WPCL √ 2BJ|x╒!╨ x ╨╨╨№Ё ф ╪                                                                    ╨╨КВ╨╚╨┴`┴SOURCES й page !╒╨ ░x ╨╨╨░дШМ А t                                                                    ░╨╨КВ╨╚╨ ├ ├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. ╨╨дШМ А t                                                                      ░╨╨G└└bor J. Sz└)└kely. Paradoxes in Probability Theory and Mathematical Statistics. Akad└)└miai Kiad└;└, Budapest and Reidel, Dordrecht, 1986. [Revised translation of: Paradoxonok a V└)└letlen atematik└└ban; M└┐└szaki K└?└nvkiad└;└, Budapest, nd.] Translated by M└└rta Alp└└r and └(└va 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'arithm└)└tique morale, section 23. 1777. (Contained in the fourth volume of the supplement to his Histoire Naturelle, pp.а101-104??) = Oeuvres Compl└/└tes de Buffon; annotated by M. Flourens; Garnier Fr└/└res, Paris, nd [c1820?], pp. 180-185. (Also in Miles & Serra, pp. 10-11.) Laplace. Th└)└orie Analytique des Probabiliti└)└s. 1812. Pp. 359-360. ??NYS. (In Miles & Serra, p. 12.) 3rd ed, Courcier, Paris, 1820, pp. 359й362. Finds the answer to Buffon's problem with needle length 2r and line spacing a. Then solves the case of two perpendicular sets of lines, with possibly different spacings. M. E. Barbier. Note sur le probl└/└me 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 └!└. Messenger of Mathematics 2 (1873) 113-114. ??NYS. Tissandier. R└)└cr└)└ations Scientifiques. 1880? 2nd ed., 1881, 139й145 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 └!└ = 3.1421. ┴┴┴┴= 5th ed., 1888, pp. 204й208. c= Popular Scientific Recreations, 1890? pp.а729-731, but the needle is 2 in on a floor of spacing 2└└ 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 └!└. 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─ ─ ╨╨░дШМ А t                                                                    ░╨╨ ┴┴How many people are required before there is an even chance that some two have the same birthday? ╨╨дШМ А t                                                                      ░╨╨ 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. 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 йй 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. =аMentor, NY, 1953, pp.а204-206. Says 24 or more. Oswald Jacoby. How to Figure the Odds. Doubleday, NY, 1947. The birthday proposition, pp. 108й109. Gives answer of 23 or more. William Feller. An Introduction to Probability Theory and Its Applications: Vol 1. Wiley, 1950, pp. 29й30. Uses an approximation to obtain 23 or more. The 2nd ed., 1957, pp.а31-32 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. 32й36, gives a variety of other forms of the problem. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨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 └s└ .52; for 11, └s└ .68; for а12, а└s└а.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. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Howard P. Dinesman. Superior Mathematical Puzzles. Op. cit. in 5.B.1. 1968. No. 58: The birthday puzzle, pp. 76, 116й117 & 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 йй 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.а(rаdistinctаbirthdays)аа=ааBC(365,аr)/BC(364+r, r), rather than P(365, r)/365├├r──. But the unordered selections with repetitions are not equally likely events йй 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 └└ 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, kа=ааа1,ааа2,ааа3, 4, 5, 6, 7, 8, 9, 10, the breakeven numbers are pа=а23,а14,а11,а9,а8,а8,а7,а7, 6, 6. L. E. Clarke & Eric S. Langford. Note 3298 йй I & II: On Note 3244. MG 55 (No. 391) (Febа1971) 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 └└ 23, the classical case. With w = 1, it requires kа└└ 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 └└ 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) 183й188. 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. S. Ejaz Ahmed & Richard J. McIntosh. An asymptotic approximation for the birthday problem. CM 26:2 (Apr 2000) 151й155. Without using Stirling's approximation, they show that for a calendar of n days and a desired probability p, 0 < p < 1, then the minimum class size to produce a probability └└ p of two people having the same birthday is asymptotically └└[2n log {1/(1йp)}]. For p = .5, n = 365, this gives 22.494. ┴┴├ ├8.C.┴┴PROBABILITY THAT A TRIANGLE IS ACUTE─ ─ ╨╨░дШМ А t                                                                    ░╨╨┴┴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. ╨╨дШМ А t                                                                      ░╨╨ 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. 480й481. ??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. E. Lemoine. 1882й1883. See entry in 8.G. L. Carroll. Pillow Problems. 1893. ??NYS. 4th ed., (1895). = Dover, 1958. Problem 58, pp.а14, 25, 83-84. Prob (acute) = .639. C. O. Tuckey. Note 1408: Why do teachers always draw acute-angled triangles? MG 23 (No.а256) (1939) 391-392. He gets Prob(obtuse) varying between .57 and .75. E. H. Neville. Letter: Obtuse angling йй 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) 22й27 & 60й61. Survey on geometric probability. Asks for the probability of an acute triangle when one takes three points at random on a circle and gets └└. Richard K. Guy. There are three times as many obtuseйangled triangles as there are acuteкangled ones. MM 66:3 (Jun 1993) 175й179. Gives 12 different approaches, five of which yield Prob(acute) = └└, with other values ranging from 0 to .361. He tracked down the Sylvester reference йй see above. In Mar 1996, I realised that the two approaches sketched in 6.BR give probabilities of acuteness as 0 and 2 й └!└/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 Probabliti└)└s (A Philosophical Essay on Probabilities). c1819, ??NYS. Translated from the 6th French ed. by F. W. Truscott & F.аL. 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 └!└ M. de M... [Montmort] du 9 Septembre 1713. IN: Pierre R└)└mond de Montmort; Essai d'analyse sur les jeux de hazards. (1708); Seconde edition revue & augmentee de plusieurs lettres, (Quillau, Paris, 1713 (reprinted by Chelsea, NY, 1980)); 2nd issue, Jombert & Quillau, 1714. Pp.а401й402. Quatri└/└me Probl└/└me & Cinqui└/└me Probl└/└me, p. 402. [See note to Euler; Vera aestimatio sortis in ludis; op. cit. below, pp.а459-461.] In the 4th problem, he proposes payйoffs of 1, 2, 3, 4, ... if the player first throws a six with a die on throw 1, 2, 3, 4, ... and asks for the expectation. He does not compute it, but I get 6. In the 5th problem, he asks what happens in the same situation if the payйoffs are 1,а2, 4, 8, ... or 1,а3,а9,а27, ... or 1, 4, 9, 16, ... or 1, 8, 27, 64, ... etc. Again, he doesn't give and results, but the first two give divergent series, while the later two are convergent. Daniel Bernoulli. Specimen theoriae novae de mensura sortis. Comm. Acad. Sci. Imp. Petropol. 5 (1730й31(1738)) 175й192, ??NYS. IN: Die Werke von Daniel Bernoulli; ed. by L. P. Bouckaert & B. L. van der Waerden; Birkh└└user, 1982; pp. 223й234 and notes by van der Waerden, pp. 195 & 197й200. English translation in Econometrica 22 (1954) 23й36, ??NYS. Lewis Carroll. Lionel Tollemache, Reminiscences of Lewis Carroll, Literature (5 Feb 1898), ??NYS, quoted inаCarrollйWakeling II, prob. 28: A good prospect, pp. 44 & 72. (Tollemache was at Balliol College, Oxford, in 1856й1860. He then entered Lincoln's Inn, London, so this must refer to c1858. ┴┴┴┴He says that Carroll gave the problem with a coin and payйoffs of 0, 1, 3, 7, 15, ... if the player first throws a tail on throw 1, 2, 3, 4, 5, .... Neither Tollemache nor Wakeling give any reference to any other version of the problem. I would compute the expectation as ┴┴┴┴└$└├├n=0── (2├├n──й1)(1/2)├├n+1── = 0/2 + 1/4 + 3/8 + 7/16 + 15/32 + .... ┴┴Wakeling does it by rewriting the sum as ┴┴┴┴1/4 + (1+2)/8 + (1+2+4)/16 + (1+2+4+8)/32 + ..., and regrouping as: ┴┴┴┴1/4 + 1/8 + 1/16 + 1/32 + ... + 2/8 + 2/16 + 2/32 + ... + 4/16 + 4/32 + ... + 8/32 + ... = 1/2 + 1/2 + 1/2 + 1/2 + ... and says "Hence, his prospects are └└d. for every successive throw of a head." Basically, making the payoff be a sum means that the expectation is a sum over half a quadrant. The normal process would be to first sum the finite rows and see that each is at least 1/4. Wakeling is first summing the columns and finding each column sum is 1/2. I wonder if Carroll carefully chose the peculiar payкoffs in order to make the latter process work. Leonhard Euler. Vera aestimatio sortis in ludis. [Op. postuma 1 (1862) 315й318.] = Op. Omnia (I) 7 (1923) 458й465. Tissandier. R└)└cr└)└ations Scientifiques. 1880? 2nd ed., 1881, p. 140 gives a brief unlabelled description, saying this "probl└/└me de P└)└tersbourg" was discussed by Daniel Bernoulli in "M└)└moires de l'Acad└)└mie de Russie". Not in the 5th ed. of 1888. Tissandier. Popular Scientific Recreations. 1890? Pp. 727й729 discusses the idea and says D. Bernoulli presented his material on this in "Memoires [sic] de l'Acad└)└mie 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 (Octа1887) 276й278. 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 log├├2── of the bank's funds. Dan Pedoe, The Gentle Art of Mathematics, op. cit. in 5.C, 1958, 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) 13й39. ??NYR. Nick Mackinnon & 5Ma. Note 74.9: A lesson on the St. Petersburg paradox. MG 74 (No.а467) (1990) 51-53. Suppose a maximum is put on the payment йй 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─ ─ ╨╨░дШМ А t                                                                    ░╨╨ ┴┴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 M└)└r└)└ 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, ...). ╨╨дШМ А t                                                                      ░╨╨ ┴┴┴┴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) йй 5 : 3; Cardanа(1539) йй 6 : 1; Tartaglia (1539) йй 2 : 1; Pascal (1654) йй 7 : 1; Fermatа(1654)акйа7а: 1; Huygens (1657) йй like Pascal. He also describes the approaches of Pascal, Fermat and Huygens to the three person case and generalizations due to Montmort (1713) and de Moivre (1718). Pacioli. Summa. 1494. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Ff. 197rй197, prob. 50 (?йnot 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). ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨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 йй (8; 5, 3), (14; 10, 8, 5) (??NYS). But there is no proper mathematics until Pascal & Fermat. Calandri, Raccolta. c1495. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨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.) ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Cardan. Practica Arithmetice. 1539. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Chap. 61, section 13, f. T.iii.r (p. 113). (10; 9, 7). He divides as 1 + 2 + 3 : 1. (10;а3,а6)ааййааdivided 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 й QQ.viii.r (p. 214). This chapter is on errors of Pacioli. Mentions (6;а5, 2) and many other examples. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨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 йй described in: Livia Giacardi & Clara Silvia Roero; Bibliotheca Matematica Documenti per la Storia della Matematica nelle Biblioteche Torinese; Umberto Allemandi & C., Torino, 1987, pp. 117й118. 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: 41й52, esp. 45й50; 1708: 37й48, esp. 42-45. Prob. 13, 1725:а123-130. Prob. 3, 1778: 117й121; 1803: 116й120; 1814: 102й106; 1840: 54й55. 