Challenge Problems

Before starting with the formal problems in this book, here are several challenge problems for you to think about as you read. As you work your way through the formal problems you may pick up ideas and mathematical techniques that will serve you well in attacking the challenge problems. These are not necessarily easy problems (you’ll have to do some integrals), and for most of them, you will have to think long and hard. All do have exact solutions, and you will find complete discussions (including Monte Carlo simulations) of each at the back of the book. But I strongly encourage you to work hard at these problems before you look there.

(1) Imagine all possible triangles in which two of the sides are independently and randomly (uniformly) assigned some length from 0 to 1. From all these triangles, consider only the set S in which the third side has the exact length of 1. If we select a triangle at random from S, what’s the probability it is obtuse?

(2) In the 1980s, variations of the following puzzle, called the glass rod problem, made several appearances in the British math journal The Mathematical Gazette. It was originally posed and solved in the March 1981 issue as follows.

A glass rod drops and breaks into three pieces. What is the probability that a triangle can be formed from the pieces?

Consider an equilateral triangle whose height is the length of the rod (say 1). It is easily proved that the sum of the lengths of the perpendiculars from the three sides to any interior point is 1 [no proof was offered, but this is often called Viviani’s theorem, after the Italian mathematician Vincenzo Viviani (1622–1703), to whom it is attributed; it is not in Euclid’s Elements]. Further, given any triple of non-negative numbers whose sum is 1, there is just one interior point whose distances from the three sides (in some prescribed order) are precisely those three numbers [these distances are the lengths of the three pieces]. Hence there is a 1:1 correspondence between the interior points of the triangle and the possible lengths of the three pieces of the broken rod (see (a) in the figure). If the perpendicular from the base extends into the shaded region in (b), then its length is over 1/2, and so the point shown corresponds to a failure to form a triangle [to form a triangle, the lengths of all three pieces must, of course, be less than 1/2]. But the same applies to the other two perpendiculars and so there are three failure regions, each 1/4 of the triangle (as shown in (c)). So the probability of being able to form a triangle with the three pieces is 1/4(= 1 − 3/4).

Figure C.1. The glass rod problem.

Well, that’s certainly slick! But are you convinced? By taking the ratio of the unshaded area to the area of the entire triangle, the central assumption of geometric probability is being invoked: each tiny patch of area in the triangle is as likely as any other tiny patch (of the same area) to be selected as the location of an interior point. Is that the case here? Now, don’t read too much into my question: the above analysis may be faulty or it may be okay—I’m simply asking you to think carefully about both the statement of the problem and the possible unstated assumptions being made in the analysis. This is important, because it will be quite useful in a later challenge problem.

(3) After you’ve thought about the previous problem for a while, here’s what seems to be the same problem, except that the details of the breaking of the glass rod are (much) more carefully specified. The rod is of unit length, with the left end at 0 and the right end at 1. We break it as follows: we independently and randomly (uniformly) pick two breakpoints at distances x and y, in the interval 0 to 1. This gives us two possibilities for the length of the broken pieces, one possibility for x < y and another for x > y. If x < y, then the three pieces have lengths x, y − x, and 1 − y. If x > y, then the three pieces have lengths y, x − y, and 1 − x. What is the probability the three pieces can form a triangle?

(4) As a continuation of the previous problem, suppose a triangle is formed. Given that, what’s the probability it’s an obtuse triangle?

(5) One last broken rod. First break the rod at a random point selected from the interval 0 to 1. Then take the longer of the two resulting pieces (there will always be a short piece and a long piece after the first break) and break it at some point randomly (uniformly) selected along its length. What is the probability the three resulting pieces can form a triangle?

(6) Getting tired of breaking rods and making triangles? Let’s toss darts instead, then. Suppose you have a circular dartboard with unit radius, and you toss two darts at it. Both darts hit the board at independent, random locations. As usual, assume the randomness is uniform. What’s the probability the two darts are at least unit distance apart?

(7) Given a circle centered on the origin, pick three points on its circumference at random. By random, I mean use the following selection process for the three points. First, pick any point on the circumference, then rotate the circle so as to put that point on the positive x-axis. That sets the first point, and there is no loss in generality in doing this because of the symmetry of the circle. Then, independently pick two angles, α and β, each uniform from 0 to 2π. The second point is located from the first point by angle α (going counterclockwise), and the third point is angle β from the first point (also going counterclockwise). Now, using those points as the vertices of a triangle, what’s the probability the center of the circle (the origin) will be inside the triangle? Hint: You might want to simulate this process first, and then do a theoretical analysis.

(8) Six sports teams, all of equal ability, form a league and play a long series of games each year to determine the overall champion. Each year the championship team receives a glorious trophy. If the same team wins the trophy three straight years, that team gets to retire the trophy permanently. How many years do these six teams have to play before the probability is at least 1/2 that the trophy is retired? How does the answer change if the league consists of ten teams? (Hint: This problem can be formulated as a difference equation that can be solved analytically, but you may find a computer approach to be vastly easier.)

(9) Imagine that a golfer has hit a long drive that has randomly placed his ball on a perfectly flat, square green, with the hole in the center of the green. What is the probability that the ball is closer to the hole than it is to any edge of the green?

(10) Imagine an urn that initially contains b black balls and w white balls, where b, w > 0. You randomly draw a ball, note its color, and then discard it. You then randomly draw more balls, one after the other, and, as long as their color matches that of the first ball, you discard each of them, too. Once you draw a ball of the other color, however, you put it back into the urn. You then repeat this entire process (the color of the next ball drawn is noted, that ball is discarded, …). Finally, there is just one ball left in the urn. When you draw it, what’s the probability it is black?

(11) This book has already included a fair amount of historical discussion dealing with probability problems involving the tossing of dice, so here’s one more for you as the penultimate challenge problem. The Swiss mathematician James (aka Jacob) Bernoulli (1654–1705) formulated the following question in 1685, and discussed it in 1690 on the pages of the early scientific journal Acta Eruditorum (Reports of the Scholars). Two players, A and B, take turns tossing a fair die. The first one to throw an ace (the face with a single dot) wins. They proceed as follows. On the first turn, A throws once, then B throws once if A didn’t throw an ace. For the second turn, if B didn’t throw an ace, then A gets the die back and throws twice, and then B throws twice if A didn’t get an ace. And so on, with each new turn allowing each player one additional toss. That is, on the kth turn with the die, A gets k tosses, and then so does B if A didn’t throw an ace. What’s the probability that A wins? (Bernoulli found an infinite series for the answer, but didn’t evaluate it.)

(12) For the final challenge problem, here’s one that is at the level of a final exam question in a first-year college course in probability. It’s easy to understand and very easy to simulate, but to do it analytically you’ll really have to understand the fundamentals of probability theory. So, with that big buildup, here it is. Suppose we pick two numbers independently, with each uniformly distributed over the interval −1 to 1. Call them A and B. What’s the probability that A^2/3 + B^2/3 < 1? To simulate this is duck soup, and here’s a MATLAB^® code that does the job:

When run several times, the code produced estimates for the probability ranging from 0.2937 to 0.2957. Calculate the exact value of this probability. (By the way, you’ll notice that after finding A and B, the code first squares them and then takes the cube root. That is, of course, the two-thirds power, but why not just compute A^2/3 and B^2/3 directly? I did the calculations this way because if A (or B) is less than zero, that number raised to the two-thirds power is positive, and doing the calculations as done in the code is guaranteed to produce a positive result. Writing A^2/3 and B^2/3 directly will not work because 2/3 in a computer is 0.66666666 … to a finite number of digits, with the result being complex results.)