8. PROBABILITY RECREATIONS
Most of the recreations in probability are connected with some paradoxical feature. A good exposition of most of these appears in the following.
Gábor J. Székely. Paradoxes in Probability Theory and Mathematical Statistics. Akadémiai Kiadó, Budapest and Reidel, Dordrecht, 1986. [Revised translation of: Paradoxonok a Véletlen atematikában; M_szaki Könvkiadó, Budapest, nd.] Translated by Márta Alpár and Éva Unger. ??NYR.
8.A. BUFFON'S NEEDLE PROBLEM
R. E. Miles & J. Serra. En Matiere d'introduction. In: Geometrical Probability and Biological Structures: Buffon's 200th Anniversary. Lecture Notes in Biomathematics, No. 23, Springer, 1978, pp. 3‑28. Historical survey, reproduces main texts.
Buffon. (Brief commentary). Histoire de l'Acad. des Sci. Paris (1733 (1735)) 43‑45. Discusses problem of a disc meeting a square lattice and then the stick (baguette) problem, but doesn't give the answer.
Buffon. Essai d'arithmétique morale, section 23. 1777. (Contained in the fourth volume of the supplement to his Histoire Naturelle, pp. 101‑104??) = Oeuvres Complètes de Buffon; annotated by M. Flourens; Garnier Frères, Paris, nd [c1820?], pp. 180‑185. (Also in Miles & Serra, pp. 10‑11.)
Laplace. Théorie Analytique des Probabilitiés. 1812. Pp. 359‑360. ??NYS. (In Miles & Serra, p. 12.)
M. E. Barbier. Note sur le problème de l'aiguille et le jeu du joint couvert. J. Math. pures appl. (2) 5 (1860) 273‑286. Gives result for arbitrary curves and considers several grids. Also gives his theorem on curves of constant width.
M. W. Crofton. On the theory of local probability. Philos. Trans. Roy. Soc. 158 (1869) 181‑199. (Excerpted in Miles & Serra, pp. 13‑15.)
A. Hall. On an experimental determination of π. Messenger of Mathematics 2 (1873) 113‑114. ??NYS.
Tissandier. Récréations Scientifiques. 1880? 2nd ed., 1881, 139-145 describes the result and says 10,000 tries with a 50 mm needle on a floor with spacing 63.6 mm, produced 5009 successes giving π = 3.1421.
= 5th ed., 1888, pp. 204-208. c= Popular Scientific Recreations, 1890? pp. 729‑731, but the needle is 2 in on a floor of spacing 2½ in and 5009 is misprinted as £5000 (sic) but 5009 is used in the calculation.
J. J. Sylvester. On a funicular solution of Buffon's "Problem of the needle" in its most general form. Acta Math. 14 (1890‑1891) 185‑205.
N. T. Gridgeman. Geometric probability and the number π. SM 25 (1959) 183‑195. Debunks experimental results which are often too good to be true, although they are frequently cited.
J. G. L. Pinhey. The Comte de Buffon's paper clip. MG 54 (No. 389) (Oct 1970) 288. Being caught without needles, he used paperclips. He derives the probability of intersection assuming a paper clip is a rectangle with semi‑circular ends.
Jack M. Robertson & Andrew F. Siegel. Designing Buffon's needle for a given crossing distribution. AMM 93 (1986) 116‑119. Discusses various extensions of the problem.
8.B. BIRTHDAY PROBLEM
How many people are required before there is an even chance that some two have the same birthday?
George Tyson was a retired mathematics teacher when he enrolled in the MSc course in mathematical education at South Bank in about 1980 and I taught him. He once remarked that he had known Davenport and Mordell, so I asked him about these people and mentioned the attribution of the Birthday Problem to Davenport. He told me that he had been shown it by Davenport. I later asked him to write this down.
George Tyson. Letter of 27 Sep 1983 to me. "This was communicated to me personally by Davenport about 1927, when he was an undergraduate at Manchester. He did not claim originality, but I assumed it. Knowing the man, I should think otherwise he would have mentioned his source, .... Almost certainly he communicated it to Coxeter, with whom he became friendly a few years later, in the same way." He then says the result is in Davenport's The Higher Arithmetic of 1952. When I talked with Tyson about this, he said Davenport seemed pleased with the result, in such a way that Tyson felt sure it was Davenport's own idea. However, I could not find it in The Higher Arithmetic and asked Tyson about this, but I have no record of his response.