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 R└)└mond de Montmort. Essai d'analyse sur les jeux de hazards. (1708); Seconde edition revue & augmentee de plusieurs lettres, (Quillau, Paris, 1713 (reprinted by Chelsea, NY, 1980)); 2nd issue, Jombert & Quillau, 1714. Avertissement (to the 1st ed.), pp. xxiйxxiv and (to the 2nd ed.) xxvйxxxvii discusses the history of the problem, the work of Fermat and Pascal and de Moivre's assertion that Huygens had solved it first. Chr. Mason, proposer; Rob. Fearnside, solver. Ladies' Diary, 1732й33 = T. Leybourn, I: 223, quest. 168. (15; 10, 8, 5). I haven't checked the solution, but the procedure is correct and another solver got the same results. Editor cites De Moivre. Pearson. 1907. Part II, no. 98, pp. 134 & 210й211. (3; 2, 1). He divides correctly as 3 : 1. Hummerston. Fun, Mirth & Mystery. 1924. Marbles, Puzzle no. 26, pp. 71 & 176. (4; 3, 2). He divides correctly as 3 : 1. ┴┴├ ├8.G.┴┴PROBABILITY THAT THREE LENGTHS FORM A TRIANGLE─ ─ ┴┴See also 8.C. E. Lemoine. Sur une question de probabilit└)└s. Bull. Soc. Math. France 1 (1872-1873) 39-40. Obtains └└ by considering that the stick can be broken at m equidistant points and then letting m increase. ? Halphen. Sur un probl└/└me de probabilit└)└s. Ibid., pp. 221-224. Extends Lemoine to n pieces, getting 1 - n/2├├n-1──, by an argument similar to homogeneous coordinates and by integration. Camille Jordan. Questions de probabilit└)└s. 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. [Laqui└/└re; Note sur un probl└/└me de probabilit└)└; ibid. 8 (1879й80) 79й80 gives a simple argument.] M. Laqui└/└re. Rectification d'une formule de probabilit└)└. Bull. Soc. Math. France 8 (1879й80) 74й79. Treats the first problem of Jordan. Observes that Jordan's formula can give a probability greater than one! Says Jordan has a confusion between 'some n' and 'a given n' and he gives a corrected version. E. Lemoine. Quelques questions de probabilit└)└s r└)└solues g└)└om└)└triquement. Bull. Soc. Math. France 11 (1882й83) 13й25. Refers to his article in vol. 1 and the many resulting works. Takes a point in a triangle and asks for the probability that the three lengths to the sides form a triangle (an acute triangle). Then says that breaking a stick corresponds to using an equilateral triangle, giving probabilities └└ and log 8 й 2 = .0794415.... Makes various generalizations. Takes a point, M, in an equilateral triangle ABC and asks the probability that MA, MB, MC form an acute triangle, getting 4 й 2└!└/└└3 = .3718.... E. Fourrey. Curiositi└)└s G└)└ometriques, op. cit. in 6.S.1. 1907. Partа3, chap. 1, section 5: Application au calcul des probabiliti└)└s, pp. 360-362. Break a stick into three pieces. Gets P = └└. 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 2аlog 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) 228й230. Take a stick and break it at two random points йй or йй break once at random and then break the longer part at random. Prob(triangle) = └└ 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 й 1 = .38..., as in Rushton. ┘┘ ┴┴├ ├8.H.┴┴PROBABILITY PARADOXES─ ─ ┴┴├ ├8.H.1.┴┴BERTRAND'S BOX PARADOX─ ─ J. Bertrand. Calcul des Probabilit└)└s. 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. 40й41 & 96. Deranged matchboxes of red and black beans йй see 5.K.1. The problem continues by unlabelling the boxes йй 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) 149й154. 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 └└, └@└ and └└. 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─ ─ ╨╨░дШМ А t                                                                    ░╨╨ ┴┴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. ╨╨дШМ А t                                                                      ░╨╨ 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. Funйtoйdo. Op. cit. in 5.C. 1948. Prob. 14: The absentйminded professor, pp. 25 & 184. Birtwistle. Math. Puzzles & Perplexities. 1971. Fifth, pp. 146 & 197. ┴┴├ ├8.J.┴┴CLOCK PATIENCE OR SOLITAIRE─ ─ ╨╨░дШМ А t                                                                    ░╨╨ ┴┴The patience or solitaire game of Clock has 13 piles of four faceйdown 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 йй if it has value n, you place it under the nйth 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. ╨╨дШМ А t                                                                      ░╨╨ John Reade, proposer; editorial solution. Problem 46.5 йй 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 йй The last 1. MM 52 (1979) 113 & 53 (1980) 184й185. 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) 42й43. 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 'fйcycles' and a distribution comes out if and only if every fйcycle contains the initial value. ┴┴├ ├8.K.┴┴SUCKER BETS─ ─ ╨╨░дШМ А t                                                                    ░╨╨ ┴┴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 M└)└r└)└ 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. ╨╨дШМ А t                                                                      ░╨╨The classic carnival game of ChuckйaйLuck is an excellent example of this category. This is the game with three die. You bet on a number йй 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 -125а+а75а+а30а+а3аа=аа-17, giving a 7.9% profit to the operator. Collins. Fun with Figures. 1928. How figures can cheat йй The tinйhorn gambler, pp. 34й35. 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 20/6├├6── = 1/2332.3, so the fair odds should be $2332 to $1. This is not quite right. The true probability is 1 й (1 й 1/6├├6──)├├20── = .00042858 = 1/2333.27. The fair odds depend on whether the winnings include the punter's $1 or not йй 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 6├├6── = 46656, the promoter gets an average of └ └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 └A└ or └└, 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(Rй1) + B(Bй1)]/[(R+B)(R+Bй1)] which is always less than └└. ┴┴├ ├8.L.┴┴NONTRANSITIVE GAMES─ ─ ╨╨░дШМ А t                                                                    ░╨╨ ┴┴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 nonйtransitivity 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 generally 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. ╨╨дШМ А t                                                                      ░╨╨ In about 1932, the following golf scores were sent to The Scotsman. ┴┴┴┴A 4 5 5 6 9 4 5 4 4 - 46 ┴┴┴┴B 5 6 6 7 3 5 6 3 5 - 46 ┴┴┴┴C 6 7 7 8 6 3 3 3 3 - 46 ┴┴A beats B by 4 and 3, B beats C by 5 and 4, C beats A by one hole. A correspondent, J. C. Smith, suggested the following series for three holes. ┴┴┴┴A 1 2 3 - 6 A one up on B. ┴┴┴┴B 2 3 1 - 6 B one up on C. ┴┴┴┴C 3 1 2 - 6 C one up on A. ┴┴Similar results can be obtained for four men playing four holes, and so on. ┴┴┴┴Reported by J. W. Stewart as Gleaning 854: Golf Scores; MG 16 (No. 218) (May 1932) 115. Walter Penney, proposer and solver. Problem 95 йй Penneyйante. 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 йй Penneyйante. JRM 2:4 (Oct 1969) 241 & 8:1 (1975) 62й65. 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 └└. 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 └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 └└. 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 nontransitive behaviour back to midй20C. 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 └A└. The bibliography in Time Travel is extensive and Gardner notes that some of the items give many further references. Richard L. Tenney & Caxton C. Foster. Nonйtransitive dominance. MM 49:3 (May 1976) 115й120. Gives three dice with odds for the second chooser being 5/9. Gives proofs for the results stated by Efron in Gardner, above, observing that they arise from results known about voting paradoxes. ├ ├9.┴┴LOGICAL RECREATIONS─ ─ ┴┴Many combinatorial recreations can be considered as logical. ┴┴├ ├9.A.┴┴ALL CRETANS ARE LIARS, ETC.─ ─ Diogenes La└-└rtius. 3C. De Clarorum Philosophorum Vitis, Dogmatibus et Apophtegmatibus, II, Life of Euclides. Ed. by C. G. Cobet; Paris, 1888, p. 108, ??NYS. Translated by C.аD. Yonge; Bell, London, 1894, pp. 97-98. Translated by R. D. Hicks; Loeb Classical Library; vol. 1, pp. 236-237. Refers to Eubulides (c-330) as the source of "The Lying One" or "The Liar" йй └└ └0└└ └└└└ └└└└└└ └└└└└└%└ (O Pseudomenos). According to: I.аM.аBochenski; Ancient Formal Logic; North Holland, Amsterdam, 1951, p. 100; Eubulides also invented: "the swindler", "the concealed", "the heap" (how many grains make a heap?), "the Electra" and "the horned" (equivalent to "Have you stopped beating your wife?"). Bochenski, Ancient Formal Logic, pp. 101-102, says the liar paradox was unknown to Plato, but is quoted by Aristotle in his Libro de Sophisticis Elenchis 25, 180 b 2-7, ??NYS, (See: M. Wallies, ed.; Topica cum Libro De Sophisticis Elenchis; Leipzig, 1923; ??NYS; and: Ethica Nicomachea, H3, 1146 a 21-27; ed. by Fr. Susemihl; Leipzig, 1887; ??NYS.) It is also given in Cicero; Ac. Pr. II, 95, 96 (=? Topica, 57); ??NYS (In:ааG. Friedrich, ed.; Opera Rhetorica; Leipzig, 1893; ??NYS) and in many later writers. Athenaeus Naucratica. c200. The Deipnosophists, Book 9 (end of c.64). Translated by C.аD.аYonge, Bohn, London, 1854, vol. 2, p.а633. Epitaph of Philetas of Cos (c-340/c-285). "Traveller, I am Philetas; the argument called the Liar and deep cogitations by night, brought me to death." Sadly, there is no indication where he died or was buried. (Bochenski; Ancient Formal Logic,; p. 102 gives the Greek of Athenaeus. I.аM.аBochenski; History of Formal Logic; corrected ed., Chelsea, 1970, p.а131; gives the English.) The Stoics. c-280. Bochenski; Ancient Formal Logic; pp. 100-102 says they invented several paradoxes, including "the crocodile" who takes a baby and says he will return it if the mother answers his question correctly. He then asks "Will I return the baby?" She answers "No". Anon. History of the Warring States. [The Warring States period is -475/-221 and this history may be -2C.] The Elixir of Death. Translated in: Herbert A. Giles; Gems of Chinese Literature; op.аcit. in 6.BN, p. 43. Chief Warden swallows an elixir of immortality which he was supposed to convey to the Prince. The Prince orders the Warden's execution, but the Warden argues that if the execution succeeds, then the elixir was false and he is innocent of crime. The Prince pardons him. Tung-Fang So (-2C, see Giles, ibid., p. 77) is said to have been in the same situation as the Warden and argued: "If the elixir was genuine, your Majesty can do me no harm; if it was not, what harm have I done?" St. Paul. Epistle to Titus, I, 12. c50? "One of themselves, even a prophet of their own, said, The Cretans are always liars, .... This witness is true." M. Cervantes. Don Quixote. 1605. Book II, chap. 51. Translated by Thomas Shelton, 1612-1620, reprinted by the Navarre Society, London, 1923, vol. 2, pp. 360-362. Sentinel paradox: truthtellers pass; liars will be hanged. "I will be hanged." Henri Decremps. Codicile de J└)└r└=└me Sharp, .... Op. cit. in 4.A.1. 1788. Avant Propos, pp.а19й20: Sentinel paradox: truthtellers pass; liars will be thrown in the river. "You will throw me in the river." Author says he will give the answer in another volume. Henri Decremps. Les Petits Aventures de Jerome Sharp. Professor de Physique Amusante; Ouvrage contenant autant de tours ing└)└nieux que de le└'└ons utiles, avec quelques petits portraits └!└ la mani└)└re noire. Brussels, 1789; also 1790, 1793. Toole Scott records an English edition, Brussels, 1793. Sentinel paradox. ??NYS. Cited by Dudeney; Some much-discussed puzzles; op. cit. in 2; 1908; as the first appearance of this paradox. The Sociable. 