Lewis Carroll's Obtuse Problem

KEYWORDS:
Geometric probability;
Obtuse triangle;
Infinite uniformity;

Summary
Carroll’s apparently impeccable solution to one of his probability problems is shown to answer another problem that is based on reasonable assumptions. His original assumptions, however, are self contradictory, hence entailing paradoxical results.

Address Correspondence to Ruma Falk, 3 Guatemala St., Apt. 718. Jerusalem, 96704, Israel.

Lewis Carroll's Obtuse Problem
INTRODUCTION
Lewis Carroll (1832-1898) suffered from sleeping problems. Although certainly unpleasant, this experience had also a favourable effect: The sleepless nights resulted in a collection entitled Pillow problems: Thought out during wakeful hours. The burden of sleepless hours is now transferred to the readers of this charming book (Carroll, 1895/1958).
One pillow-problem, in particular, interfered with our sleep.  In this problem (No. 58), we are asked:
Three Points are taken at random on an infinite Plane. Find the chance of their being the vertices of an obtuse-angled Triangle. (Carroll, 1958, p. 14)
A PROBLEMATIC PROBLEM
The disturbing element in the problem's text is the random sampling of points on an infinite plane. How could this be? This is a practically and conceptually impossible procedure. Put explicitly, Carroll assumed (1) that the sample-space for his "statistical experiment" is infinite, and (2) that the probability-density function on that space is uniform. However, these are two contradictory assumptions that are sooner or later bound to entail paradoxical results (see Falk & Konold, 1992).
A probability (density) distribution - whether discrete or continuous, univariate or bivariate - can be uniform only in the finite case, which is tantamount to saying that random variables with an infinite domain cannot be uniformly distributed. Presuming a uniform and infinite probability space is self contradictory.
Nevertheless, on the face of it, Carroll's solution of the obtuse-angled-triangle problem appears flawless (Wells, 1992, pp. 248-249, for example, reproduced this solution in his collection without protest):
It may be assumed that the 3 Points form a Triangle, the chance of their lying in a straight Line being (practically) nil. Take the longest side of the Triangle, and call it 'AB': and, on that side of it, on which the Triangle lies, draw the semicircle AFB. Also, with centres A, B, and distances AB, BA, draw the arcs BDC, AEC, intersecting at C.
Then it is evident that the vertex of the Triangle cannot fall outside the Figure ABDCE.
Also, if it fall inside the semicircle, the Triangle is obtuse-angled: if outside it, acute-angled. (The chance, of its falling on the semicircle, is practically nil.)
area of semicircle

Hence required chance = area of fig. ABDCE
Now let AB = 2a: then area of semicircle = and area of Fig. ABDCE = 2 ' sector ABDC - Triangle ABC
chance = Q. E. F.

Carroll (1958, pp. 83-84)

Completing this calculation yields a probability of 0.6394.
A closer look at Carroll's solution reveals that he indeed made use of uniformity. However, he applied that assumption only to the interior of the region bounded by Figure ABDCE. The computed value of about 0.64 provides the correct answer to a different question. The correct problem for Carroll's solution would be:
Given two arbitrary points A and B, a triangle is constructed by choosing a third point at random from the region for which AB is the triangle’s longest side. What is the probability that this triangle is obtuse-angled?
Getting back to the original problem, where three points have to be chosen at random on an infinite plane, the error in the above solution to this problem lies in accepting the existence of a triangle formed by this procedure as taken for granted. That starting point was not even questioned. Obviously, a self-contradictory basic assumption invalidates the subsequent analysis and numerical calculation.
The illegitimacy of Carroll's premises would become more conspicuous if a different answer to the same problem could be derived from the same underlying (faulty) assumptions, thereby creating a paradox.
Following Carroll, we assume that those notorious three points have been selected and they form a triangle. Let AB be one side of the triangle. Without loss of generality we may consider only the upper half of the plane (above the straight-line l) as the sample-space for choosing at random the triangle's third point. Choice of the third point, C (unmarked as yet), will form an obtuse triangle if either D ACB or DCAB or DCBA is obtuse. These three events are (pairwise) mutually exclusive so that we need only add their individual probabilities in order to obtain the probability of their union. We go on steadily assuming uniformity over the entire infinite plane. The probability that D ACB is obtuse may be ignored because this will occur if point C falls in the semicircle built on AB as diameter, the area of which is negligible relative to the infinite area of the sample-space. However, (DCAB> 900) = 1/2, because DCAB is obtuse whenever C lies to the left side of line a. By the same token, P (DCBA> 900) = 1/2, because this is the probability of the random point lying on the right side of line b.
The result is P (Obtuse triangle) = 1/2+1/2 = 1, in blunt contradiction to the previous outcome of 0.64. Admittedly, the computation that resulted in P=1 appears nonsensical. Yet, this is a valid consequence of nonsensical assumptions. Undoubtedly, many other conflicting conclusions may be drawn from assuming random uniform choice of points on an unbounded region of any number of dimensions (For example, the referee of this paper argues that if the probability of CAB being obtuse is 1/2, then by symmetry, the same is true for all three angles. By exclusiveness, the required probability is thus 1.5!).
The derivation of different answers to the same probability problem is reminiscent of Bertrand's geometrical paradox (Szekely, 1986, pp. 43-48) that obtained different probabilities as answers to a question concerning a random chord of a given circle by applying different methods for choosing that chord. Bertrand's instruction to apply a "uniform random choice" did not determine a" unique procedure, but could be implemented in several different ways. Each of these legitimate methods of choosing a random chord was well defined and resulted, in turn, in its unique answer to the chord problem.
In contrast, in Caroll's case, the computations above result from the instruction to choose three points "at random on an infinite plane". Although they seem workable, these instructions are impossible to fulfill. This leads to different procedures that apparently follow the directions, but yield different, bizarre answers for the probability of an obtuse triangle.
What, then, is the probability of a "random obtuse triangle"? Consider, instead of the infinite plane, a uniform distribution on a very large n'n sided square. A little reflection yields that the answer cannot depend on the size of the area from which the points are to be sampled, but just on its geometric form: here, a square (equal change of the size of both sides of the square can be considered just as a change in the unit of measurement). Therefore, we may let n=1. On the other hand, we can let n increase indefinitely without affecting the probability. Presumably, this will come closest to Carroll's intentions.
An analytic answer to this question is very difficult to obtain. But computer simulations can be done easily. Let the computer pick randomly X1, X2, X3, Y1, Y2, Y3 all independent and uniformly distributed on the interval [0,1]. Create the three points (X1, Y1), (X2, Y2), (X3, Y3) in the plane. These are randomly chosen points on the unit square. Now, denote by L1, L2, and L3 the respective lengths of the sides of the triangle formed by the randomly chosen points.