Anne Davenport. Letter of 23 Feb 1984 to me in response to my writing her about Tyson's report. "I once asked my husband about this. The impression that both my son and I had was that my husband did not claim to have been the 'discoverer' of it because he could not believe that it had not been stated earlier. But that he had never seen it formulated."
I have discussed this with Coxeter (who edited the 1939 edition of Ball in which the problem was first published) and C. A. Rogers (who was a student of Davenport's and wrote his obituary for the Royal Society), and neither of them believe that Davenport invented the problem. I don't seem to have any correspondence with Coxeter or Rogers with their opinions and I think I had them verbally.
Richard von Mises. Ueber Aufteilungs‑ und Besetzungs‑ Wahrscheinlichkeiten. Rev. Fac. Sci. Univ. Istanbul (NS) 4 (1938‑39) 145‑163. = Selected Papers of Richard von Mises; Amer. Math. Soc., 1964, vol. 2, pp. 313‑334. Says the question arose when a group of 60 persons found three had the same birthday. He obtains expected number of repetitions as a function of the number of people have that birthday. He finds the expected number of pairs with the same birthday is about 1 when the group has 29 people, while the expected number of triples with the same birthday is about 1 when there are 103 people. He doesn't solve the usual problem, contrary to Feller's 1957 citation of this paper.
Ball. MRE, 11th ed., 1939, p. 45. Says problem is due to H. Davenport. Says "more than 23" and this is repeated in the 12th and 13th editions.
P. R. Halmos, proposer; Z. I. Mosesson, solver. Problem 4177 _ Probability of two coincident birthdays. AMM 52 (1945) 522 & 54 (1947) 170. Several solvers cite MRE, 11th ed. Solution is 23 or more.
George Gamow. One, Two, Three ... Infinity. Viking, NY, 1947. = Mentor, NY, 1953, pp. 204‑206. Says 24 or more.
Oswald Jacoby. How to Figure the Odds. Doubleday, NY, 1947. The birthday proposition, pp. 108-109. Gives answer of 23 or more.
William Feller. An Introduction to Probability Theory and Its Applications: Vol 1. Wiley, 1950, pp. 29-30. Uses an approximation to obtain 23 or more. The 2nd ed., 1957, pp. 31‑32 erroneously cites von Mises, above.
J. E. Littlewood. A Mathematician's Miscellany. Op. cit. in 5.C. 1953. P. 18 (38) mentions the problem and says 23 gives about even odds.
William R. Ransom. Op. cit. in 6.M. 1955. Birthday probabilities, pp. 38‑42. Studies usual problem and graphs probability of coincidence as a function of the number of people, but he doesn't compute the break‑even point. He then considers the probability of two consecutive birthdays and gets upper and lower estimates for this.
Gamow & Stern. 1958. Birthdays. Pp. 48‑49. Says the break‑even point is "about twenty‑four".
C. F. Pinzka. Remarks on some problems in the American Mathematical Monthly. AMM 67 (1960) 830. Considers number of people required to give greater than 50% chance of having 3, 4 or 5 with the same birthday. He gets 88, 187, 314 respectively, using a Poisson approximation. He gives the explicit formula for having 3 with the same birthday.
Charlie Rice. Challenge! Op. cit. in 5.C. 1968. Probable probabilities, pp. 32-36, gives a variety of other forms of the problem.
A chooses five letters of the alphabet; B tries to guess at least one of them in five guesses. Author says odds are two to one in favor of B, though I get four to one. This is the same as getting five distinct items from a set of 21.
Get people to think of cards. For 9 or more people, the probability of two the same is _ .52; for 11, _ .68; for 12, _ .75.
Get people to count their change. For a moderate number of people, it is likely that two have the same amount (or the same number of coins). Likewise, with a moderate number of people, two are likely to have fathers (or mothers) with the same given name.
He gives magic numbers, i.e. the size of the set to be selected from, for different sizes of group in order that a duplication is more likely than not. For 6 people, the magic number is 23; for 8, 43; for 12, 99.