1858. Prob. 43: The Grecian paradox, pp. 299 & 317. Protagoras suing his pupil who had promised to pay for his tuition when he won his first case. = Book of 500 Puzzles, 1859, prob. 43, pp. 17 & 35. c= Magician's Own Book (UK version), 1871, pp.а26й27. c= Mittenzwey, 1895?: prob. 157, pp. 33 & 81; 1917: 157, pp. 30 & 79, which claims the teacher wins. Leske. Illustriertes Spielbuch f└G└r M└└dchen. 1864? Prob. 564й21, pp. 253 & 395. Form of the sentinel paradox. To enter a garden, one must make a statement; if true, one pays 3 marks; if false, one pays 6 marks. "I will pay 6 marks." CarrollйWakeling II. c1890? Prob. 32: Bag containing tickets, pp. 50 & 73. This is one of the problems on undated sheets of paper that Carroll sent to Bartholomew Price. Wakeling reproduces the MS. ┬┬┬┬A bag contains 12 tickets, 3 marked 'A', 4 'B', 5 'C'. One is drawn in the presence of 12 witnesses of equal credibility: three say it was 'A', four 'B', five 'C'. What is the chance that it was 'A'? ┴┴There is no answer on the Carroll MS. Wakeling gives an answer. ┬┬┬┬┴┴Let the credibility of a witness be "a" when telling the truth. Hence, the credibility of a witness when telling a lie is "1 й a". ┬┬┬┬┴┴If it was A, then 3 tell the truth, and 9 lie; hence the credibility is 3 in 12, or 1 in 4. ┴┴┴┴┴┴Therefore the chance that it is A, and no other, is: ┴┴┴┴┴┴┴┴3/12 x 1/4 + 4/12 x 3/4 + 5/12 x 3/4 = 5/8 ┴┴I cannot see how this last formula arises. Wakeling writes that he was assisted in this problem by a friend who has since died, so he does not know how the formula was obtained. ┴┴┴┴Assuming a is the probability of a person telling the truth, this probability depends on what the chosen ticket is and is not really determined by the information given йй e.g., if the ticket was A, then any value of a between 0 and 1 is possible. The value 3/12 is an estimate of a, indeed the maximum likelihood estimate. If k of the 12 people are telling the truth, I would take the situation as a binomial distribution. There are BC (12, k) ways to select them and the probability of having k liars is then BC (12, k) a├├k── (1йa)├├12йk──. Now it seems that Bayes' Theorem is the most appropriate way to proceed. Our basic events can be denoted A, B, C and their a priori probabilities are 3/12, 4/12, 5/12. Taking a = k/12, the a posteriori probabilities are proportional to ┴┴k/12 x BC (12, k) (k/12)├├k── ({12йk}/12)├├12йk──, for kа=а3,а4, 5. Dropping the common denominator of 12├├13──, these expressions are 6.904, 8.504, 10.1913 times 10├├12──. Dividing by the total gives the a posteriori probabilities of A, B, C as 27.0%, 33.2%, 39.8%. ┴┴┴┴This is really a probability problem rather than a logical problem, but it illustrates that the logical problem seems to have grown out of this sort of probability problem. Lewis Carroll. Diary entry for 27 May 1894. "I have worked out in the last few days some curious problems on the plan of 'lying' dilemma. E.g. 'A says B lies; B says C lies; C says A and B lie.' Answer: 'A and C lie; B speaks truly'. The problem is quoted in CarrollйGardner with his discussion of the result, pp. 22й23. Gardner says this was printed as an anonymous leaflet in 1894. ┴┴┴┴CarrollйWakeling. Prob. 9: Who's telling the truth?, pp. 11 & 65. Wakeling says "This is a puzzle based on a piece of logic that appears in his diary. ┴┴┴┴The Dodo says the Hatter tells lies. ┴┴┴┴The Hatter says that the March Hare tells Lies. ┴┴┴┴The March Hare says that both the Dodo and the Hatter tell lies. ┴┴┴┴Who is telling the truth?" ┴┴┴┴In a letter of 28 May 2003, Wakeling quotes the text from the Diary as above, but continues with it: "And today 'A says B says C says D lies; D says two lie and one speaks true.' Answer: 'D lies; the rest speak truly.' Wakeling adds that the Diary entry for 2 Jun 1896 (not given in the Lancelyn Green edition) says: "Finished the solution of the hardest 'TruthйProblem' I have yet done", but Carroll gives no indication what it was. Lewis Carroll. The problem of the five liars. In his unpublished Symbolic Logic, Part II. He was working on this after Part I appeared in 1896 and he had some galley proofs when he died in 1898. Published in Lewis Carroll's Symbolic Logic, ed. by William Warren Bartley III; Clarkson N. Potter, NY, 1977, ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Pp. 352-361, including facsimiles of several letters to John Cook Wilson (not in Cohen). Each of five people make two statements, e.g. A says "Either B or D tells a truth and a lie; either C or E tells two lies." When analysed, one gets contradictions because a form of the Liar Paradox is embedded. Pp. 423й444 is a survey of logical paradoxes with some variations by Carroll. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Cesare Burali Forti. Una questione sui numeri transfiniti. Rendiconti del Circolo Matematico di Palermo 9 (1897) 154й164. ??NYS. This was the first published antinomy of modern set theory. The set of all ordinal numbers is itself an ordinal! However, Cantor had observed the paradox in 1895 and communicated it to Hilbert in a letter in 1896. Irving Anellis. The first Russell paradox. Paper given at AMS meeting, Chicago, Mar 1985. ??NYS йй abstract given in HM 12 (1985) 380. Says it is usually believed that Russell discovered his paradox in Jun 1901, but he sent a version of it to Couturat on 8 Dec 1900 (unpublished MS in the Russell Archives, McMaster Univ.). Gregory H. Moore. A house divided against itself: The emergence of firstйorder logic as the basis for mathematics. IN: Esther R. Phillips, ed.; Studies in the History of Mathematics; MAA, 1987, pp. 98й136. On pp. 114й115, he dates Russell's paradox to May 1901 and says Russell wrote about it to Frege on 16 Jun 1902. The first publications are in Russell's Principles of Mathematics and Frege's Fundamental Laws, vol. 2, both in 1903. B. Russell. The Principles of Mathematics. CUP, 1903. ??NYS йй cited in Garciadiego. He discusses Russell's paradox and also Cantor's paradox of the greatest cardinal and Burali Forti's paradox of the greatest ordinal. I won't consider these much further, but this may have inspired the development of the more verbal paradoxes described in this section. G. G. Berry. Letter to Russell on 21 Dec 1904. In the Russell Archives, McMaster University. Quoted in Garciadiego. "... the least ordinal which is not definable in a finite number of words. But this is absurd, for I have just defined it in thirteen words." The paradox of Jules Richard (late Jun 1905) is very similar and similar versions were found by J. K└?└nig and A. C. Dixon about the same time, though these all use Zermelo's wellйordering axiom. Sometime earlier, Berry had introduced himself to Russell with a note saying "The statement on the other side of this paper is true" with the other side reading "The statement on the other side of this paper is false", and consequently is also considered the inventor of the "visiting card paradox". B. Russell. Les paradoxes de la logique. Revue de M└)└taphysique et de morale 14 (1906) 627-650. ??NYS йй cited by Garciadiego. First publication of a modified version of Berry's paradox. B. Russell. Mathematical logic as based on the theory of types. Amer. J. Math. 30 (1908) 222-262. On p. 223, he first gives Berry's paradox: "the least integer not nameable in fewer than nineteen syllables". He also reformulates K└?└nig & Dixon as "the least indefinable ordinal". Kurt Grelling & Leonard Nelson. Bemerkungen zu den Paradoxien von Russell und Burali-Forti. Abhandlungen der Fries'schen Schule (NS) 2 (1908) 301-344. ??NYS. Grelling's paradox: "Is heterological heterological?" A. N. Whitehead & B. Russell. Principia Mathematica. CUP, 1910. Vol.а1, pp. 63-64. Discusses several paradoxes and repeats Berry's paradox. F. & V. Meynell. The Week-End Book. Op. cit. in 7.E. 1924. 2nd ed., prob. five, p. 275; 5th? ed., prob. nine, p. 408, gives Russell's paradox as a problem йй and gives no solution! Hummerston. Fun, Mirth & Mystery. 1924. The bridge, pp. 68й69. Sentinel paradox йй "I am going to be hanged on that gallows!" Author says "It is impossible to answer ... satisfactorily. Perhaps the best plan is to throw the varlet in the river." John van Heijenoort. Logical paradoxes. Encyclopedia of Philosophy 5 (1967) 45-51. Excellent survey of the paradoxes of logic and set theory, but only a mention of pre-19C paradoxes. Alejandro R. Garciadiego. The emergence of some of the nonlogical paradoxes of the theory of sets, 1903й1908. HM 12 (1985) 337й351. Good survey. He has since extended this to a book: Bertrand Russell and the Origins of the Setйtheoretic "Paradoxes"; Birkh└└user, 1992, ??NYS At the 19th International Puzzle Party in London, 1999, Lennart Green told the story of a friend of his who was such a failure in life that he decided to wrote a book on "How to be a Failure". But if this failed, he would be successful and if it succeeded, he would have failed. ┴┴├ ├9.B.┴┴SMITH йй JONES йй ROBINSON PROBLEM─ ─ ┴┴See also 5.K.2 for a special form of these problems. Dudeney. PCP. 1932. Prob. 49: "The Engine-Driver's Name", pp. 24 & 132. = 536; prob.а521: "The Engineer's Name", pp. 214 & 411. The driver is Smith, but the other two names are not determined. Phillips. Week-End. 1932. Time tests of intelligence, no. 39, pp. 22 & 39. Same as Dudeney. Phillips. Brush. 1936. Prob. K.2: The Engine-driver, pp. 36 & 96. Same as Dudeney. Rudin. 1936. No. 183, pp. 65 & 119. Similar to Dudeney, but Americanized, somewhat simplified(?) and asking for the brakeman's (= guard's) name. HaldemanйJulius. 1937. No. 4: RobinsonйSmithйJones problem, pp. 3 & 20. Similar to Dudeney, but Americanized differently than in Rudin, asking for the engineer's name. Says it was sent by J. C. Furnas. James Joyce. Finnegans Wake. Viking Press, NY, 1939. P. 302, lines 23й24: "SmithйJonesкOrbison". J. G. Oldroyd. Mathematicians in the army. Eureka 5 (Jan 1941) 6 & 6 (May 1941) 10. Six men of different ranks from three different schools, colleges and faculties (i.e. subjects). Irving Adler. Thinking Machines. Dobson, London, 1961. Pp. 111й116: Who is the engineer? Essentially identical to Rudin, but asks for the engineer's name. Gives a systematic solution via boolean algebra. Doubleday й 2. 1971. Flight plan, pp. 153й154. Same as Dudeney, slightly reordered. Liz Allen. Brain Sharpeners. Op. cit. in 5.B. 1991. Who's who?, pp. 86 & 134й135. Spaceship version, similar to Dudeney, but more precise. ┴┴├ ├9.C.┴┴FORTY UNFAITHFUL WIVES─ ─ ┴┴I now realise that this is an extension of 9.D. ┴┴See Littlewood, 1953, in 9.D. Gamow & Stern. 1958. Forty unfaithful wives. Pp. 20-23. (Communicated by V. Ambarzuminian.) Michael Spivak. Calculus. 2nd ed., Publish or Perish. ??date, place. P. 35, probs. 27 & 28. Uri Leron & Mike Eisenberg. On a knowledgeйrelated paradox and its resolution. Int. J. Math. Educ. Sci. Technol. 18 (1987) 761й765. Ed Barbeau. Fallacies, flaws and flimflam. CMJ 22:4 (Sep 1991) 307. Gives a brief discussion and the reference to Spivak and to Leron & Eisenberg. The paradox has to do with what information has been provided by the stranger. ┴┴├ ├9.D.