Sources page probability recreations

A chooses five letters of the alphabet; B tries to guess at least one of them in five guesses. Author says odds are two to one in favor of B, though I get four to one. This is the same as getting five distinct items from a set of 21.

Get people to think of cards. For 9 or more people, the probability of two the same is  .52; for 11,  .68; for  12,   .75.

Get people to count their change. For a moderate number of people, it is likely that two have the same amount (or the same number of coins). Likewise, with a moderate number of people, two are likely to have fathers (or mothers) with the same given name.

He gives magic numbers, i.e. the size of the set to be selected from, for different sizes of group in order that a duplication is more likely than not. For 6 people, the magic number is 23; for 8, 43; for 12, 99.

Howard P. Dinesman. Superior Mathematical Puzzles. Op. cit. in 5.B.1. 1968. No. 58: The birthday puzzle, pp. 76, 116-117 & 122. Asks for probability of a shared birthday among 30 people. Says the problem was introduced by Gamow. Answer is .70. He adds that 24 or more will give better than even odds and then asks how many people are necessary for one of them to have his birthday on a given day -- e.g. today. Here the answer is 253.

E. J. Faulkner. A new look at the probability of a coincidence of birthdays in a group. MG 53 (No. 386) (Dec 1969) 407 409. Suggests the probability should be obtained by the ratio of the unordered selections, i.e. Prob. (r distinct birthdays)  =  BC(365, r)/BC(364+r, r), rather than P(365, r)/365^r. But the unordered selections with repetitions are not equally likely events -- see Clarke & Langford, below. For his approach, the breakeven number is r = 17.

Morton Abramson & W. O. J. Moser. More birthday surprises. AMM 77 (1970) 856 858. If a year has n days, k  1 and p people are chosen at random, what is the probability that every two people have birthdays at least k days apart? For n = 365, k =   1,   2,   3, 4, 5, 6, 7, 8, 9, 10, the breakeven numbers are p = 23, 14, 11, 9, 8, 8, 7, 7, 6, 6.

L. E. Clarke & Eric S. Langford. Note 3298 -- I & II: On Note 3244. MG 55 (No. 391) (Feb 1971) 70 72. Note the non equally likely events in Faulkner.

W. O. J. Moser. It's not a coincidence, but it is a surprise. CM 10 (1984) 210 213. Determines probability P that in a group of k people, at least two have birthdays at most w days apart. This turns out to have a fairly simple expression. To get P > .5 with w = 0 requires k  23, the classical case. With w = 1, it requires k  14, ....

Tony Crilly & Shekhar Nandy. The birthday problem for boys and girls. MG 71 (No. 455) (Mar 1987) 19 22. In a group of 16 boys and 16 girls, there is a probability greater than ½ of a boy and a girl having the same birthday and 16 is the minimal number.

Roger S. Pinkham. Note 72.25: A convenient solution to the birthday problem for girls and boys. MG 72 (No. 460) (Jun 1988) 129 130. Uses an estimate to obtain the value 16 of Crilly and Nandy.

M. Lawrence Clevenson & William Watkins. Majorization and the birthday inequality. MM 64:3 (1991) 183-188. Do the numbers necessary for P > .5 get bigger if birthdays are not random? Answer is "no" and it is a result in majorization theory, but they give an elementary treatment.

S. Ejaz Ahmed & Richard J. McIntosh. An asymptotic approximation for the birthday problem. CM 26:2 (Apr 2000) 151-155. 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/(1-p)}]. For p = .5, n = 365, this gives 22.494.
8.C. PROBABILITY THAT A TRIANGLE IS ACUTE
See also 6.BR, esp. the Mathematical Log article and my comments. I have a large number of similar results, mostly by myself, in a file. I estimate there are about 20 possible answers, ranging from 0 to 1.
J. J. Sylvester. On a special class of questions in the theory of probabilities. Birmingham British Association Report (1865) 8. = The Collected Mathematical papers of James Joseph Sylvester, (CUP, 1908); Reprinted by Chelsea, 1973, item 75, pp. 480-481. ??NYS, but Guy (below) reports that it is a discursive article with no results. Attributes problem for three points within a circle or sphere to Woolhouse but feels the problem is not determinate.

C. Jordan. 1872 1873. See entry in 8.G.

E. Lemoine. 1882-1883. See entry in 8.G.

L. Carroll. Pillow Problems. 1893. ??NYS. 4th ed., (1895). = Dover, 1958. Problem 58, pp. 14, 25, 83 84. Prob (acute) = .639.

C. O. Tuckey. Note 1408: Why do teachers always draw acute angled triangles? MG 23 (No. 256) (1939) 391 392. He gets Prob(obtuse) varying between .57 and .75.

E. H. Neville. Letter: Obtuse angling -- a catch. MG 23 (No. 257) (1939) 462. In response to Tuckey, he shows Prob(obtuse) = 0 and deduces that Prob(acute) = 0 (!!!).

Nikolay Vasilyev. The symmetry of chance. Quantum 3:5 (May/Jun 1993) 22-27 & 60-61. Survey on geometric probability. Asks for the probability of an acute triangle when one takes three points at random on a circle and gets ¼.

