Mathematical Games: Probability and Ambiguity

Editor’s Note: Published in 1959, this article comes from Martin Gardner’s legendary Scientific American column Mathematical Games. Read more classic Mathematical Games articles in our 2024 special digital issue, Fun and Games.

Charles Sanders Peirce once observed that in no other branch of mathematics is it so easy for experts to blunder as in probability theory. History bears this out. Leibniz thought it just as easy to throw 12 with a pair of dice as to throw 11. Jean Ie Rond d’Alembert, the great 18th-century French mathematician, could not see that the results of tossing a coin three times are the same as tossing three coins at once, and he believed (as many amateur gamblers persist in believing) that after a long run of heads, a tail is more likely.

Today probability theory provides clear, unequivocal answers to simple questions of this sort but only when the experimental procedure involved is precisely defined. A failure to do this is a common source of confusion in many recreational problems dealing with chance. A classic example is the problem of the broken stick. If a stick is broken at random into three pieces, what is the probability that the pieces can be put together in a triangle? This cannot be answered without additional information about the exact method of breaking to be used.

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One method is to select, independently and at random, two points from the points that range uniformly along the stick, then break the stick at these two points. If this is the procedure to be followed, the answer is ¹⁄₄, and there is an elegant way of demonstrating it with a geometrical diagram. We draw an equilateral triangle, then connect the midpoints of the sides to form a smaller shaded equilateral triangle in the center [see illustration below]. If we take any point in the large triangle and draw perpendiculars to the three sides, the sum of these three lines will be constant and equal to the altitude of the large triangle. When this point, like point A, is inside the shaded triangle, no one of the three perpendiculars will be longer than the sum of the other two. Therefore, the three line segments will form a triangle. On the other hand, if the point, like point B, is outside the shaded triangle, one perpendicular is sure to be longer than the sum of the other two, and consequently no triangle can be formed with the three line segments.

We now have a neat geometrical analogy to the problem of the broken stick. The sum of the three perpendiculars corresponds to the length of the stick. Each point on the large triangle represents a unique way of breaking the stick, with the three perpendiculars corresponding to the three broken pieces. The probability of breaking the stick favorably is the same as the probability of selecting a point at random and finding that its three perpendiculars will form a triangle. As we have seen, this happens only when the point is inside the shaded triangle. Since this area is one fourth the total area, the probability is ¹⁄₄.

Suppose, however, that we interpret in a different way the statement “break a stick at random into three pieces.” We break the stick at random, we select randomly one of the two pieces, and we break that piece at random. What are the chances that the three pieces will form a triangle? The same diagram will provide the answer. If, after the first break, we choose the smaller piece, no triangle is possible. What happens when we pick the larger piece? Let the vertical perpendicular in the diagram represent the smaller piece. In order for this line to be smaller than the sum of the other two perpendiculars, the point where the lines meet cannot be inside the small triangle at the top of the diagram. It must range uniformly over the lower three triangles. The shaded triangle continues to represent favorable points, but now it is only one third the area under consideration. The chances, therefore, are ¹⁄₃ that when we break the larger piece, the three pieces will form a triangle. Since our chance of picking the larger piece is ¹⁄₂, the answer to the original question is the product of ¹⁄₂ and ¹⁄₃, or ¹⁄₆.

Geometrical diagrams of this sort must be used with caution because they, too, can be fraught with ambiguity. For example, consider this problem discussed by Joseph Bertrand in a famous 19th-century French work on probability. What is the probability that a chord drawn at random inside a circle will be longer than the side of an equilateral triangle inscribed in the circle?

We can answer as follows. The chord must start at some point on the circumference. We call this point A, then draw a tangent to the circle at A, as shown in the top illustration above. The other end of the chord will range uniformly over the circumference, generating an infinite series of equally probable chords, samples of which are shown on the illustration as colored lines. It is clear that only those chords that cut across the triangle are longer than the side of the triangle. Since the angle of the triangle at A is 60 degrees, and since all possible chords lie within a 180-degree range, the chances of drawing a chord larger than the side of the triangle must be ⁶⁰⁄₁₈₀, or ¹⁄₃.

