## Assumptions in Geometric Probability ☆

| September 12, 2014

Geometric probability makes use of measurement to determine probabilities. For example, the probability of hitting a bullseye could be found by computing the ratio of the area of the bullseye (4π in^2) and the area of the entire dart board (64π in^2), which is 4/64 = 0.0625; or the probability of breaking a stick randomly and one of the pieces being less than 3 in could be found by the ratio of the length(s) that would make the break point result in one of the pieces being less than 3 in (3 in + 3 in) and the total length of the stick (10 in), which is 6/10 = 0.60 (see figures below). This geometric approach using measurements parallels the most common approach to probability more generally: P(E) = number of outcomes in the event / number of outcomes in the sample space. (Despite the fact that in the scenarios below the number of outcomes – where the dart or the break could land – is (uncountably) infinite.)

So let’s take a look at another problem. (Note: this problem was inspired by conversations with a colleague, Bill Zahner.) Let’s assume that we are going to construct an isosceles triangle. The base length is 10 in. The lengths of the other two sides will be a randomly chosen real number between 5 and 10 in (note, the side length being greater than 5 in guarantees forming a triangle, and anything under 10 in makes it non-equilateral). What is the probability that the resultant triangle is acute? obtuse? right? The figure below provides some insight – where along the perpendicular bisector of the isosceles triangle determines the classification of the triangle. So, using geometric probability, we could determine the likelihood of forming an obtuse triangle by computing the ratio of lengths: namely, the length up to the semicircle, which is 5 in, divided by the entire length, which is 5√3. So P(obtuse)=5/5√3 ≈ 0.577. Similarly, P(acute)=(5√3 – 5)/5√3 ≈ 0.423, and P(right)=0. Dealing with the probability of forming a right triangle being zero, despite in fact being possible, is difficult enough; but dealing with the fact that long-term simulations of the problem put the probability of being obtuse as less than the probability of being acute, indicating that these probabilities are, in fact, incorrect, takes additional insight into and understanding about the underlying assumptions of geometric probabilities.

Using geometric probability assumes that every point in the space is equally likely – in other words, that the distribution of outcomes is uniform. In this case, the “space” is the line segment (along the perpendicular bisector) for where the third vertex of the triangle could be. Although it is possible for the third vertex to fall anywhere along this line, how these points fall on that line is, in fact, not uniformly distributed. The video below models the situation in motion, indicating that the third vertex being near the base is much less likely than other places – in other words, the possible vertex points are not uniformly distributed. Using geometric measurements to determine the probability is inappropriate. 