┴┴SPOTS ON FOREHEADS─ ─ ┴┴See also 9.C. 7.AP is somewhat related. William Wells Newell. Games and Songs of American Children. Harper and Brothers, (1883); 2nd ed. 1903; reprinted with Editor's Note of 1883 and new Introduction and Index, Dover, 1963. Chap. IX, No. 77: Laughter games, pp. 136-137. "In a Swiss game, .... Each child pinches his neighbor's ear; but by agreement, the players blacken their fingers, keeping two of the party in ignorance. Each of the two victims imagines it to be the other who is the object of the uproarious mirth of the company." The Notes on p. 274 indicate that this probably comes from: E. L. Rochholz; Alemannisches Kinderleid und Kinderspiel, Leipzig, 1857, ??NYS Phillips. Week-End. 1932. Time tests of intelligence, no. 13, pp. 14 & 188. Two boys fall down, one gets a dirty face, the other washes his own face. Phillips. The Playtime Omnibus. Op. cit. in 6.AF. 1933. Section XVII, prob. 11: Odd, pp. 55 & 237. Identical to Week-End. W. E. Buker, proposer; Robert Wood and O. B. Rose, solvers. Science Question 686. SSM 35 (1935) 212 & 429. 3 persons. A. A. Bennett, proposer; E. P. Starke and G. M. Clemence, solvers. Problem 3734. AMM 42 (1935) 256 & 44 (1937) 333-334. n persons with smudges on foreheads. Says the 3 person case was suggested by Dr. Church of Princeton and cites the Buker problem. Phillips. Brush. 1936. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Prob. A.2: The green crosses, pp. 1-2 & 73. Three men with green crosses. Prob. R.1: The roof, pp. 58 & 112. Same as in Week-End, above. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Rudin. 1936. No. 145, pp. 51й52 & 110. 3 persons with black crosses on foreheads, who whistle when they see a black cross. HaldemanйJulius. 1937. No. 76: The circle problem, pp. 10 & 24. Three persons with red crosses, who put up their hands if they see a red cross. "This problem promises to become famous. It has been going the rounds during the past few weeks йй .... We printed it several years ago, but believe it deserves reprinting." McKay. At Home Tonight. 1940. Prob. 36: The five disks, pp. 70 & 83. Three mandarins and five disks, two black and three white. Emperor puts white on each forehead. Answer argues using the number of blacks. M. Kraitchik. Mathematical Recreations. Op. cit. in 4.A.2. 1943. Chap. 1. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Prob. 3: The problem of the three philosophers, p. 15. Three painted faces. Prob. 4, pp. 15-16. 3 white and 2 black discs йй three whites placed on backs. All realize simultaneously. (Neither is in Math. des Jeux.) ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Harold Hart. The World's Best Puzzles. Op. cit. in 7.AS. 1943. The problem of the marked foreheads, pp. 23 & 55. Three students with blue and green crosses. Leopold. At Ease! 1943. Short cut to chevrons, pp. 23й24 & 199. Three men with smudged foreheads. A. K. Austin. A calculus for know/don't know problems. MM 49:1 (Jan 1976) 12й14. He develops a setйtheoretic calculus for systematically solving problems involving spots on foreheads, etc., including problems with knowing sums, see 7.AP. His typical problem has a man with four red and three blue stamps and he sticks three on the foreheads of two boys, telling them they each have at least one red. The first says he doesn't know what he has; then the second says he doesn't know; then the first says he does know. What stamps did the first have? Leeming. 1946. Chap. 3, prob. 9: The three small boys, pp. 21-22 & 153-154. Three boys with smudged foreheads. Henry Cattan. The Garden of Joys. (An anthology of oriental anecdotes, fables and proverbs.) Namara Publications, London, 1979. How he won the office of Grand Vizier, pp.а107-109 & note 89 on p. 114. Same as Kraitchik's problem 4, but with the simpler solution based on symmetry. The note says: "This story is anonymous and was heard by the author in Palestine." A letter from the author says he heard the story in this form in Palestine before 1948. Jerome S. Meyer. Funйtoйdo. Op. cit. in 5.C. 1948. Prob. 51: Red hats and green hats, pp. 46 & 190й191. 2 red hats & 3 green hats. Answer gives the symmetry solution and the logical solution without clearly recognizing the distinction. Max Black. Critical Thinking. 1952. Op. cit. in 6.F.2. Prob. 11, pp. 12 & 432. Three sons with white marks. J. E. Littlewood. A Mathematician's Miscellany. Op. cit. in 5.C. 1953. Pp. 3-4 (25-26). Three dirty faces. Mentions that this can be extended to n dirty faces, which "has not got into the books so far as I know". This may be the origin of 9.C? The Little Puzzle Book. Op. cit. in 5.D.5. 1955. Pp. 34й35: Clean and dirty. Two men fall through a roof. Man with clean face goes to wash. T. J. Fletcher. The n prisoners. MG 40 (No. 332) (May 1956) 98-102. Considers Kraitchik's problem with n persons, n white discs and n-1 black discs. Also studies various colour distributions and assignments, including < n white discs and use of three colours. Gamow & Stern. 1958. Three soot-smeared faces. Pp. 77-79. Birtwistle. Math. Puzzles & Perplexities. 1971. Hats Off!, pp. 108 & 193. A, B, C are seated in a row, so B can see A while C can see both B and A. They all know that there are three white and two red hats in a bag. A hat is taken out and put on A's head, but he can't see it. Similarly, hat are taken and put onto B and C. C is now asked it he knows what colour his hat is and replies that he does not. B, having heard C's response, is asked if he knows what colour his hat is and he replies that he does not. Is A, having heard these, able to know what colour his hat is? Extends to various other combinations and to n people. The answer is "The answer to all the questions is, yes, it is possible." ┴┴├ ├9.E.┴┴STRANGE FAMILIES─ ─ ╨╨░дШМ А t                                                                    ░╨╨ ┴┴Complicated questions of kinship have arisen due to religious taboos on incest. Most religions have a list of kinship relations which are not permitted to marry. These get a bit more involved than I want to go into. See the items in the first section below for some typical material. ┴┴The second section deals with marrying a deceased wife's sister. ┴┴The third section deals with general strange families riddles and puzzles, but 'That man's father ...' are collected in 9.E.1. ┴┴Ripley's Believe It or Not! books give a number of examples of strange families. I will enter these under the date of the persons involved. BIONйxx denotes the xxйth series of Believe It or Not! ┴┴Problems of this type are generally put in the form of a riddle, and many of these are collected in the following. ╨╨дШМ А t                                                                      ░╨╨Mark Bryant. Dictionary of Riddles. Routledge, 1990. (Based on his Riddles Ancient and Modern; Hutchinson, 1983.) ┴┴┴┴├ ├GENERAL STUDIES OF KINSHIP RELATIONS─ ─ J. Cashdan & Martin D. Stern. Forbidden marriages from a woman's angle. MG 71 (No. 456) (1987). ??NYS йй cited by Stern, 1990. Martin D. Stern. Consanguinity of witnesses йй a mathematical analysis. Teaching Mathematics and Its Applications 6:2 (1987). ??NYS йй cited by Stern, 1990. Martin D. Stern. Mathematical motivation through matrimony. MM 63:4 (Oct 1990) 231-233. ??NYS йй reproduced in Robert L. Weber; Science with a Smile; Institute of Physics, Bristol, 1992, pp. 314й318. Presents a notation for kinship relations and uses it to see that the table of prohibited degrees of marriage given in the Book of Common Prayer is symmetric with respect to sex and hence there are no unexpected prohibitions. However, the Jewish restrictions on marriage and on testimony by consanguineous relatives are not symmetric йй cf the above items. Marcia Ascher. Ethnomathematics. Op. cit. in 4.B.10. 1991. Chapter Three: The logic of kinship relations, pp. 66й83. Gives a number of folk puzzles and then analyses several complicated kinship systems. Some references. Martin Stern. Discrete avoidance of marital indiscretion. Mathematics Review (Univ. of Warwick) 2:3 (Feb 1992) 8й11. He presents a notation for kinship relations and uses it to describe the prohibited relations in Christian, Jewish and Islamic traditions. The Jewish prohibitions are not symmetric between male and female. Helen Cooper. A little more than kin [Review of Elizabeth Archibald; Incest and the Medieval Imagination; OUP, 2001]; The Times Literary Supplement (26 Oct 2001) 27. This notes that the taboos on incest are very variable. Egyptian Pharaohs indulged in brotherйsister marriages and the Macedonians did not understand why Oedipus was upset when he discovered he was married to his mother. Medieval Christianity extended the family to include godparents, who were spiritual siblings, and even inйlaws of inйlaws, as well as all the relatives of lovers, since sex made the lovers 'one flesh'. On the other hand, Henry VIII married his deceased brother's wife and later divorced her in order to marry the sister of his mistress. Archibald notes that since Christ is God, Mary is "Maid and mother, daughter of thy son!" [Chaucer's translation of Dante]. The medieval legend of Judas makes him a version of Oedipus йй he unknowingly killed his father and married his mother. The medieval period introduced the double incest version where the son of a brotherйsister relation is sent away and returns as an adult and unknowingly marries his mother. In 13C French versions of the Arthurian legend, Mordred, the nephew of Arthur, is made into his son, the result of a liaison between Arthur and his halfйsister, who did not know of their relationship. Mordred attempted to marry Arthur's wife. Luther urged that if a wedded couple were later discovered to be brother and sister or halfйsister, or even mother, that the knowledge should be suppressed lest it drive them to the ultimate sin of despair. ┴┴┴┴├ ├DECEASED WIFE'S SISTER, ETC.─ ─ E. S. Turner. Roads to Ruin йй The shocking history of social reform. Michael Joseph, London, 1950. Chap. 5: Two wives, one motherйinйlaw, pp. 98й121. This surveys the British preoccupation with the legality of marrying a deceased spouse's sibling. Since a couple were considered to become 'one flesh' (Ephesians 5:31), such a marriage was considered incestuous by the Church. Leviticus 18:6 & 16 were interpreted as prohibiting such marriage, but Leviticus 18:18 was interpreted as saying that the previous verses stated that a man should not have sisters as wives at the same time [which is the Islamic interpretation], while Deuteronomy 25:5й10 not only permits, but even commands, that a man should marry his brother's widow. ┴┴┴┴The English preoccupation with the problem dates from Henry VIII's marriage to his brother's widow, Catherine of Aragon. [In fact, he then divorced her to marry his mistress's sister.] This particular question is mentioned in Shakespeare's Henry VIII and the general question is the basis of Hamlet, whose mother marries her dead husband's brother. There was at least one execution, in early 18C Scotland, of a woman who had sex with her sister's widower. ┴┴┴┴Up to 1835, marriage to a deceased wife's sister was permitted, but it could be voided and the children declared bastards, if an action was brought. But if an action was brought and dropped, further actions were prevented. In 1835, the Duke of Beaufort, who had married his deceased wife's halfйsister, persuaded the Lord Chancellor to introduce a bill to legitimize such marriages up to date. The Bishops managed to amend this to prohibit such marriages in the future. However, such a couple could go to Europe to be married and such marriages remained legal in places like Jersey, though they were not legitimate in England. The Catholic Church generally gave dispensation for such marriages. From 1841 onwards, bills to remove the prohibition were introduced in almost every Parliament. Marriage to a deceased husband's brother or to a deceased spouse's nephew/niece was not sufficiently common to be considered by the reformers. The question was mentioned by Gilbert & Sullivan (near the end