The reader should convince her/himself that Z=1 if and only if the (random) triangle is obtuse.
We ran the above experiment one million (106) times. The proportion of obtuse triangles obtained was 0.7249, with a standard error of 0.00045, quite a bit larger than the value 0.6394 obtained from Carroll's calculation! The time needed for the experiment was negligible.
We ran similar simulations (106 times) over various other geometric shapes as sample spaces: circle, equilateral triangle, and rectangle where the ratio of width to length is 1:k, and k goes from 2, through every third number, to 20. Note that when expanding the dimensions of these shapes indefinitely, they will cover any given three points in the infinite plane. Table 1 summarizes our findings. We see that the shape of the area largely influences the (estimated) probability of obtaining an obtuse triangle. For rectangles with a ratio of 1:20 the (estimated) probability is virtually one. In all these cases the probability is greater than Carroll's answer.
[Insert Table 1 here]
CONCLUSION
Our didactic lesson from this class of geometric-probability problems is for increased awareness and caution in examining our assumptions. Ambiguous and unstated assumptions concerning the conditions of the problem, on which the solvers sometimes rely unawares, have often led to difficulties and conflicting conclusions in probabilistic reasoning (Nickerson, 1996). In the case of Carroll's pillow-problem No. 58, the assumptions were not even implicit or unstated; they were openly posited at the beginning. The author, and apparently many readers as well, just did not stop to check these assumptions' tenability.
References
Carroll, L. (1958). Pillow Problems and a Tangled Tale. New York: Dover (Pillow Problems - originally published in 1895).
Falk, R., and Konold, C. (1992). The psychology of learning probability. In F.S. Gordon and S.P. Gordon (Eds.), Statistics for the Twenty-First Century (pp. 151-164). Washington, DC: The Mathematical Association of America.
Nickerson, R.S. (1996). Ambiguities and unstated assumptions in probabilistic reasoning. Psychological Bulletin, 120, 410-433.
Szekely, G. J. (1986). Paradoxes in Probability Theory and Mathematical Statistics. Dordrecht: Reidel.
Wells, D. (1992). The Penguin Book of Curious and Interesting Puzzles. Harmondsworth, Middlesex, England: Penguin Books.

Acknowledgements
This study was partially supported by the Struman Center for Human Development, Hebrew University, Jerusalem. We thank Israel Einot for carrying out the simulations and Raphael Falk for good advice concerning this problem.
After completing our simulations, Dr. Uwe Saint-Mont gracefully called our attention to the existence of analytic derivations of the probabilities in question. The case of randomly sampling points within a rectangle of varying ratios between length and width was analysed in: Langford, E. (1969). The probability that a random triangle is obtuse. Biometrika, 56, 689-690; and the case of sampling within a circle, in: Hall, G. R. (1982). Acute triangles in the n-ball. Journal of Applied Probability, 19, 712-715. The numerical results obtained from these analyses are extremely close to our simulation outcomes (accurate to the third decimal place, they are all identical). Though the mathematical formulas seem to obviate the need for simulations, the value of the latter lies in demonstrating the feasibility of the statistical experiment of uniformly sampling points within a bounded region, in contrast to the impossibility of Carroll's procedure.

Table 1: Estimated Probability of Obtaining an Obtuse Triangle, by Shape of Sample Space.

 Shape Estimated probability Standard error Circle .7201 .00045 Square .7249 .00045 Equilateral triangle .7484 .00043 Rectangle 1:k k=2 .7987 .00040 k=5 .9324 .00025 k=8 .9660 .00018 k=11 .9795 .00014 k=14 .9861 .00012 k=17 .9899 .00010 k=20 .9924 .00009

Ruma Falk and Ester Samuel-Cahn
The Hebrew University of Jerusalem, Israel
Center for Rationality and Interactive Decision Theory
e-mail: rfalk@cc.huji.ac.il

January 2001