Math--the application of reasonable logic to reasonable assumptions--usually produces reasonable results. But sometimes math generates astonishing paradoxes--conclusions that seem completely unreasonable or just plain impossible but that are nevertheless demonstrably true. Did you know that a losing sports team can become a winning one by adding worse players than its opponents? Or that the thirteenth of the month is more likely to be a Friday than any other day? Or that cones can roll unaided uphill? In *Nonplussed!*--a delightfully eclectic collection of paradoxes from many different areas of math--popular-math writer Julian Havil reveals the math that shows the truth of these and many other unbelievable ideas.

in formulae for Dn , and perhaps the nicest example of its type is Dn = n! +m e and so pn = 1 n! +m , n! e where m is any number such that 1 3 m 1 2. DERANGEMENTS 59 (Here, the · is the ﬂoor function deﬁned by x = the greatest integer less than or equal to x.) To see this, write 1 1 1 1 1 = 1− + − + · · · + (−1)n e 1! 2! 3! n! + (−1)n+1 1 1 + (−1)n+2 + ··· , (n + 1)! (n + 2)! which means that Dn = n! 1 − = n! 1 1 1 1 + − + · · · + (−1)n 1! 2! 3! n! 1 1 1 − (−1)n+1 + (−1)n+2 + ··· e

Table 1.3. Outcome of an all-plays-all tournament between the various teams. TB 10 2 TW 3 7 5 B B B 1 B 8 B W W W W 9 B W W W W 6 B W W B W 4 B W W B B B Logical calm is restored if we look at the expected number of wins. With FTF it is EF = 0 × (1 − f )(1 − t)(1 − f ) + 1 × {f (1 − t)(1 − f ) + (1 − f )t(1 − f ) + (1 − f )(1 − t)f } + 2 × {f t(1 − f ) + f (1 − t)f + (1 − f )tf } + 3 × f tf = 2f + t and a similar calculation for TFT yields ET = 2t + f . Since f > t, 2f − f >

us to our problem: A smuggler, travelling as fast as possible in a straight line, is being pursued and caught up by a coastguard when a fogbank engulfs them and each becomes invisible to the other. The smuggler’s boat is too small for electronic detection or to leave an appreciable wake to follow, yet in spite of the coastguard not knowing where the smuggler is or in which direction he is then travelling, he can steer a course that guarantees capture of the smuggler. One of the crucial words

calculations are Γ ( 12 n + 1) 12 ln π e(n/2) ln π − e(n/2) ln π 12 Γ ( 12 n + 1) dVn (1) = dn [Γ ( 12 n + 1)]2 = Γ ( 12 n + 1) 12 ln π π n/2 − π n/2 12 Γ ( 12 n + 1) [Γ ( 12 n + 1)]2 . The requirement that dVn (1)/dn = 0 means that 1 1 1 1 Γ ( 2 n + 1) 2 ln π π n/2 − π n/2 2 Γ ( 2 n + 1) = 0 and so ln π − Γ ( 12 n + 1) Γ ( 12 n + 1) =0 and 1 Ψ ( 2 n + 1) = ln π . HYPERDIMENSIONS 141 3 (x) 2 ln π 1 5 10 15 20 x Figure 12.7. The Digamma function. Figure 12.7 shows a plot of

SG = P P1 + (r − x tan β tan γ). From ﬁgures 2.4 and 2.6 we have P P1 = SG1 = a + x tan α. Therefore, y = (a + x tan α) + (r − x tan β tan γ) = a + r + x(tan α − tan β tan γ). THE UPHILL ROLLER 19 The path of the centre of mass of the cone is, then, the straight line y = a + r + x(tan α − tan β tan γ), which has gradient tan α − tan β tan γ and for the motion to be possible this gradient has to be negative, which means that the deﬁning condition for the paradox to exist is that tan α < tan β