Philippians 3 13 14

SOURCES -page 18.PROBABILITY RECREATIONSMost of the recreations in probability are connected with some paradoxical feature. A good exposition of most of these appears in the following.Gábor J. Székely. Paradoxes in Probability Theory and Mathematical Statistics. Akadémiai Kiadó, Budapest and Reidel, Dordrecht, 1986. [Revised translation of: Paradoxonok a Véletlen atematikában; Műszaki Könvkiadó, Budapest, nd.] Translated by Márta Alpár and Éva Unger. ??NYR.8.A.BUFFON'S NEEDLE PROBLEMR. 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 thenthe stick (baguette) problem, but doesn't give the answer.Buffon. Essai d'arithmétique morale, section 23. 1777. (Contained in the fourth volume of the supplement to his Histoire Naturelle, pp.101-104??) = Oeuvres Complètes de Buffon; annotated by M. Flourens; Garnier Frères, Paris, nd [c1820?], pp. 180-185. (Also in Miles & Serra, pp. 10-11.)Laplace. Théorie Analytique des Probabilitiés. 1812. Pp. 359-360. ??NYS. (In Miles & Serra, p. 12.) 3rd ed, Courcier, Paris, 1820, pp. 359-362. 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 problème de l'aiguille et le jeu du joint couvert. J. Math. pures appl. (2) 5 (1860) 273-286. Gives result for arbitrary curves and considers several grids. Also gives his theorem on curves of constant width.M. W. Crofton. On the theory of local probability. Philos. Trans. Roy. Soc. 158 (1869) 181-199. (Excerpted in Miles & Serra, pp. 13-15.)A. Hall. On an experimental determination of π. Messenger of Mathematics 2 (1873) 113-114. ??NYS.Tissandier. Récréations Scientifiques. 1880? 2nd ed., 1881, 139-145 describes the result and says 10,000 tries with a 50 mm needle on a floor with spacing 63.6 mm, produced 5009 successes giving π = 3.1421.= 5th ed., 1888, pp. 204-208. c= Popular Scientific Recreations, 1890? pp.729-731, but the needle is 2 in on a floor of spacing 2½ in and 5009 is misprinted as £5000 (sic) but 5009 is used in the calculation.J. J. Sylvester. On a funicular solution of Buffon's "Problem of the needle" in its most general form. Acta Math. 14 (1890-1891) 185-205.N. T. Gridgeman. Geometric probability and the number π. SM 25 (1959) 183-195. Debunks experimental results which are often too good to be true, although they are frequently cited.J. G. L. Pinhey. The Comte de Buffon's paper clip. MG 54 (No. 389) (Oct 1970) 288. Being caught without needles, he used paperclips. He derives the probability of intersection assuming a paper clip is a rectangle with semi-circular ends.Jack M. Robertson & Andrew F. Siegel. Designing Buffon's needle for a given crossing distribution. AMM 93 (1986) 116-119. Discusses various extensions of the problem.8.B.BIRTHDAY PROBLEMHow many people are required before there is an even chance that some two have the same birthday?George Tyson was a retired mathematics teacher when he enrolled in the MSc course in mathematical education at South Bank in about1980 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
SOURCES -page 2he 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 becausehe 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 --Probability of two coincident birthdays. AMM 52 (1945) 522 & 54 (1947) 170. Several solvers cite MRE, 11th ed. Solution is 23 or more.George Gamow. One, Two, Three ... Infinity. Viking, NY, 1947. =Mentor, NY, 1953, pp.204-206. Says 24 or more.Oswald Jacoby. How to Figure the Odds. Doubleday, NY, 1947. The birthday proposition, pp. 108-109. Gives answer of 23 or more.William Feller. An Introduction to Probability Theory and Its Applications: Vol 1. Wiley, 1950, pp. 29-30. Uses an approximation to obtain 23 or more. The 2nd ed., 1957, pp.31-32 erroneously cites von Mises, above.J. E. Littlewood. A Mathematician's Miscellany. Op. cit. in 5.C. 1953. P. 18 (38) mentions the problem and says 23 gives about even odds.William R. Ransom. Op. cit. in 6.M. 1955. Birthday probabilities, pp. 38-42. Studies usual problem and graphs probability of coincidence as a function of the number of people, but he doesn't compute the break-even point. He then considers the probability of two consecutive birthdays and gets upper and lower estimates for this.Gamow & Stern. 1958. Birthdays. Pp. 48-49. Says the break-even point is "about twenty-four".C. F. Pinzka. Remarks on some problems in the American Mathematical Monthly. AMM 67 (1960) 830. Considers number of people required to give greater than 50% chance of having 3, 4 or 5 with the same birthday. He gets 88,187, 314 respectively, using a Poisson approximation. He gives the explicit formula for having 3 with the same birthday.Charlie Rice. Challenge! Op. cit. in 5.C. 1968. Probable probabilities, pp. 32-36, gives a variety of other forms of the problem. A chooses five letters of the alphabet; B tries to guess at least one of them in five guesses. Author says odds are two to one in favor of B, though I get four to one. This is the same as getting five distinct items from a set of 21.Get people to think of cards. For 9 or more people, the probability of two the same is .52; for 11, .68; for 12, .75.