Paradox

248 A BRIEF HISTORY OF THE PARADOXFig. 17.3of an equilateral triangle. The small circle has an area ¼ ofthe big circle. Therefore, the answer is ¼.Third solution: Consider a line bisecting the triangle andcircle, as in figure 17.3. The chords longer than the sides oftriangles have their midpoints closer to the center than halfthe radius, i.e., below H and above I. If the midpoints aredistributed uniformly over the radius (instead of over thearea, as was the case in the second solution), the probabilitybecomes ½.Bertrand has presented us with an embarrassment ofriches. Each answer is acceptable on its own, both in itsreasoning process and its conclusion. The paradox lies in theincompatibility between the deductions rather than withinthe deductions themselves. The deductions are individuallyplausible but jointly inconsistent.The paradoxes of geometrical probability are trouble fortheorists who classify each paradox in terms of an argumentwith an unacceptable conclusion. Bertrand’s three-armedantinomy has three individually acceptable conclusions.Those who identify paradoxes with surprising conclusionsmight reply as follows: Bertrand’s paradox uses theseparate calculations as the basis for separate subconclusions,which are then recruited into a superargument with three