net area of points satisfying the condition of forming a triangle is the shaded potion.</p> <p><a href="http://s20.postimg.org/ocm01vbe5/cube.png" rel="nofollow">Shaded Area</a>(Click on the link to see the shaded area)</p> <p>Since the points $J,K,I$ are the midpoints of the sides of the of the triangle ACE.The probability = $\dfrac{\text{Area of }\Delta JKI}{\text{Area of }\Delta ACE}=\dfrac{1}{4}$</p>