Howard P. Dinesman. Superior Mathematical Puzzles. Op. cit. in 5.B.1. 1968. No. 58: The birthday puzzle, pp. 76, 116-117 & 122. Asks for probability of a shared birthday among 30 people. Says the problem was introduced by Gamow. Answer is .70. He adds that 24 or more will give better than even odds and then asks how many people are necessary for one of them to have his birthday on a given day _ e.g. today. Here the answer is 253.
E. J. Faulkner. A new look at the probability of a coincidence of birthdays in a group. MG 53 (No. 386) (Dec 1969) 407‑409. Suggests the probability should be obtained by the ratio of the unordered selections, i.e. Prob (r distinct birthdays) = BC(365, r)/BC(364+r, r), rather than P(365, r)/365r. But the unordered selections with repetitions are not equally likely events _ see Clarke & Langford, below. For his approach, the breakeven number is r = 17.
Morton Abramson & W. O. J. Moser. More birthday surprises. AMM 77 (1970) 856‑858. If a year has n days, k ³ 1 and p people are chosen at random, what is the probability that every two people have birthdays at least k days apart? For n = 365, k = 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, the breakeven numbers are p = 23, 14, 11, 9, 8, 8, 7, 7, 6, 6.
L. E. Clarke & Eric S. Langford. Note 3298 _ I & II: On Note 3244. MG 55 (No. 391) (Feb 1971) 70‑72. Note the non‑equally likely events in Faulkner.
W. O. J. Moser. It's not a coincidence, but it is a surprise. CM 10 (1984) 210‑213. Determines probability P that in a group of k people, at least two have birthdays at most w days apart. This turns out to have a fairly simple expression. To get P > .5 with w = 0 requires k ³ 23, the classical case. With w = 1, it requires k ³ 14, ....
Tony Crilly & Shekhar Nandy. The birthday problem for boys and girls. MG 71 (No. 455) (Mar 1987) 19‑22. In a group of 16 boys and 16 girls, there is a probability greater than ½ of a boy and a girl having the same birthday and 16 is the minimal number.
Roger S. Pinkham. Note 72.25: A convenient solution to the birthday problem for girls and boys. MG 72 (No. 460) (Jun 1988) 129‑130. Uses an estimate to obtain the value 16 of Crilly and Nandy.
M. Lawrence Clevenson & William Watkins. Majorization and the birthday inequality. MM 64:3 (1991) 183-188. Do the numbers necessary for P > .5 get bigger if birthdays are not random? Answer is "no" and it is a result in majorization theory, but they give an elementary treatment.
8.C. PROBABILITY THAT A TRIANGLE IS ACUTE
See also 6.BR, esp. the Mathematical Log article and my comments. I have a large number of similar results, mostly by myself, in a file. I estimate there are about 20 possible answers, ranging from 0 to 1.
J. J. Sylvester. On a special class of questions in the theory of probabilities. Birmingham British Association Report (1865) 8. = The Collected Mathematical papers of James Joseph Sylvester, (CUP, 1908); Reprinted by Chelsea, 1973, item 75, pp. 480-481. ??NYS, but Guy (below) reports that it is a discursive article with no results. Attributes problem for three points within a circle or sphere to Woolhouse but feels the problem is not determinate.
C. Jordan. 1872‑1873. See entry in 8.G.
L. Carroll. Pillow Problems. 1893. ??NYS. 4th ed., (1895). = Dover, 1958. Problem 58, pp. 14, 25, 83‑84. Prob (acute) = .639.
C. O. Tuckey. Note 1408: Why do teachers always draw acute‑angled triangles? MG 23 (No. 256) (1939) 391‑392. He gets Prob(obtuse) varying between .57 and .75.
E. H. Neville. Letter: Obtuse angling _ a catch. MG 23 (No. 257) (1939) 462. In response to Tuckey, he shows Prob(obtuse) = 0 and deduces that Prob(acute) = 0 (!!!).
Nikolay Vasilyev. The symmetry of chance. Quantum 3:5 (May/Jun 1993) 22-27 & 60-61. Survey on geometric probability. Asks for the probability of an acute triangle when one takes three points at random on a circle and gets ¼.
Richard K. Guy. There are three times as many obtuse-angled triangles as there are acute-angled ones. MM 66:3 (Jun 1993) 175-179. Gives 12 different approaches, five of which yield Prob(acute) = ¼, with other values ranging from 0 to .361. He tracked down the Sylvester reference _ see above.