of the first act of Iolanthe (1882), the Queen, referring to Strephon, says "He shall prick that annual blister, / Marriage with deceased wife's sister;"). The journal Moonshine commented: "To be able to marry two wives at the cost of but one motherйinйlaw is something to fight for." In 1906, the Colonial Marriages Bill legitimized such marriages made in the colonies. In 1907, the Deceased Wife's Sister Bill was passed. Canon Law was later changed to accept this. One man who had married his deceased wife's sister sued a Canon who refused him Communion and won, with his win being confirmed by the Court of Appeals and the House of Lords in 1912. However, marriage to a divorced wife's sister was not permitted while the exйwife lived. Marriage to a deceased husband's brother was permitted in 1921. A number of other marriages were permitted in 1921 and all these acts are consolidated in the Marriage Act of 1949. Turner is not clear whether marrying a divorced spouse's sibling was permitted, and I don't know the further history. A 2000 article says marrying a deceased wife's aunt or niece was permitted by the Marriage Act of 1931. A discussion of Strawberry Hill House says marriage to a deceased husband's brother was prohibited by the 1835 act, so that Frances, the widow of John Waldegrave, had to go to Scotland to marry his halfйbrother George Waldegrave, 7th Earl Waldegrave. Susan Kelz Sperling. Tenderfeet and Ladyfingers. Viking, NY, 1981, p. 98й99. She gives some details of the Hebrew view. The Hebrew law of yibbum declares that if a man dies without heir, his brother or nearest relative is obliged to marry the widow (i.e. she is marrying her dead husband's brother). However, he could decline the duty by a ceremony called halitzah, as specified in ├ ├Deuteronomy─ ─, by putting on a special shoe which the widow removed and then she spat in front of him to break the contract. This takes place in ├ ├Ruth─ ─, allowing her to marry Boaz. BIONй11 cites an American example where a woman successively married the widowers of ├├two── of her sisters. William Holman Hunt, the Victorian painter, married his deceased wife's sister in 1866, in Switzerland [Judi Culbertson & Tom Randall; ├ ├Permanent Londoners─ ─; Robson Books, London, 1991, p. 140]. See Dudeney, AM, prob. 52, below, for a complication of this situation. HaldemanйJulius. 1937. No. 88: Marriage problem, pp. 11 & 25. How can a man have married his widow's sister? (Also entered under General Family Riddles.) Mindgames. Frontiers (a UK popular science magazine) No. 1 (Spring 1998) 107. "My wife's sister is also her cousin. How can this be?" Solution is that her father married his dead wife's sister and had another daughter. ┴┴┴┴├ ├GENERAL FAMILY RIDDLES─ ─ The riddles which the Queen of Sheba proposed to Solomon are not recorded in the biblical account of their meeting (IаKings 10 & II Chronicles 9), which would be cй960. Josephus' History of the Jews only mentions that Hiram and Solomon traded riddles, without giving any of them. Bryant, p. 19, says the Queen's riddles are given are given in the 2nd Targum to the Book of Esther and elsewhere in the rabbinical literature. The Targums are commentaries on biblical books, created after the Babylonian Captivity of -587/й538 and written down from 100 onwards. One of these is a strange family riddle which occurs in the first few entries below. If this is really due to the Queen of Sheba, or even actually in the Targums, it would be by far the earliest strange families riddle known. A variant of the riddle is given by Yachya Ben Sulieman, c1430, qv below. Ms Zimmels at the library of the London School of Jewish Studies told me that there is an 1893 German translation: Targum Shennai(?) zum Buch Esther and that the riddles occur in Ginzberg. However, Rappoport gives more precise information. Louis Ginzberg. The Legends of the Jews. Translated from the German Manuscript. Vol. IV: Bible Times and Characters From Joshua to Esther. Jewish Publication Society of America, Philadelphia, (1913), 5th ptg, 1947, pp. 142й149. He gives 22 riddles. P. 146, no. 2: 'Then she [the Queen of Sheba] questioned him [Solomon] further: "A woman said to her son, thy father is my father, and thy grandfather my husband; thou art my son, and I am thy sister." "Assuredly," said he, "it was the daughter of Lot who spake thus to her son."' However, Ginzberg gives no source or date for this. Angelo S. Rappoport. Myth and Legend of Ancient Israel. Vol. III. Gresham Publishing Co., London, 1928. The riddles of the Queen of Sheba, pp. 125й130. P. 127: ┴┴┴┴'Said she: "I will ask thee another question. A woman once said unto her son: Thy father is my father, thy grandfather my husband; thou art my son but I am thy sister." ┴┴┴┴To which Solomon made answer: "It must surely have been one of Lot's daughters who thus spoke to her son." ┴┴┴┴The similarity of the text with Ginzberg's makes it clear that they are both taken from the same source. Fortunately Rappoport is specific as to his sources. He says the second Targum to Esther (citing Targum Sheni to the Book of Esther; ed. P. Cassel, Leipzig, 1885; ed. E. David, Berlin, 1898) contains three riddles (the last three in Ginzberg) and then says that the Midrash Mishle, or Midrash to the Proverbs (citing Midrash Mishle, ed. S. Buber, Vilna, 1893 and A. W└G└nsche, Midrash Mishle, Leipzig, 1885), gives four riddles, which are the first four in Ginzberg, hence include our riddle. For our riddle, he also gives another reference: J. Lightfoot, Hor└%└ Hebraica, Rotterdam, 1686, II, 527; see also Yalkut, II, └└1085. After these four riddles, he says the Midrash Hachefez (ed. and translated by S. Schechter, Folklore, No. 1, pp. 349й358) gives 19 riddles, which are the first 19 of Ginzberg, so again include our riddle. However, Rappoport gives no indication of the dates of these Midrashs. The Exeter Book Riddles. 8й10C (the book was owned by Leofric, first Bishop of Exeter, who mentioned it in his will of 1072). Translated and edited by Kevin CrossleyйHolland. (As: The Exeter Riddle Book, Folio Society, 1978, Penguin, 1979.) Revised ed., Penguin, 1993. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨No. 43, pp. 47 & 103. Body and soul both have the earth as their mother and sister. Their mother because they are made from dust; their sister because all are made by the same heavenly father. No. 46, pp. 50 & 104. ┴┴A man sat sozzled with his two wives, ┴┴his two sons and his two daughters, ┴┴darling sisters, and with their two sons, ┴┴favoured firstborn; the father of that fine ┴┴pair was in there too; and so were ┴┴an uncle and a nephew. Five people ┴┴in all sat under that same roof. ┴┴┴┴The solution is given in Genesis 19:30й38, which describes Lot and his two daughters who bore sons by him. "The first use of this incestuous story for the purpose of a riddle is attributed to the Queen of Sheba; she tried it on Solomon." Cf above entries and Yachya Ben Sulieman, c1430, below. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Alcuin. 9C. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Prob. 11: Proposito de duobus hominibus singulas sorores accipientibus. Two men marry each other's sister. Prob. 11a (in the Bede text): Propositio de duobus hominibus singulas matres accipientibus. Two men each marrying the other's mother. This is the classic "I'm my own grandfather" situation. Prob. 11b (in the Bede text): Propositio de patre et filio et vidua ejusque filia. Father and son marrying daughter and mother. This is like 11a. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Abbot Albert. c1240. P. 335. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Prob. 11. Two widows and sons marry. This is the same as Alcuin/Bede 11a. He says the sons of the unions are each other's paternal uncles. Prob. 12. Two widowers and daughters marry. This is the same as the previous except for a sex change. Latin distinguishes paternal uncle from maternal uncle. Prob. 13. Complex situation with a man and three wives. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Cooper (above under General Studies of Kinship Relations) says the medieval period introduced the double incest situation, where the son of a brotherйsister relation is sent away and returns as an adult and unknowingly marries his mother. Cf Pacioli, c1500, and the Martham tombstone of 1730 for a triple incest. Dialogue of Salomon and Saturnus. 14C. ??NYS. Given in Bryant, p. 12. "Tell me, who was he that was never born, was then buried in his mother's womb, and after death was baptised?" Answer: Adam. Cf: Adevineaux Amoureux, 1478; Vyse, 1771?, prob. 2. In about 1380, the Duke of Gloucester, uncle of Richard II, opposed the marriage of his brother's son to his wife's younger sister. [John Kinross; ├ ├Discovering Castles 1. Eastern England─ ─; Shire Discovering Series No. 23, 1969, p. 12.] Yachya Ben Sulieman. Hebrew text, c1430. ??NYS. Quoted in Folk-Lore (1890) ??NYS. Quoted in Tony Augarde, op. cit. in 5.B, p. 3. A riddle attributed to the Queen of Sheba. "A woman said to her son, thy father is my father, and thy grandfather my husband; thou art my son, and I am thy sister." "Assuredly," said he [Solomon], "it was the daughter of Lot who spake thus to her son." Bryant, no. 1116, pp. 259 & 346 gives the same wording, with an extra level of quotation marks, and attributes it to the Queen of Sheba with no further details. Cf Queen of Sheba and Exeter Book above. Adevineaux Amoureux. Bruges, 1478. ??NYS йй quoted by Bryant, no. 6, pp. 67й68 & 333. "Je fus nez avant mon pere / Et engendr└)└ avant ma mere, / Et ay occis le quart du monde, / Ainsi qu'il gist a la reonde, / Et si despucelay ma taye. / Or pensez se c'est chose vraie." (Bryant's translation: "I was born before my father, begotten before my mother and have slain a quarter of the world's population. How can this be?" Answer: Cain. Cf: Dialogue of Salomon and Saturnus, 14C; Vyse, 1771?, prob. 2. Chuquet. 1484. Prob. 166. Same as Alcuin/Bede 11a. FHM 233 mentions it briefly without giving the relationships. In 1491, the 14 yearйold Duchess Anne of Brittany married Charles VIII, King of France in 1491. This was slightly complicated because both of them were married already, indeed Charles was married to the daughter of Anne's husband, the future Emperor Maximilian of Austria, so he was marrying his own stepйmotherйinйlaw. Fortunately, as was often the case in those days, both marriages were unconsummated йй indeed the couples had probably not yet seen each other and such proxy marriages were more like engagements йй so a little influence at Rome got both marriages dissolved. Somewhat surprisingly, as the marriage was more or less forced by Charles' siege of Rennes, the couple got on very well and developed a definite affection. Pacioli. De Viribus. c1500. Part 3. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨F. 263v, fourth item. = Peirani 377. Mother says her son is also her nephew and her brother. A man impregnated his mother and this yielded a daughter, who is the speaker. The man then impregnated her and this produced a son. He is the son of her brother, hence her nephew, and the son of her father, so her brother. F. 266r, middle. = Peirani 383. Two widows marrying the other's son. = Alcuin 11a. A woman is carrying the son of the other couple, so it is her grandson and the brother of her husband, as well as the son of her motherйinйlaw. This is followed by a variation which I cannot understand. F. 267v, first item. = Peirani 385. Tomb holds mother and son, man and wife, sister and brother, but only two people. Son is offspring of fatherйdaughter incest and later marries his mother. F. 287v, no. 191. = Peirani 416. Two fathers and two sons are only three people. Ff. 291r й 291v, no. 218. = Peirani 421й422. Quotes a Roman epitaph from S. Bartolomeo which seems to say that Hersillus is buried with Maralla who was his mother, sister and wife. Pacioli's comment seems irrelevant. