Sources page probability recreations

Prob. 8. Messengers, O-(50, 80), T = 15.

Prob. 16. One mason working faster than the other: O (140, 218), T = 39. [The 140 is misprinted as 218!]

Mahavira. 850. Chap VI, v. 320 327, pp. 177 179.

V. 320: O-(3, 8; 21, 0).

V. 321: MR-(5, 3; 216).

V. 323: O-(4, 8; 10, 2).

V. 325: O-(5, 8; 45, -8).

V. 327: O-(9, 13), D = 100.

Alcuin. 9C. Prob. 26: Propositio de campo et cursu canis ac fuga leporis. Hound catching hare, hound goes 9 to hare's 7, hare has 150 head start. = O-(7, 9), D = 150. (H&S 72 gives Latin and English.) The actual rates are not given, only their ratio.

Sridhara. c900.

V. 65 67(i), ex. 81 83, pp. 52 53 & 95. The verses give rules for various cases.

Ex. 81 82. O-(8/(5   ½), 3), T = 6   ¼.

Ex. 83. Slower goes 2, faster goes 8 and then returns from 100 away. When and where do they meet?

V. 96 98, ex. 111 112, pp. 78 79 & 96.

Ex. 111: O-(3, 1; 10, 0).

Ex. 112: O-(a, b; c, 2), T = 6. Determine a, b, c so that they will meet twice. Answer assumes a = 1, b = 6 and first meeting is at n = 10, (which gives c = 67) and then asserts there is a second meeting at n = 18.

al Karkhi. c1010. Sect. I, no. 5 9, p. 82.

5: O-(1, 1; 10, 0).

6: O-(1, 1; 11, 0), T = -5.

7: O-(1, 2; 10, 0).

8: O-(2, 2; 10, 0).

9: 10 + 15 + 20 + ... + 5(n+2) = 325. This could be considered as MR (10, 5; 0, 0; 325).

Tabari. Miftāh al-mu‘āmalāt. c1075. Pp. 103f. ??NYS -- quoted in Tropfke 593.

No. 4. O-(6, 9), T = 4.

No. 5. O-(1, 1; 30, 0).

No. 6. M-(5, 7).

Ibn Ezra, Abraham (= ibn Ezra, Abraham ben Meir = ibn Ezra, Abu Ishaq Ibrahim al Majid). Sefer ha Mispar. c1163. Translated by Moritz Silberberg as: Das Buch der Zahl ein hebräisch-arithmetisches Werk; J. Kauffmann, Frankfurt a. M., 1895. P. 56. Brothers meeting, MR-(17, 19; 100). (H&S 72 gives English.) Silberberg's note 121 (p. 109) says a similar problem occurs in Elia Misrachi, c1500.

Fibonacci. 1202. He has many examples. I give a selection.

P. 168 (S: 261-262). O-(1, 1; 20, 0).

P. 168 (S: 262). O-(1, 2; 21, 0).

P. 168 (S: 262). O-(2, 2; 30, 0).

Pp. 168-169 (S: 262). O-(3, 3; 60, 0).

P. 169 (S: 262). O-(5, 5; 60, 0).

P. 169 (S: 263). O-(3, 3; 10, 0). This has a non integral solution. He computes n = 5 2/3 days from the equation and then considers the travel on the 6th day to be at constant rates 10 and 18, with the first starting 5 ahead, so the overtaking is at 5 5/8 days.

Pp. 177 178 (S: 274): De duobus serpentibus [On two serpents]. Serpents approaching, MR (1/3   1/4,  1/5   1/6;  100).

Pp. 179 180 (S: 276): De cane et vulpe [On a dog and a fox]. O-(6, 9), D = 50.

P. 182 (S: 280): De duabus formicis quorum una imittatur aliam [On two ants, one of which follows the other]. An ant pursuing another. O (1/3   1/4, 1/5   1/6; 100).

P. 182 (S: 280): De duabus navibus se se invicem coniungentibus [On two ships that meet]. Two ships approaching, M (5, 7).

Yang Hui. Supplements to the Analysis of the Arithmetical Rules in the Nine Sections (=?? Chiu Chang Suan Fa Tsuan Lei). 1261. Repeats first problem of the Chiu Chang Suan Ching.

BR. c1305.

No. 24, pp. 42 43. Ship goes 380, starting 24 days after another and overtakes in 85 days. How fast does the slower ship go? This is O-(a, 380), T = 24 -- determine a such that the solution is 85.

General form: a, b = velocities of slower and faster ships; n  = time the faster ship sails; n + T = time the slower ship sails, i.e. the slower ship has T days headstart or the faster ship is delayed by T. This gives a(n+T) = bn. The above problem has b = 380, T = 24, n = 85 and asks for the other value, namely a. I will denote this by (a, 380, 24, 85), etc.

No. 41, pp. 60 61. (20, 25, 4, n) = O-(20, 25), T = 4.

No. 42, pp. 60 61. (20, b, 4, 16).

No. 43, pp. 62 63. (20, 25, d, 16).

No. 44, pp. 62 63. (a, 50, 5, 10).

No. 45, pp. 62 63. (15, 20, 10, n) = O-(15, 20), T = 10.

No. 46, pp. 64 65. MR-(31½, 21; 105).

No. 87, pp. 106 107. O-(24, 30), T = 4.

No. 88, pp. 106 107. Hare is 40 leaps ahead but hound's leap is 13/11 of hare's.

No. 94, pp. 114 117. MR-(18, 22; 240).

Gherardi?. Liber habaci. c1310.

P. 144. O-(1, 1; 25, 0).

P. 144. Couriers, M-(20, 30), D = 200.

Lucca 1754. c1330. F. 59r, p. 134. Couriers, M-(20, 30) with D = 200, though this is not used.

Munich 14684. 14C. Prob. VIII & XX, pp. 78 & 81. Discusses O (1, 1; k, 0).

Folkerts. Aufgabensammlungen. 13-15C. 18 sources for O-(1, 1; c, 0). Cites AR for extensions; Chiu Chang Suan Ching, Alcuin, Fibonacci.

See Smith, op. cit. in 3.

Bartoli. Memoriale. c1420. Prob. 28, f. 78r (= Sesiano, pp. 144 & 149-150. Fox is 121 (fox-)steps ahead of a dog. 9 dog-steps = 13 fox-steps. He computes 9/13 of 121. Sesiano notes that this assumes the fox stands still and determines how many dog-steps they are apart. The correct answer, which assumes that both steps take the same time, is that the dog gains 4 fox-steps for every 9 steps he makes, so he has to make 121 · 9/4 = 272 1/4 dog-steps, which are equal to 121 · 13/4 = 393 1/4 = 121 + 272 1/4 fox-steps.

Pseudo-dell'Abbaco. c1440.

Prob. 91, p. 78 with plate on p. 79 showing hound chasing fox holding a chicken. Fox is 40 fox steps ahead and 3 dog steps = 5 fox steps (assuming both steps take the same time). I have a colour slide of this.

Prob. 108, pp. 89 91 with plate on p. 90. M-(8, 5). Asks for time to meet, but also gives D = 60.

Prob. 118, pp. 96 97. O-(2, 2; 30, 0). = Fibonacci, p. 168.

Prob. 119, pp. 97. O-(3, 3; 18, 0).

AR. c1450. Several problems with arithmetic progressions which I omit.

Prob. 33, p. 37, 164 165, 177, 223. Hound and hare. Hare 100 ahead and goes 7 for each 10 of the hound.