In Mar 1996, I realised that the two approaches sketched in 6.BR give probabilities of acuteness as 0 and 2 - π/2 = .429.
8.D. ATTEMPTS TO MODIFY BOY‑GIRL RATIO
This is attempted, e.g. by requiring families to stop having children after a girl is born.
Pierre Simon, Marquis de Laplace. Essai Philosophique sur les Probablitiés (A Philosophical Essay on Probabilities). c1819, ??NYS. Translated from the 6th French ed. by F. W. Truscott & F. L. Emory, Dover, 1951. Chap. XVI, pp. 160‑175, especially pp. 167‑169. Discusses whether the excess of boys over girls at birth is due to parents stopping having children once a son is born.
Gamow & Stern. 1958. A family problem. Pp. 17‑19.
8.E. ST. PETERSBURG PARADOX
Nicholas Bernoulli. Extrait d'une Lettre de M. N. Bernoulli à M. de M... [Montmort] du 9 Septembre 1713. IN: Pierre Rémond de Montmort; Essai d'analyse sur les jeux de hazards; 2nd ed., (Quillau, Paris, 1713 (reprinted by Chelsea, NY, 1980)); Jombert & Quillau, 1714 [The 1714 may differ from the 1713 ??], pp. 401-402. Cinquième Probléme, p. 402. [See note to L. Euler; Vera aestimatio sortis in ludis; Opera Omnia (1) 7 (1923) 458‑465, esp. pp. 459‑461.] In earlier problems, he mentions pay-offs of 1, 2, 3, 4, .... Here he just asks what happens if the payoffs are 1, 2, 4, 8, ... or 1, 3, 9, 27, ... or 1, 4, 9, 16, ... or 1, 8, 27, 64, ... etc.
Daniel Bernoulli. Specimen theoriae novae de mensura sortis. Comm. Acad. Sci. Imp. Petropol. 5 (1730-31(1738)) 175-192, ??NYS. IN: Die Werke von Daniel Bernoulli; ed. by L. P. Bouckaert & B. L. van der Waerden; Birkhäuser, 1982; pp. 223-234 and notes by van der Waerden, pp. 195 & 197-200. English translation in Econometrica 22 (1954) 23-36, ??NYS.
Leonhard Euler. Vera aestimatio sortis in ludis. [Op. postuma 1 (1862) 315-318.] = Op. Omnia (I) 7 (1923) 458-465.
Tissandier. Récréations Scientifiques. 1880? 2nd ed., 1881, p. 140 gives a brief unlabelled description, saying this "problème de Pétersbourg" was discussed by Daniel Bernoulli in "Mémoires de l'Académie de Russie". Not in the 5th ed. of 1888.
Tissandier. Popular Scientific Recreations. 1890? Pp. 727-729 discusses the idea and says D. Bernoulli presented his material on this in "Memoires [sic] de l'Académie de Russie". This is somewhat longer than the material in the 2nd French ed of 1881.
Anon. [presumably the editor, Richard A. Proctor]. Strange chances. Knowledge 10 (Oct 1887) 276-278. Brief discussion of "the famous Petersburg problem".
C. S. Jackson. Note 438: The St. Petersburg problem. MG 8 (No. 116) (Mar 1915) 48. Notes that the value of the game is not more than log2 of the bank's funds.
Dan Pedoe, The Gentle Art of Mathematics, op. cit. in 5.C, p. 55, says D. Bernoulli published it in the Transactions (??) of of the St. Petersburg Academy. ??NYS.
Pedoe, ibid, p. 57, also says that Buffon tested this with 2048 games and he won 10,057 in them.
Jacques Dutka. On the St. Petersburg paradox. Archive for the History of the Exact Sciences 39 (1988) 13-39. ??NYR.
Nick Mackinnon & 5Ma. Note 74.9: A lesson on the St. Petersburg paradox. MG 74 (No. 467) (1990) 51‑53. Suppose a maximum is put on the payment _ i.e. the game stops if n heads appear in a row. How does this affect the expected value? They find n = 10 gives expected value of 6.