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Ivan Morris. Foul Play and Other Puzzles of all Kinds. (Bodley Head, London, 1972); Vintage (Random House), NY, 1974. Prob. 21: No incest, pp. 39 & 93. Quotes Dudeney (??NYS) who gave an authentic 1538 epitaph describing the situation of Alcuin/Bede 11a with each couple having a child. Tartaglia. General Trattato, 1556, art. 135, p. 256r. 2 fathers and 2 sons make only 3 people. 16C(?) riddle in Mantuan dialect. Given in: Franco Agostini & Nicola Alberto De Carlo; Intelligence Games; (As: Giochi della Intelligenza; Mondadori, Milan, 1985); Simon & Schuster, NY, 1987; p. 69. Two fathers and two sons make three people. The discussion is a bit unclear as to the date of this riddle. Book of Merry Riddles. 1629? ╨ дx ╨╨╨ШМ А th                                                                      д╨╨The 71 Riddle, pp. 42й43. Two ladies and two boys are met and they say as follows. ┴┴┴┴The sons of our sons they be certain, ┴┴┴┴Brothers to our husband they be I wis, ┴┴┴┴And each of them unto the other Uncle is ┴┴┴┴Begotten and born in wedlock they be, ┴┴┴┴An we are their mothers we tell you truly. ┴┴I.e., the same as Alcuin/Bede 11a. The 73 Riddle, p. 44. 2 fathers and 2 sons make only 3 people. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Tombstone in the church at Martham, Norfolk. 1730. ??NYS йй quoted in a letter from Judith Havens (Norwich, Norfolk) in Challenging Centipede; The Guardian, section 2, (1аDecа1994) 6. ┬┬┬┬┴┴"Here Lyeth / The Body of Christ. / Burraway, who departed / this life ye 18 day / of October, Anno Domini 1730 / Aged 59 years / And their Lyes / Alice, who by hir Life / was my Sister, my Mistress, / My Mother, and my Wife. / Dyed Feb ye 12, 1729 / Aged 76 years." ┴┴┴┴┴┴This was a response to a vague description of the epitaph in: Centipede; Famous last words; The Guardian, section 2 (24 Nov 1994) 4. There it is stated that Burraway was the result of an incestuous union between a man and his daughter. The baby was sent away to be brought up and years later happened to return to his native village where he met a older woman and became her lover, then her husband! (Shades of Oedipus!) Consequently he was his own uncle, stepйfather and brotherйinйlaw! Cf Cooper, above for the idea of double incest йй this seems to be a triple incest. ┴┴┴┴If Burraway's union with his sister/mother/wife produced a child, then it would have only three grandparents and six greatйgrandparents (as is the case for the offspring of any halfйsiblings). Vyse. Tutor's Guide. 1771? ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Prob. 1, 1793: p. 305; 1799: p. 318 & Key p. 360. "Suppose two Women, and each a Son, were walking together, and were met by another Person, who asked the Boys in what Relation they stood to each other? They replied, We are Sons and Grandsons by the Fathers; Brothers and first Cousins by the Mothers; who also are Aunts to each of us. This Combination of Kinship once happened, but in what Manner? See Gen. xix. ver. 31." This is the same situation as in the Exeter Book Riddles, no. 46, above. Prob. 2, 1793: p. 305; 1799: p. 318 & Key p. 360. "Who was he that was begot before his Father, born before his Mother, and had the Maidenhead of his Grandmother?" The answer notes that Adam was made "of the Dust of the Ground", etc. and then runs: "Now Abel ... was murdered by his Brother Cain; therefore he got the Maidenhead of his Grandmother (the Earth); and was got before his Father (Adam, who was made of the Earth, therefore was not begotten; and was born before his Mother (Eve), who was made of Man, therefore was not born." I think the meaning is that Abel was buried so he was the first man in the Earth. Cf: Dialogue of Salomon and Saturnus, 14C; Adevineaux Amoureux, 1478. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Jackson. Rational Amusement. 1821. Arithmetical Puzzles. No. 6, pp. 2 & 52. Father, mother, son, grandson, brother and daughter comprise only 3 people in the situation of Alcuin/Bede 11a when a couple has a son. Curiosities for the Ingenious selected from The most authentic Treasures of Nature, Science and Art, Biography, History, and General Literature. 2nd ed., Thomas Boys, London, 1822. Singular intermarriage, p. 100. Man and daughter marry daughter and father. "My father is my son, and I am my mother's mother; / My sister is my daughter, and I'm grandmother to my brother." Richard Breen. Funny Endings. Penny Publishing, UK, 1999, p. 25. Gives the following. ┴┴┴┴"Here, beneath this stone, / Lie buried alone / The father and his daughter, / ┴┴┴┴The brother and his sister, / The man and his wife, / And only two bodies. // ┴┴┴┴┴┴Work it out... ┴┴┴┴Early 19th century, Erfunt Cemetery, Germany." ┴┴I suspect 'Erfunt' is a misprint for 'Erfurt' and I'm a bit suspicious as to the authenticity of this, but it's a good puzzle problem. Judge Leicester King (1789й1856) of Akron, Ohio, and his son married sisters, so he was his son's brotherйinйlaw. BIONй11. Illustrated Boy's Own Treasury. 1860. Arithmetical and Geometrical Problems, No. 32, pp.а430 & 434. Will involving 6 relatives who turn out to be just 3 due to the situation of Alcuin/Bede 11a. Charades, Enigmas, and Riddles. 1862: prob. 40, pp. 138 & 144й145; 1865: prob. 584, pp.а110 & 157й158. "How can a man be his own grandfather?" Mother and daughter marry son and father. Mother and son produce a son, Tom. Notes that perhaps Tom is only his own grandfatherйinйlaw. Leske. Illustriertes Spielbuch f└G└r M└└dchen. 1864? Prob. 564й12, pp. 253 & 395. 25 relationships among only 7 people. Answer is a couple, with their son and his wife, with their son and two daughters. This omits a grandchild, so there really ought to be 26 relationships here. Dudeney, AM 54, gives the same grouping, but with a different list of 23 relationships. However neither counts the relationships properly йй e.g. both count 4 children and 2 sons and 2 daughters. I find 23 reasonable relationships йй Dudeney's 23 less 4 repeated children plus 2 husbands and 2 wives that he omitted. Leske has the husbands and wives, but omits the grandmother and a grandchild. ┴┴┴┴In theory, n people can have n(nй1) possible relationships, but not all of these relationships have distinct names. E.g. in the above, the son of the first couple is a son to both parents, so two distinct relationships are both denoted by 'son'. However, in the classic problem of man, son and grandson, we actually have 2 fathers, 2 sons, 1аgrandfather and 1 grandson, giving a full 6 relationships among just 3 people. Extending this to a string of n generations gives the full n(nй1) relationships among n people. One might ask if one can compact this a bit by using fewer generations for the n people. E.g., Leske's problem has 3 generations. So I pose the following problem: for n people in g generations, how many of the n(nй1) relationships are distinctly named in English (where 'distinctly named' is a bit vague!). I now realise that the same relationship may have different names, indeed several different names! See Alcuin, above, and Carroll, below. Mittenzwey. 1880. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Prob. 25, pp. 3 & 59; 1895?: 30, pp. 9 & 63; 1917: 30, pp. 9 & 58. Two fathers and two sons are three people. Prob. 26. pp. 3й4 & 59; 1895?: 31, pp. 10 & 63; 1917: 31, pp. 9 & 58. 26 relationships among only 7 people, as in Leske. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨E. S. Turner. Op. cit. above, p. 109. [Retold in his: Amazing Grace; Michael Joseph, London, 1975; pp. 279й280.] The 7th Duke of Marlborough described to the House of Lords, c1880, the supposedly real case of a father and son marrying a daughter and mother. The son was his own grandfather and he became so confused that he committed suicide. In a footnote, Turner quotes a letter in the Welwyn NewsйChronicle of 1949 from a man who married his stepйmother's sister, i.e. widower and son married sisters. J. M. Letter: Genealogical puzzle. Knowledge 3 (6 Jul 1883) 13, item 865. + Answer to genealogical puzzle in our last. Ibid. (13 Jul 1883) 29. Two unrelated persons have the same brother. Editorial note to the Answer says there are several ways to solve the puzzle йй how many? Lewis Carroll. A Tangled Tale. (1885) = Dover, 1965. The dinner party. Knot II: Eligible apartments. Pp. 7й8 & 84й85. (In the answers, this part of the Knot is denoted └└1. The dinner party.) =аCarrollйWakeling II, prob. 41: Who's coming to dinner?, pp. 59 & 75. A man's: father's brotherйinйlaw; brother's fatherйinйlaw; fatherйinйlaw's brother and brotherйinйlaw's father are all the same person. This involves several marriages between cousins. ┴┴┴┴But brotherйinйlaw denotes both sister's husband and wife's brother! With an earlier marriage between cousins, we can have this person being both the man's father's wife's brother and father's sister's husband. His wife's brother's father is just his fatherкinйlaw, but another marriage between cousins makes this person also the man's brotherкinйlaw's fatherйinйlaw. E. W. Cole. Cole's Fun Doctor. The Funniest Book in the World. Routledge, London & E.аW. Cole, Melbourne, nd [HPL dates it 1886 and gives the author's name as Arthur C. Cole]. P. 57: A smart cutйout & Genealogy are two stories of widow and daughter marrying son and father. = Alcuin 11b. Lemon. 1890. How is this?, no. 725, pp. 83 & 123. 33 relatives who are only 8 people. Hoffmann. 1893. Chap. IX, no. 42: The family party, pp. 321 & 328-329 =аHoffmann-Hordern, p. 214. A man is his father's brother-in-law, his brother's father-in-law, his father-in-law's brother-in-law and his brother-in-law's father-in-law! C. C. Bombaugh. Facts and Fancies for the Curious. Lippincott, 1905, ??NYS. A Mr. Harwood and John Cosick, both widowers, married each other's daughter, at Durham in eastern Canada. Quoted in: George Milburn; A Book of Puzzles and Brainteasers; Little Blue Book No. 1103, Haldeman-Julius, Girard, Kansas, nd [1920s?], pp. 33-34. Pearson. 1907. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Part I, no. 25: A family party, pp. 120 & 184. Two widows and sons marry and each couple has a daughter, causing 24 apparent people to be only 6. Part I, no. 36: Quite a family party, pp. 123 & 186. Same as Hoffmann. Part I: A table of affinity, p. 127. Reports that M. de Lesseps and his son were to marry sisters and discusses the complications that would ensue. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Mr. X [cf 4.A.1]. His Pages. The Royal Magazine 26:6 (Oct 1911) 569. The mean Duke. Same as Hoffmann. Henry Edwards Huntington (1850й1927) married Arabella Duval Huntington ( й1924), the widow of his uncle Collis P. Huntington ( й1900) in 1913. Henry and Arabella were the same age. Dudeney. AM. 1917. Several examples, including the following. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Prob. 52: Queer relationships, pp. 8 & 153. Discusses two brothers who married two sisters. One man and one woman died and the survivors married and had a child. The man married his deceased wife's sister which was legal, so he is married to the woman and his child is legitimate. But the woman married her deceased husband's brother, which was not legal at the time, so she is not married to the man and her child is illegitimate!! Mentions a man who married his widow's sister and a man who has a nephew, but the nephew is not the nephew of his sister. Prob. 54: A family party, pp. 8 & 153. 