Now let us approach the same problem a bit differently. The chord we draw must be perpendicular to one of the circle's diameters. We draw the diameter, then add the triangle as shown in the middle illustration above. All chords perpendicular to this diameter will pass through a point that ranges uniformly along the diameter. Samples of these chords are again shown in color. It is not hard to prove that the distance from the center of the circle to A is half the radius. Let B mark the midpoint on the other side of the diameter. It is now easy to see that only those chords crossing the diameter between A and B will be longer than the side of the triangle. Since AB is half the diameter, we obtain an answer to our problem of ¹⁄₂.

Here is a third approach. The midpoint of the chord will range uniformly over the entire space within the circle. A study of the bottom illustration above will convince you that only chords whose midpoints lie within the smaller shaded circle are longer than the side of the triangle. The area of the small circle is exactly one fourth the area of the large circle, so the answer to our problem now appears to be ¹⁄₄.

Which of the three answers is right? Each is correct in reference to a certain mechanical procedure for drawing a chord at random. The problem as originally stated is ambiguous. It has no answer until the meaning of “draw a chord at random” is made precise by a description of the procedure to be followed. Apparently nothing resembling any of the three procedures is actually adopted by most people when they are asked to draw a random chord. In an interesting unpublished paper entitled “The Human Organism as a Random Mechanism,” Oliver L. Lacey, professor of psychology at the University of Alabama, reports on a test that showed the probability to be much better than ¹⁄₂ that a subject would draw a chord longer than the side of the inscribed triangle.

Another example of ambiguity arising from a failure to specify the randomizing procedure appeared in this department last May. Readers were told that Mr. Smith had two children, at least one of whom was a boy, and were asked to calculate the probability that both were boys. Many readers correctly pointed out that the answer depends on the procedure by which the information “at least one is a boy” is obtained. If from all families with two children, at least one of whom is a boy, a family is chosen at random, then the answer is ¹⁄₃. But there is another procedure that leads to exactly the same statement of the problem. From families with two children, one family is selected at random. If both children are boys, the informant says, “At least one is a boy.” If both are girls, he says, “At least one is a girl.” And if both sexes are represented, he picks a child at random and says, “At least one is a...,” naming the child picked. When this procedure is followed, the probability that both children are of the same sex is clearly ¹⁄₂. (This is easy to see because the informant makes a statement in each of the four cases—BB, BG, GB, GG—and, in half of these cases, both children are of the same sex.) That the best of mathematicians can overlook such ambiguities is indicated by the fact that this problem, in unanswerable form, appears in one of the best of recent college textbooks on modern mathematics.

A wonderfully confusing little problem involving three prisoners and a warden, even more difficult to state unambiguously, is now making the rounds. Three men—A, B and C—were in separate cells under sentence of death when the governor decided to pardon one of them. He wrote their names on three slips of paper, shook the slips in a hat, drew out one of them and telephoned the warden, requesting that the name of the lucky man be kept secret for several days. Rumor of this reached prisoner A. When the warden made his morning rounds, A tried to persuade the warden to tell him who had been pardoned. The warden refused.

“Then tell me,” said A, “the name of one of the others who will be executed. If B is to be pardoned, give me C’s name. If C is to be pardoned, give me B’s name. And if I’m to be pardoned, flip a coin to decide whether to name B or C.”

“But if you see me flip the coin,” replied the wary warden, “you’ll know that you’re the one pardoned. And if you see that I don’t flip a coin, you’ll know it’s either you or the person I don’t name.”

“Then don’t tell me now,” said A. “Tell me tomorrow morning.”

The warden, who knew nothing about probability theory, thought it over that night and decided that if he followed the procedure suggested by A, it would give A no help whatever in estimating his survival chances. So next morning he told A that B was going to be executed.

After the warden left, A smiled to himself at the warden’s stupidity. There were now only two equally probable elements in what mathematicians like to call the “sample space” of the problem. Either C would be pardoned or himself, so by all the laws of conditional probability, his chances of survival had gone up from ¹⁄₃ to ¹⁄₂.

The warden did not know that A could communicate with C, in an adjacent cell, by tapping in code on a water pipe. This A proceeded to do, explaining to C exactly what he had said to the warden and what the warden had said to him. C was equally overjoyed with the news because he figured, by the same reasoning used by A, that his own survival chances had also risen to ¹/⁄₂.

Did the two men reason correctly? If not, how should each calculate his chances of being pardoned? An analysis of this bewildering problem will be given next month. [Editor’s Note: The solution from the November 1959 issue of Scientific American appears below.]