8.F. PROBLEM OF POINTS
A game consists of n points. How do you divide the stake if you must quit when the score is a to b to ....? This problem was resolved by Fermat and Pascal in response to a question of the Chevalier de Méré and is generally considered the beginning of probability theory. The history of this topic is thoroughly covered in the first works below, so I will only record early or unusual occurrences.
NOTATION: Denote this by (n; a, b, ...).
GENERAL HISTORIES
Florence Nightingale David. Games, Gods and Gambling. The origins and history of probability and statistical ideas from the earliest times to the Newtonian era. Griffin, London, 1962.
Anthony W. F. Edwards. Pascal's Arithmetical Triangle. Griffin & OUP, London, 1987. This corrects a number of details in David.
David E. Kullman. The "Problem of points" and the evolution of probability. Handout from talk given at MAA meeting, San Francisco, Jan 1991. Uses (6; 5, 3) as a common example and outlines approaches and solutions of: Pacioli (1494) _ 5 : 3; Cardan (1539) _ 6 : 1; Tartaglia (1539) _ 2 : 1; Pascal (1654) _ 7 : 1; Fermat (1654) _ 7 : 1; Huyghens (1657) _ like Pascal. He also describes the approaches of Pascal, Fermat and Huyghens to the three person case and generalizations due to Montmort (1714) and de Moivre (1718).
Pacioli. Summa. 1494.
Ff. 197r-197, prob. 50 (?-not printed). (6; 5, 2). Discusses it and says there are diverse opinions. His discussion is confusing, but he divides as 5 : 2 (should be 15 : 1). Also says losing cardplayers at (5; 4, 3) offer to divide in the ratio 3 : 2 (should be 3 : 1). His discussion is again confusing, but may be saying that one more game makes either a win or an even situation, and averaging these gives the right division as 3 : 1
F. 197v, prob. 51. (6; 4, 3, 2). Divides as 4 : 3 : 2 (I get 451 : 195 : 83).
Oystein Ore. Pascal and the invention of probability theory. AMM 67 (1960) 409‑419. Ore says Pacioli is the first printed version of the problem. He translates parts of the texts. Ore says he has seen the problem in Italian MSS as early as 1380, but he doesn't give details (???). He opines that the problem is of Arabic origin. He discusses Cardan and Tartaglia and gives some examples from Forestiani, 1603 _ (8; 5, 3), (14; 10, 8, 5) (??NYS). But there is no proper mathematics until Pascal & Fermat.
Calandri, Raccolta. c1500.
Prob. 12, pp. 13‑14. (6; 4, 3). Divides in ratio 3 : 2, but says this may not be the exact truth. (Answer should be 11 : 5.)
Prob. 43, pp. 39‑40. (3: 2, 1, 0). He says there are two ways to do this, based on the numbers of points won or needed to win. He then says 3/7 of the game has been played and distributes 3/7 of the stake in the proportion 2 : 1 : 0 and then distributes the remaining 4/7 equally, giving a final distribution in the proportion 10 : 7 : 4. (Should be 19 : 6 : 2.)
Cardan. Practica Arithmetice. 1539.
Chap. 61, section 13, f. T.iii.r (p. 113). (10; 9, 7). He divides as 1 + 2 + 3 : 1. (10; 3, 6) _ divided as 1+ 2 + 3 + 4 : 1 + ... + 7. Section 14, f. T.iii.v (p.113) may be discussing this problem.
Chap. 68 (ultimo), section 5, ff. QQ.iv.r - QQ.viii.r (p. 214). This chapter is on errors of Pacioli. Mentions (6; 5, 2) and many other examples.
Tartaglia. General Trattato, 1556, art. 206, pp. 265r‑265v. (6; 5, 3). Criticises Luca del Borgo (= Pacioli) and gives another method.
Giovanni Francesco Peverone. Due brevi e facili trattati, il primo d'Arithmetica, l'altro di Geometria ..., Gio. Tournes, Lyons, 1558. Reissued as: Arithmetica e geometria del Sig. Gio. Francesco Peverone di Cuneo, ibid., 1581. ??NYS _ described in: Livia Giacardi & Clara Silvia Roero; Bibliotheca Matematica Documenti per la Storia della Matematica nelle Biblioteche Torinese; Umberto Allemandi & C., Torino, 1987, pp. 117-118. They say he includes correct solutions of some problems of games of chance, in particular the 'divisione della posta', i.e. the problem of points.