23 apparent people are only 7. See Leske. Prob. 55: A mixed pedigree, pp. 8й9 & 153. A is B's father's brotherйinйlaw, brother's fatherйinйlaw and fatherйinйlaw's brother. Prob. 56: Wilson's poser, pp. 9 & 153. A is B's uncle and nephew. This is due to two men marrying the mother of the other and both couples producing йй the results are A and B. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Ahrens. A&N. 1918. Pp. 105-122, esp. 111-122. Describes an 11C report of a person with only three grandparents, but it turns out to be erroneous. However, Cleopatra had only three grandparents, since her parents were half-siblings йй I had believed they were full siblings which would give her only two grandparents. See also the 1730 entry about Christopher Burraway. ┴┴┴┴For convenience in the following, let n-parents denote one's ancestors n generations back, so 1-parents are parents, 2-parents are grandparents, 3-parents are great-grandparents, etc. Normally one has 2├├n── n-parents. ┴┴┴┴Prince Don Carlos of Spain (1545-1568) had only 4 3-parents, 6 а4-parents, 12аа5-parents and 20 6-parents. Ahrens also gives more extended examples, e.g. the 12 generations of ancestors of Kaiser Wilhelm II comprise only 1549 people instead of the expected 8190, and one person occurs in 70 places. Smith. Number Stories. 1919. Pp. 114-116 & 142. Tartaglia's problem. Also a more complex version where 27 apparent people are only 7. Ackermann. 1925. Pp. 92-93. Three mothers, each with two daughters, require only 7 beds. Loyd Jr. SLAHP. 1928. A puzzling estate, pp. 68 & 111. Three fathers and three sons are only four people. Collins. Fun with Figures. 1928. He was his own grandfather, pp. 231й232. Claims to quote a Pittsburgh newspaper story of a resident who committed suicide when he found he was his own grandfather. Son and father married widow and daughter. Dr. Th. Wolff. Die l└└chelnde Sphinx. Academia Verlagsbuchhandlung, Prague, 1937. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Prob. 4, pp. 188 & 197. 'My grandfather is only five years older than my father.' Prob. 5, pp. 188 & 197. 'My father and my grandfather are twins.' ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨HaldemanйJulius. 1937. No. 88: Marriage problem, pp. 11 & 25. How can a man have married his widow's sister? (Also entered under Deceased Wife's Sister.) Depew. Cokesbury Game Book. 1939. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Three ducks, p. 201. Two fathers and two sons make three people. Separate Room, p. 216. Three mothers, each with two daughters, make seven people. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Joseph Leeming. Riddles, Riddles, Riddles. Franklin Watts, 1953; Fawcett Gold Medal, 1967. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨P. 50, no. 5: "If your uncle's sister is not your aunt, just what relation is she to you? P. 53, no. 32: "If Dick's father is Tom's son, what relation is Dick to Tom?" P. 91, no. 3: "A doctor had a brother who went out West. But the man who went out West had no brother. How can this be?" P. 91, no. 4: "Two men, with their two wives and two sons, are related to each other as follows: The men are each other's fathers and sons, their wives' fathers and husbands, and their children's fathers and grandfathers. The women are the children's mothers and sisters; and the boys are uncles to each other. How can this be?" Same as Abbot Albert, prob. 12. P. 92, no. 8. Schoolteacher and his daughter, the minister's wife and the minister are just three people. P. 109, no. 11: "Sisters and brothers have I none, but that man's father is my father's son. Who am I looking at?" P. 113, no. 44: "What relation is that child to its father who is not its father's own son?" P. 113, no. 47: "Two Indians are standing on a hill, and one is the father of the other's son. What relation are the two Indians to each other?" P. 153, no. 30: "It wasn't my sister, not my brother, But it still was the child of my father and mother. Who was it?" ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Ripley's Believe It Or Not! 6th series, Pocket Books, NY, 1958. P. 145. Jacob van Nissen, of Zwolle, Holland, and his son married a girl and her mother. W. Leslie Prout. Think Again. Frederick Warne & Co., London, 1958. Catch Quiz, No. 6, pp. 12 & 115. "Said one boy to another: "My mother's sister is your sister's mother." What relation were the two boys?" Kathleen Rafferty. Dell Pencil Puzzles & Word Games. Dell, NY, 1975. Noodle Nudger 3, pp. 106 & 127. "Mary's husband's fatherйinйlaw is Mary's husband's brother's brotherкinйlaw, and Mary's sisterйinйlaw is Mary's brother's stepmother. HOW COME?" "Mary's father married her husband's sister." Scot Morris. The Book of Strange Facts and Useless Information. Doubleday, NY, 1979, p.а98. In order to give his divorced mother some benefit from his father's estate, Robert Berston adopted his mother in 1967. Patrick Donovan. Peculiar People. Fontana, 1984. P. 76 reports that Dave Woodhouse of Wolverhampton divorced his wife and married her mother in 1983 at a double wedding where his exйwife was also married to a new man. Ripley's Believe It or Not! й Strange Coincidences. Tor (Tom Doherty Associates), NY, 1990, p. 21. George Clark Cheever, of Warsaw, Indiana, three of his sons and one of his daughters all married siblings. Marcia Ascher. Ethnomathematics. Op. cit. in 4.B.10. 1991. Chapter Three: The logic of kinship relations, pp. 66й83. Gives a number of folk puzzles and references. ┴┴┴┴Two mothers and two daughters are three people (Brazil). ┴┴┴┴Two brothers say "My brother's son is buried there"; third brother says "My brother's son is not buried there" (Ireland). ┴┴┴┴Who is the sister of my aunt, who is not my aunt? (Puerto Rico). ┴┴┴┴His mother is my mother's motherйinйlaw (Russia). ┴┴┴┴Who is my mother's brother's brotherйinйlaw? (Wales). "Smallweed" diary column. The Guardian (10 Apr 1993) 18. Bill Wyman of the Rolling Stones married and later divorced Mandy Smith. Mandy's mother Patsy is about to marry Stephen Wyman, Bill's son from his first marriage. Item on front of Society section of The Guardian (2 Apr 1997) 1. "A 46йyearйold Bedfordshire woman has married the father of an 18йyearйold girl who eloped with her husband." David Singmaster. Some years before I read a description of some English aristocrat which said that he only had four grandparents while most people had eight. This was later corrected йй he had only four greatйgrandparents while most people had eight. Of course this puzzled me for a bit and I worked out a reasonable way this could have happened. I later realised that there is another way it could have happened, though I think the second method is slightly less likely. I posed the problem of finding both ways as a Brain Jammer: Four GreatйGrandparents; ├├Weekend Telegraph── (12 Dec 1998) 21 & (2 Jan 1999) 17. At the time I didn't know of any real examples, but Ahrens (1918) says Prince Don Carlos of Spain (1545-1568) had only four greatйgrandparents. I also used this on a Puzzle Panel program. When cousins marry, their children have six greatйgrandparents. Mindgames. Frontiers (a UK popular science magazine) No. 1 (Spring 1998) 107. "My wife's sister is also her cousin. How can this be?" Solution is that her father married his dead wife's sister and had another daughter. ┴┴├ ├9.E.1.┴┴THAT MAN'S FATHER IS MY FATHER'S SON, ETC.─ ─ ╨╨░дШМ А t                                                                    ░╨╨┴┴New section йй I have just found the 17C example, but there must be other and older examples?? But see Proctor, 1883. The problem continues to perplex people йй see Fairon, 1992. ┴┴For the 'Blind beggar' type of problem see: Boy's Own Book, 1828; Rowley, 1866; Rowley, 1875; Neil, 1880s; Lemon, 1890; Clark, 1897; Home Book, 1941; Charlot, c1950s. Again one thinks this should be older. ┴┴Some of the items given above, could go in here. ╨╨дШМ А t                                                                      ░╨╨ Tom└)└ Pinheiro da Veiga. Fastiginia o fastos geniales. (This work is a chronicle of courtly life in Castilla, from 1601 to 1606. Translated & edited by Narciso Alonso Cort└)└s. Imprenta del Colegio de Santiago, Valladolid, 1916, pp. 155bй156a, ??NYS.) French translation in: Augustin Redondo; Le jeu de l'└)└nigme dans l'Espagne du XVII├├e── si└/└cle. Aspect ludique et subversion; IN: Les Jeux └!└ la Renaissance; Actes du XXIII├├e── Colloque International d'└(└tudes Humanistes (Tours, 1980); J. Vrin, Paris, 1982; pp. 445й458, with the problem being on p. 453. There are two brothers, born of the same father and mother. One is my uncle, but the other isn't. How is this possible? The Book of Merry Riddles. London, 1629. ??NYS йй Santi 235 gives this and says it is reprinted in J. O. Halliwell, The literature of the sixteenth and seventeenth centuries, London, 1851, pp.а67-102, ??NYS, and as pp. 7й29 in Alois Brandl, Shakespeares Book of Merry Riddles und die anderen R└└thselb└G└cher seiner Zeit, Jahrbuch der deutschen ShakespeareйGesellschaft 42 (1906) 1й64, ??NYS. Bryant, pp.а100й102, quotes from: A Booke of Merrie Riddles, Robert Bird, London, 1631 and says it is also known as Prettie Riddles. Santi 237 gives Booke of Merrie Riddles, London, 1631, and says it is reprinted as pp. 53й63 in Brandl, ??NYS. Santi 307 gives The Booke of Merry Riddles, London, 1660, reprinted by J. O. Halliwell in 1866 in an edition of 25 copies, of which 15 were destroyed!, ??NYS. In Bryant, no. 275, pp. 102 & 336: "I know a child borne by my mother, / naturall borne as other children be, / that is neither my sister nor my brother. / Answer me shortly: what is he?" Boy's Own Book. Conundrums. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨"What kin is that child to its own father who is not its father's own son?" 1828:аNo.а70,аpp.а432а& 440. 1828-2:аNo.а70,аpp.а436 & 444; 1829а(US):аNo.а70, pp. 238 & 258; 1855:аNo.а102, pp. 583 & 595. (I think this is the forerunner of the Blind Beggar problem.) "What relation is your uncle's brother to you who is not your uncle?" 1828:аNo.а77,аpp.а433а&а440. 1828-2:аNo.а77,аpp.а437а&а444; 1829а(US):аNo.а77, pp. 238 & 258; 1855:аNo.а109, pp. 584 & 595. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨The Riddler. 1835. Conundrums Nos. 66 and 73, p. 16, no answers in my copy. These are identical to Nos. 70 and 77 in Boy's Own Book. Boy's Own Book. 1843 (Paris): 437 & 441, no. 11. Woman says: 'Your mother was my mother's only daughter.' = Boy's Treasury, 1844, pp. 425 & 429. = de Savigny, 1846, pp. 354 & 358, no. 8: 'sa m└/└re └!└ lui est la seule fille de sa m└/└re └!└ elle', which doesn't seem quite right to me. Child. Girl's Own Book. 1848: Enigma 46, p. 236; 1876: Enigma 37, p. 199. "His mother was my mother's only child." ┴┴┴┴= Fireside Amusements, 1850: No. 36, pp. 111 & 181; 1890: No. 21, p. 99. Fireside Amusements. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨1850: No. 17, pp. 134 & 184; 1890: No. 16, p. 110. "If Dick's father be John's son, what relation is Dick to John?" 1850: No. 29, pp. 135 & 185; 1890: No. 25, p. 110. "What kin is that child to its own father, who is not its own father's own son?" c= Boy's Own Book. 1850: No. 89, pp. 138 & 186; 1890: No. 25, p. 110. "What relation is your uncle's brother to you who is not your uncle?" = Boy's Own Book. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Charades, Enigmas, and Riddles. 1860: prob. 176 & 177, pp. 21 & 44; ┴┴┴┴1862: prob. 176 & 177, pp. 73 & 111. