We consider a simple problem in probability:
A thin rod is broken at random into three pieces.What is the probability that these three piecescan be used to form a triangle?
This problem is solved without difficulty. If the rod is of unit length, aligned along the interval , we choose two points at random in , assuming a uniform distribution. Supposing the first point is at , we have either or .
Let us consider the case where the first point is ; the configuration is shown in Figure 1. The second point, , is again chosen from a uniform distribution in . Then the three line segments are formed by (if ), or (if ).
Figure 1. A rod of unit length, with the point x first point chosen being in the interval [0,1].It is clear that if ( in the blue zone), the two segments and together cover the interval , of length less than , and the remaining segment is longer than their sum. Thus, no triangle can be formed from , and .
Similarly, if ( in the left grey zone), it is clear that no triangle can be formed from , and . Finally, if ( in the right grey zone), the segment will be longer than so the sum of the two remaining segments cannot exceed it, and no triangle is possible.
The conclusion is that when , the second point must be in the green zone (), which is of length . The probability of a randomly chosen point being in this zone is . Now integrating over all possible values we get
If we now consider the case where the first point is in , we obtain the mirror image of the case above; a simple argument based on symmetry shows that the probability that a triangle can be formed is . Combining the two cases, the probability that a triangle can be formed is .
Another Method
There is a subtlety with this problem: how are the break-points chosen? Two possible methods are:
Choose two random break-points on the rod.
Break once, then break the longer piece.
The discussion above assumed the first method was used, leading to the probability that a triangle can be formed from the three segments. However, the second seems more natural, and more like what might be done in practice. Breaking the shorter piece prevents the construction of a triangle. This justifies the claim that Method 2 is more natural than Method 1, which is like throwing two darts at the rod to choose the break-points.
We examine Method 2 now. We begin by considering the case where the first point is ; the configuration is the same as in Figure 1. The second point, , is again chosen from a uniform distribution, but is now restricted to the interval , which is of length . To ensure that a triangle can be formed, it must again be in the green zone, of length , an the probability of this is . Integrating over all values of in , we get
We must now combine this with the case . By symmetry, the probability of a triangle being possible is the same as before. Thus, the overall chance of a triangle is . A numerical check was made, using Mathematica to generate pairs of points following Method 2. The results were consistently close to the theoretical value .
In summary, the solutions for the two methods are
Bertrand’s Paradox
The point of the above discussion is that the problem is not well-posed: the probability depends on the method of breaking the rod. This is reminiscent of Bertrand’s paradox, a simple result in geometric probability.
Given a circle, a chord is drawn at random.What is the probability p that the chord length is greaterthan the side of an equilateral triangle inscribed in the circle?
Bertrand (1889) considered three ways of drawing a chord in the outer circle (Figure 2):
Fix the end-points of the chord ().
Choose the chord centre on a fixed diameter ().
Fix the mid-point of the chord ().
Figure 2. Three ways of choosing a chord.
He showed that that the probability differs for the three methods of choice. This is discussed in an earlier post on thatsmaths.com. The ruler-and-triangle problem considered above is analogous to Bertrand’s Paradox, in that the method of choosing break-points must be specified for the probability to be uniquely determined.
Sources
Bertrand, Joseph, 1889: Calculs de Probabilités. Gaultier-Villars, Paris.
Des MacHale, 2023: Lateral Solutions to Mathematical Problems A K Peters/CRC Press. Taylor & Francis Group. ISBN: 978-100-3341-46-8.
ThatsMaths.com. Bertrand’s Chord Problem. Link