Richard K. Guy. There are three times as many obtuse-angled triangles as there are acute-angled ones. MM 66:3 (Jun 1993) 175-179. Gives 12 different approaches, five of which yield Prob(acute) = ¼, with other values ranging from 0 to .361. He tracked down the Sylvester reference -- see above.

In Mar 1996, I realised that the two approaches sketched in 6.BR give probabilities of acuteness as 0 and 2 - π/2 = .429.
8.D. ATTEMPTS TO MODIFY BOY GIRL RATIO
This is attempted, e.g. by requiring families to stop having children after a girl is born.
Pierre Simon, Marquis de Laplace. Essai Philosophique sur les Probablitiés (A Philosophical Essay on Probabilities). c1819, ??NYS. Translated from the 6th French ed. by F. W. Truscott & F. L. Emory, Dover, 1951. Chap. XVI, pp. 160 175, especially pp. 167 169. Discusses whether the excess of boys over girls at birth is due to parents stopping having children once a son is born.

Gamow & Stern. 1958. A family problem. Pp. 17 19.

8.E. ST. PETERSBURG PARADOX
Nicholas Bernoulli. Extrait d'une Lettre de M. N. Bernoulli à M. de M... [Montmort] du 9 Septembre 1713. IN: Pierre Rémond de Montmort; Essai d'analyse sur les jeux de hazards. (1708); Seconde edition revue & augmentee de plusieurs lettres, (Quillau, Paris, 1713 (reprinted by Chelsea, NY, 1980)); 2nd issue, Jombert & Quillau, 1714. Pp. 401-402. Quatrième Problème & Cinquième Problème, p. 402. [See note to Euler; Vera aestimatio sortis in ludis; op. cit. below, pp. 459 461.] In the 4th problem, he proposes pay-offs 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 pay-offs are 1, 2, 4, 8, ... or 1, 3, 9, 27, ... 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 (1730-31(1738)) 175-192, ??NYS. IN: Die Werke von Daniel Bernoulli; ed. by L. P. Bouckaert & B. L. van der Waerden; Birkhäuser, 1982; pp. 223-234 and notes by van der Waerden, pp. 195 & 197-200. English translation in Econometrica 22 (1954) 23-36, ??NYS.

Lewis Carroll. Lionel Tollemache, Reminiscences of Lewis Carroll, Literature (5 Feb 1898), ??NYS, quoted in Carroll-Wakeling II, prob. 28: A good prospect, pp. 44 & 72. (Tollemache was at Balliol College, Oxford, in 1856-1860. He then entered Lincoln's Inn, London, so this must refer to c1858.

He says that Carroll gave the problem with a coin and pay-offs 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

Σ_n=0 (2ⁿ-1)(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 ½d. 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 pay-offs in order to make the latter process work.

Leonhard Euler. Vera aestimatio sortis in ludis. [Op. postuma 1 (1862) 315-318.] = Op. Omnia (I) 7 (1923) 458-465.

Tissandier. Récréations Scientifiques. 1880? 2nd ed., 1881, p. 140 gives a brief unlabelled description, saying this "problème de Pétersbourg" was discussed by Daniel Bernoulli in "Mémoires de l'Académie de Russie". Not in the 5th ed. of 1888.

Tissandier. Popular Scientific Recreations. 1890? Pp. 727-729 discusses the idea and says D. Bernoulli presented his material on this in "Memoires [sic] de l'Académie de Russie". This is somewhat longer than the material in the 2nd French ed of 1881.

Anon. [presumably the editor, Richard A. Proctor]. Strange chances. Knowledge 10 (Oct 1887) 276-278. Brief discussion of "the famous Petersburg problem".

C. S. Jackson. Note 438: The St. Petersburg problem. MG 8 (No. 116) (Mar 1915) 48. Notes that the value of the game is not more than log₂ of the bank's funds.

Dan Pedoe, The Gentle Art of Mathematics, op. cit. in 5.C, 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) 13-39. ??NYR.

Nick Mackinnon & 5Ma. Note 74.9: A lesson on the St. Petersburg paradox. MG 74 (No. 467) (1990) 51 53. Suppose a maximum is put on the payment -- i.e. the game stops if n heads appear in a row. How does this affect the expected value? They find n = 10 gives expected value of 6.
8.F. PROBLEM OF POINTS
A game consists of n points. How do you divide the stake if you must quit when the score is a to b to ....? This problem was resolved by Fermat and Pascal in response to a question of the Chevalier de Méré and is generally considered the beginning of probability theory. The history of this topic is thoroughly covered in the first works below, so I will only record early or unusual occurrences.

NOTATION: Denote this by (n; a, b, ...).

GENERAL HISTORIES
Florence Nightingale David. Games, Gods and Gambling. The origins and history of probability and statistical ideas from the earliest times to the Newtonian era. Griffin, London, 1962.

Anthony W. F. Edwards. Pascal's Arithmetical Triangle. Griffin & OUP, London, 1987. This corrects a number of details in David.

David E. Kullman. The "Problem of points" and the evolution of probability. Handout from talk given at MAA meeting, San Francisco, Jan 1991. Uses (6; 5, 3) as a common example and outlines approaches and solutions of: Pacioli (1494) -- 5 : 3; Cardan (1539) -- 6 : 1; Tartaglia (1539) -- 2 : 1; Pascal (1654) -- 7 : 1; Fermat (1654) -- 7 : 1; Huygens (1657) -- 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.