Ozanam. 1694. Prob. 10, 1696: 41-52, esp. 45-50; 1708: 37-48, esp. 42‑45. Prob. 13, 1725: 123‑130. Prob. 3, 1778: 117-121; 1803: 116-120; 1814: 102-106; 1840: 54-55. Discusses the problem in general and specifically (3; 1, 0), but 1778 et seq. changes to (3; 2, 1) and adds reference to Pascal and Fermat.
Pierre Rémond de Montmort. Essai d'analyse sur les jeux de hazards. (1708??); 2nd ed., (Quillau, Paris, 1713 (reprinted by Chelsea, NY, 1980)); Jombert & Quillau, 1714. [The 1714 may differ from the 1713 ??] Avertissement (to the 1st ed.), pp. xxi-xxiv and (to the 2nd ed.) xxv-xxxvii discusses the history of the problem, the work of Fermat and Pascal and de Moivre's assertion that Huygens had solved it first.
Pearson. 1907. Part II, no. 98, pp. 134 & 198. (3; 2, 1).
8.G. PROBABILITY THAT THREE LENGTHS FORM A TRIANGLE
See also 8.C.
E. Lemoine. Sur une question de probabilités. Bull. Soc. Math. France 1 (1872‑1873) 39‑40. Obtains ¼ by considering that the stick can be broken at m equidistant points and then letting m increase.
? Halphen. Sur un problème de probabilités. Ibid, pp. 221‑224. Extends Lemoine to n pieces, getting 1 ‑ n/2n‑1, by an argument similar to homogeneous coordinates and by integration.
Camille Jordan. Questions de probabilités. Ibid, pp. 256‑258 & 281‑282. Generalizes to find the probability that n of the m parts, into which a line is broken, have length > a. He finds the probability that four points on a sphere form a convex spherical quadrilateral. Pp. 281‑282 corrects this last result.
E. Fourrey. Curiositiés Géometriques, op. cit. in 6.S.1. 1907. Part 3, chap. 1, section 5: Application au calcul des probabilitiés, pp. 360‑362. Break a stick into three pieces. Gets P = ¼. Cites Lemoine.
G. A. Bull. Note 2016: A broken stick. MG 32 (No. 299) (May 1948) 87‑88. Gets Halphen's result by using homogeneous coordinates.
S. Rushton. Note 2083: A broken stick. MG 33 (No. 306) (Dec 1949) 286‑288. Repeats Bull's arguments and then considers making one break, then breaking the larger piece, etc. (It's not clear if he takes the longer of the two new pieces when trying for a 4‑gon.) Says that the only example of this that he has seen is Whitworth's DCC Exercises in Choice and Chance, 1897, exer. 677, ??NYR, which has Prob(triangle) = _. Author says this is wrong and should be 2 log 2 ‑ 1 = .386... He gets a solution for an n‑gon.
D. N. Smith. Letter: Random triangles. MiS 19:2 (Mar 1990) 51. Suggests, as a school project, generating random triangles by rolling three dice and using the values as sides.
Joe Whittaker. Random triangles. AMM 97:3 (Mar 1990) 228-230. Take a stick and break it at two random points _ or _ break once at random and then break the longer part at random. Prob(triangle) = ¼ in the first case and appears to be _ in the second case, but the second analysis assumes an incorrect distribution. Correcting this leads to Prob(triangle) = 2 log 2 - 1 = .38..., as in Rushton.
8.H. PROBABILITY PARADOXES
8.H.1. BERTRAND'S BOX PARADOX
J. Bertrand. Calcul des Probabilités. Gauthier‑Villars, Paris, 1889. Chap. I, art. 2, pp. 2‑3.
Howard P. Dinesman. Superior Mathematical Puzzles. Op. cit. in 5.B.1. 1968. No. 26: Mexican jumping beans, pp. 40-41 & 96. Derranged matchboxes of red and black beans _ see 5.K.1. The problem continues by unlabelling the boxes _ if you draw a red bean, what is the probability that the other bean in the box is red?