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Prob. 176. "If your uncle's sister is not your aunt, what relationship does she bear to you?" Prob. 177. ┴┴My mother had a child, my very own mother, ┴┴It was not my sister nor yet was it my brother; ┴┴If you are as clever as I fancy you to be, ┴┴Pray tell me what relation that child was to me. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Boy's Own Conjuring Book. 1860. P. 381. "I've no sister or brother; / You may think I am wild; / But that man's mother / Was my mother's child." Hugh Rowley. Puniana or Thoughts Wise and Otherйwise A New Collection of the Best Riddles, Conundrums, Jokes, Sells, etc, etc. Chatto & Windus, London, 1866. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨P. 36. Brothers A and B were walking. ""I must speak to those children," said A; "they are my nephews and nieces." "Ah!" said B, "as I have no nephews and nieces, I shall walk on." How was this?" P. 88. "What kin is that child to his own father who is not his own father's son?" P. 88. "If Dick's father is Tom's son, what relation is Dick to Tom?" P. 88. "Who were your grandfather's first cousin's sister's son's brother's forefathers? Why, his aunt's sisters, of course." (This is nonйlogical, being a pun on ancestors, but it illustrates that the idea of such problems must have been well known.) P. 137. A fiddler said his brother played the doubleйbass, but the doubleйbass player denied having a brother. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Hugh Rowley. More Puniana; or, Thoughts Wise and OtherйWhy's. Chatto & Windus, London, 1875. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨P. 28. "A blind beggar had a brother, who died and went to heaven. What relation was the blind beggar to the person who went to heaven?" Answer is 'sister', but the posing of the question is defective йй it should include the assertion that the person who went to heaven had no brother. P. 134. "What relation is that child to its own father, who is not its own father's son?" P. 217. "That gentleman's mother is my mother's only child." P. 231. "What relation is your father's only brother's sisterйinйlaw to you?" P. 231. "Brothers and sisters have I none, but this man's father is my father's son." ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨[Richard A. Proctor] Letters received and short answers. Knowledge 3 (26аOctа1883) 264. Answer to Harry. "Sisters and brothers have I none But that man's father is my father's son." Implies that the puzzle is not well known. Recalls it being posed on a ship and distracting all the passengers for a day. E. W. Cole. Cole's Fun Doctor. The Funniest Book in the World. Routledge, London & E.аW. Cole, Melbourne, nd [HPL dates it 1886 and gives the author's name as Arthur C. Cole]. P. 329: A riddle. "His mother was my mother's only child." James Neil [= "A Literary Clergyman"]. Riddles: And Something New About Them. (Lang Neil & Co., London, nd [1880s?]; Simpkin Marshall & Co., London); Village Games, London, 1993. General Riddles: Relationship, p. 28. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨"A blind beggar had a brother, the brother died, deceased had no brother. What relation was the blind beggar to deceased?" "What relation is that child to its own father who is not its own father's own son?" "If your uncle's sister is not your aunt, what relation is she to you?" ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Lemon. 1890. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨Do you see it?, unnumbered section after no. 80, pp. 15: "That gentleman's mother is my mother's only child." Conundrums, no. 142(a), pp. 23 & 102 (= Sphinx, no. 470(a), pp. 65 & 113): child "who is not his own father's son." ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Fireside Amusements йй A Book of Indoor Games. Op. cit. in 7.L.1. 1890? P. 99, no. 21. "His mother was my mother's only child." William Crompton. The odd halfйhour. The Boy's Own Paper 13 (No. 657) (15 Aug 1891) 731й732. A true friend. "If your uncle's sister is not your aunt, what relationship does she bear to you?" Bennett Coll. Prove it! The Idler 2 (1892й1893, probably Dec 1892) 510й517. Man in front of a portrait says "Sisters and brothers have I none; That man's father is my father's son." Says the portrait is himself! Observes that this leads to his father being his own son and being the father of his father. Describes the difficulties people have in trying to see this answer [not surprisingly]. Various other solutions given: grandfather, brother, uncle on the mother's side. Hoffmann. 1893. Chap. IX, no. 25: The portrait, pp. 318 & 326 =аHoffmannйHordern, p.а211. "Uncles and brothers have I none, But that man's father is my father's son." He notes "This venerable puzzle forms the subject of a humorous article, entitled "Prove It," in a recent number of the Idler. Its most amusing feature is that the writer has himself gone astray, ...." [I'm not sure whether Coll has gone astray or is using the error to generate humour??] W. H. Howe. Everybody's Book of Epitaphs Being for the Most Part What the Living Think of the Dead. Saxon & Co., London, nd [c1895] (reprinted by Pryor Publications, Whitstable, 1995). P. 165 has the following entry. ┴┴"In Llanidan Churchyard, Anglesea:йй ┴┴┴┴Here lies the world's mother, ┴┴┴┴By nature my Aunt йй sister to my mother, ┴┴┴┴My grandmother йй mother to my mother, ┴┴┴┴My great grandmother йй mother to my grandmother, ┴┴┴┴My grandmother's daughter and her mother." ┴┴Could this be a real case of 'I'm my own grandmother'?? Clark. Mental Nuts. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨1897, no. 3. The beggar. "A beggar had a brother, the brother died and the man who died had no brother." 1897, no. 16. The man in jail. "Brothers and sisters have I none, but that man's father is my father's son." ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Somerville Gibney. So simple! The Boy's Own Paper 20 (No. 992) (15 Jan 1898) 252. 'That very old catch йй "If Dick's father is Tom's son, What relation is Dick to Tom?"' Dudeney. "The Captain" puzzle corner. The Captain 3:1 (Apr 1900) 1 & 90 & 3:3 (Junа1900) 193 & 279. No. 3: Overheard in an omnibus. "Was that your father." "No, that gentleman's mother was my mother's motherйinйlaw." Essentially the same as: AM; 1917; Prob. 53: Heard on the tube railway, pp. 8 & 153; "That gentleman's mother was my mother's motherйinйlaw, but he is not on speaking terms with my papa." Hummerston. Fun, Mirth & Mystery. 1924. Grandfather's problems: The portrait, p. 68. "Sisters and brothers have I none, But that man's father is my father's son." James Joyce. Ulysses. (Dijon, 1922); Modern Library (Random House), NY, 1934, apparently printed 1946. P. 692 (Gardner says the 1961 ed. has p. 708; this is about 4/5 of the way between the start of Part III and Molly's soliloquy). "Brothers and sisters had he none, Yet that man's father was his grandfather's son." This is given as a quotation, while Bloom is looking in a mirror йй otherwise it could be a cousin. [Given in Bryant, no. 782, pp. 194 & 342.] Streeter & Hoehn. Op. cit. in 7.AE. Vol. 2, 1933, p. 16, no. 10: "Brain twister". "My son's father is your father's only child. What relative of yours am I?" Dr. Th. Wolff. Die l└└chelnde Sphinx. Academia Verlagsbuchhandlung, Prague, 1937. Prob.а33, pp. 194 & 204. 'This man's mother is my mother's only child.' HaldemanйJulius. 1937. No. 44: Portrait problem, pp. 7 & 22. Woman points to a man's portrait and says to her brother: "The man's mother was my mother's motherйinйlaw." Answer is that she is his daughter, but she might be his stepйdaughter. McKay. Party Night. 1940. No. 7, p. 175. "Brothers and sisters have I none; yet this man's father was my father's son." Meyer. Big Fun Book. 1940. No. 5, pp. 175 & 756. "My father is the brother of your sister. What relative am I of yours?" Answer is nephew, but son is also possible. The Home Book of Quizzes, Games and Jokes. Op. cit. in 4.B.1, 1941. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨P. 149, prob. 8. "Sisters and brothers I have none, but that man's father is my father's son." P. 149, prob. 10. "A beggar's brother died. But the man who died had no brother." ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨John Henry Cutler. Dr. Quizzler's Mind Teasers. Greenberg, NY, 1944. ??NYS йй excerpted in: Dr. Quizzler's mind teasers; Games Magazine 16:3 (No. 109) (Jun 1992) 47 & 43, prob. 14, with additional comments in Ibid. 16:4 (No. 110) (Aug 1992) 4 and 16:6 (No. 112) (Dec 1992) 4. "What relation is a man to his mother's only brother's only niece?" Answer is her brother, but comments point out that she could be his cousin, i.e. his mother's sister's daughter, or even a kind of cousinйinйlaw, i.e. his mother's brother's wife's sibling's daughter. Yvonne B. Charlot. Conundrums of All Kinds. Universal, London, nd [c1950?]. ┴┴P. 77: "If your aunt's brother is not your uncle, who is he?" ┴┴P. 82: "What kin are those children to their own father who are not their own father's sons?" Hubert Phillips. Party Games. Witherby, London, 1952. Chap. XIII, prob. 3: Photograph, pp.а204 & 252-253. ┴┴┴┴"Though sons and brothers have I none, ┴┴┴┴Your father was my father's son." ┴┴Solution says this "is my own invention". See Ascher in 9.E for some examples. Iona & Peter Opie. I Saw Esau: The Schoolchild's Pocket Book. (Williams & Norgate, London, 1947.) Revised edition, Walker Books, London, 1992, ??NX. No. 42, p. 45, just gives the rhyme; illustration on p. 44; answer on p. 144 just gives the answer, with no historical comments. Ripley's Believe It or Not!, 8th series. Pocket Books, 1962, p. 26. Jimmy Burnthet, of Emmerdale, England, courted a girl for 40 years. She broke off the engagement and married his nephew, whereupon he married her niece. Philip Kaplan. More Posers. (Harper & Row, 1964); MacfaddenйBartell Books, 1965. Prob. 16, pp. 25 & 88. "The father of the person in the portrait is my father's son, but I have no brothers or sons." Ripley's Puzzles and Games. 1966. ╨ дx ╨╨╨ШМ А th                                                                      д╨╨P. 12. "What relation is your mother's brother's brotherйinйlaw to you. Answer: Your father". There are several more answers. My mother's brother is my maternal uncle. He could have several sisters whose husbands are his brothersйinйlaw йй one is my father, the others are brothersйinйlaw of my mother. Also, he could be married and any of his wife's brothers are also his brothersйinйlaw. P. 13. "Moab and Benйamйmi were brothers yet cousins and their father was their grandfather." These are the sons of Lot by his daughters. P. 13. Bible puzzle. "Two people died who were never born" йй Adam and Eve. "Two people were born who never died" йй Enoch and Elias both just disappeared! "The oldest man who ever lived died before his father did" йй Methusaleh, the son of Enoch. ╨ ░x ╨╨╨дШМ А t                                                                      ░╨╨Pat Fairon. Irish Riddles. Appletree Press, Belfast, 1992; Chronicle Books, SanаFrancisco, 1992; pp. 27 & 60. "Brothers and sisters have I none But this man's father Was my father's son." Answer is "Oneself". ┴┴├ ├9.E.2.┴┴IDENTICAL SIBLINGS WHO ARE NOT TWINS─ ─ ╨╨░дШМ А t                                                                    ░╨╨ ┴┴Two siblings are born on the same day to the same parents but are not twins. New section. This must be older than the example below. ╨╨дШМ А t                                                                      ░╨╨Harold Hart. The World's Best Puzzles. Op. cit. in 7.AS. 1943. The problem of the two students, pp. 4 & 50 ┴┴├ ├9.F.┴