Nicholas Falletta. The Paradoxicon. Doubleday, NY, 1983; Turnstone Press, Wellingborough, 1985. Probability paradoxes, pp. 116‑125, esp. pp. 118‑121, which describes: a three‑card version due to Warren Weaver (1950), the surprise ace paradox of J. H. C. Whitehead (1938) and a three prisoner paradox.
Ed Barbeau. The problem of the car and goats. CMJ 24:2 (Mar 1993) 149-154. A version of the problem involving three doors with a car behind one of them appeared in Marilyn vos Savant's column in Parade magazine and generated an immense amount of correspondence and articles. This article describes 4 (or more?) equivalents and gives 63 references, including Bertrand, but not Dinesman, Falletta, Weaver or Whitehead.
8.H.2. BERTRAND'S CHORD PARADOX
J. Bertrand. Op. cit. in 8.H.1. 1889. Chap. I, art 5, pp. 4‑5. Gets answers ¼, _ and ½.
F. Garwood & E. M. Holroyd. The distance of a "random chord" of a circle from the centre. MG 50 (No. 373) (Oct 1966) 283‑286. Take two random points and the chord through them. This gives an expected distance from the centre of .2532.
8.I. TAKING THE NEXT TRAIN
This is the problem where a man can board equally frequent trains going either way and takes the next one to appear, but finds himself going one way more often than the other. Why?
New section.
W. T. Williams & G. H. Savage. The Penguin Problems Book. Penguin, 1940. No. 42: Bus times, pp. 28 & 116. Two bus lines running the same route equally often, but it is twice as likely that the next bus is Red rather than Green.
Jerome S. Meyer. Fun-to-do. Op. cit. in 5.C. 1948. Prob. 14: The absent-minded professor, pp. 25 & 184.
Birtwistle. Math. Puzzles & Perplexities. 1971. Fifth, pp. 146 & 197.
8.J. CLOCK PATIENCE OR SOLITAIRE
The patience or solitaire game of Clock has 13 piles of four face-down cards arranged with 12 in a circle and the 13th pile in the centre. You turn up a card from the top of the 13th pile _ if it has value n, you place it under the n-th pile and turn up the card from the top of that pile and repeat the process. You win if you turn over all the cards. The probability of winning is precisely 1/13 since the process generates an arbitrary permutation of the 52 cards and is a win if and only if the last card is a K (i.e. a 13). In response to a recent question, I looked at my notes and found there are several papers on this.
John Reade, proposer; editorial solution. Problem 46.5 _ Clock patience. M500 46 (1977) 17 & 48 (1977) 16. What is the probability of winning? Solution says several people got 1/13 and the problem is actually easy.
Anon. proposal & solution, with note by David Singmaster. Problem 11.3. MS 11 (1978/79) 28 & 101. (a) What is the probability of winning? Solution as in my comment above. (b) If the bottom cards are a permutation of A, 2, ..., K, then the game comes out if and only if this is a cyclic permutation. Singmaster notes that the probability of a permutation of 13 cards being cyclic is 1/13, so the probability of winning in this situation is again 1/13.
Eric Mendelsohn & Stephen Tanny, proposers; David Kleiner, solver. Problem 1066 _ The last 1. MM 52 (1979) 113 & 53 (1980) 184-185. Generalizes to k copies of L ranks. Asks for probability of winning and for a characterization of winning distributions. Solution is not very specific about the distributions, just saying there is a correspondence between the cards turned up and the original piles.
T. A. Jenkyns & E. R. Muller. A probabilistic analysis of clock solitaire. MM 54 (1981) 202‑208. Says the expected number of cards played is 42.4. They consider continuing the game after the last K by restarting with the first available unturned card. They call this the 'second play' and allow you to continue to third play, etc. They determine the expected numbers of cards turned in each play and also generalize to m cards of n ranks. They show that the number of plays is determined by the relative positions of the last cards of each rank and show that the probability that the game takes p plays, but fail to note that this is │s(n,p)│/n!, where s(n,p) is the Stirling number of the first kind, so they rederive a number of properties of these numbers.
Michael W. Ecker. How to win (or cheat) in the solitaire game of "Clock". MM 55 (1982) 42-43. Shows that whether you can win is determined by the bottom cards of each pile. Though these values are not a permutation, one can define 'f-cycles' and a distribution comes out if and only if every f-cycle contains the initial value.
8.K. SUCKER BETS
This covers situations where the punter can not easily tell if the bet is reasonable or not. These are often used to lure suckers, but they have also been historically important as incentives to develop probabilistic methods. As an example, the Chevalier de Méré knew that the probability of throwing one six in four throws of a die was better than even, but he thought this should make throwing a double six in 24 throws also better than even and it is not. See the histories cited in 8.F for more on the early examples of these problems. Of course lotteries come into this category. See 8.L for some other forms.
The classic carnival game of Chuck-a-Luck is an excellent example of this category. This is the game with three die. You bet on a number _ if it comes up once, you win double your bet (i.e. your bet and the same again); if it comes up twice, you win triple; if it comes up thrice, you win quadruple. The relative frequency of 0, 1, 2, 3 of your number is 125, 75, 15, 1, so the return in 216 throws will be ‑125 + 75 + 30 + 3 = ‑17, giving a 7.9% profit to the operator.
Collins. Fun with Figures. 1928. How figures can cheat _ The tin-horn gambler, pp. 34-35. Six dice, each marked on just one face. Gambler bets $100 to $1 that the punter will not get the six marked faces all up in 20 throws. Collins asserts that the probability of winning is 20/66 = 1/2332.3, so the fair odds should be $2332 to $1. This is not quite right. The true probability is 1 - (1 - 1/66)20 = .00042858 = 1/2333.27. The fair odds depend on whether the winnings include the punter's $1 or not _ in this case, they do not, so the odds according to Collins ought to be $2331.3 to $1, or more exactly $2332.27 to $1.
In the 1980s, I saw stands at school and village fairs, where one gets one throw with six dice for £.50. If one gets six 6s, one wins a new Rover. Since 66 = 46656, the promoter gets £23,328 for each car, which was a tidy profit.
Gardner. Nontransitive dice and other probability paradoxes. SA (Dec 1970). Extended in Wheels, chap. 5. Consider two red and two black cards. Choosing two, what are the odds of getting two of the same colour? Typical naive arguments get _ or ½, but the true answer is _. In the Addendum in Wheels, S. D. Turner describes the version with R red cards and B black cards. The probability of choosing two of the same colour is then [R(R-1) + B(B-1)]/[(R+B)(R+B-1)] which is always less than ½.
8.L. NONTRANSITIVE GAMES
In the simplest form, we know that A can defeat B and B can defeat C, but this does not imply that A can defeat C, so non-transitivity is quite common in real game playing. Indeed, in the classical game of Rock, Scissors, Paper, the game situation has C defeating A. Similar phenomena occur in preferences, particularly voting and loving. This section will deal with mathematical versions, particularly where the game seems fair, but making a later choice than your opponent gives you an edge. Hence these versions can be used as the basis of sucker bets.
Walter Penney, proposer and solver. Problem 95 _ Penney-ante. JRM 2:4 (Oct 1969) 241 & 7:4 (Fall 1974) 321. Opponent picks a triple of heads and tails, then you pick a triple. A coin is thrown until the triple occurs. If he chooses HHH and you choose HTH, show your probability of winning is 3/5.
Walter Penney and David L. Silverman, proposers and solver. Problem 96 _ Penney-ante. JRM 2:4 (Oct 1969) 241 & 8:1 (1975) 62-65. As above, but the opponent and you both pick a triple without the other's knowledge. Elaborate analysis is required to obtain the optimal mixed strategy which guarantees you a probability of ½.
Gardner. Nontransitive dice and other probability paradoxes. SA (Dec 1970). Extended in Wheels, chap. 5. Describes Bradley Efron's sets of 4 nontransitive dice which give the second chooser a _ chance of winning. Efron says it had been proven that this is the maximum obtainable with four dice. For three dice, the maximum is .618, but this requires dice with more than six faces. As the number of dice increases, the maximum value approaches ¾. The Addendum in Wheels describes numerous variants developed by magicians and mathematicians.
Gardner. Nontransitive paradoxes. SA (Oct 1974) c= Time Travel, chap. 5. Discusses voting paradoxes and describes examples of naontransitive behaviour back to mid-20C. Describes Penney's game, giving it with pairs first and showing that for each choice by the opponent, you can pick a better choice, with probability of winning being at least _. The bibliography in Time Travel is extensive and Gardner notes that some of the items give many further references.
