Sources page probability recreations

Quest. 35, p. 167. O-(20, 33), T = 6.

Quest. 37, pp. 168-169. Couriers, M-(3, 5).

Quest. 38, p. 169. Couriers, M-(⅓, ½).

Riese. Die Coss. 1524.

No. 50, p. 47. Messengers, MR-(10, 8; 40).

No. 51, p. 47. O-(6, 9), D = 100.

No, 109, p. 54. Messengers, MR-(7, 9; 300).

No. 133, p. 59. Fox is 300 fox leaps ahead of hound. Each hound leap is 31/20 of a fox leap and they leap at the same rate.

Apianus. Kauffmanss Rechnung. 1527.

F. M.v.r. Couriers between Leipzig and Venice. M-(18, 24).

F. M.viii.v. O-(7, 9), D = 64.

Ff. M.viii.v - N.i.r. Messengers between Prague and Vienna. MR-(7, 9; 33).

F. N.i.r. Hare is 30 leaps ahead of a hound. Hound makes 8 leaps to hare's 6, assumed same size. I.e. O-(8, 6), D = 30.

Ff. N.i.r - N.i.v. Hare is 50 hare leaps ahead of a hound. Hare makes 4 leaps while hound makes 3, but 2 hound leaps equal 3 hare leaps in length. This can be viewed as O-(8, 9), where the units are hare leaps per period in which a hare makes 8 leaps, with a head start of 50 hare leaps.

van Varenbraken. 1532. ??NYS -- cited by the editor of King, 1795, p. 154. Hound and hare -- O-(12, 15), D = 200.

Cardan. Practica Arithmetice. 1539. Chap. 66.

Section 7, ff. BB.viii.v - CC.i.r (p. 137). 10n  =  1 + 4/3 + (6/5)(4/3) + (4/3)(6/5)(4/3) + (6/5)(4/3)(6/5)(4/3) + ....

Section 11, ff. CC.ii.v - CC.iii.r (p. 138) (iii is misprinted ii). Hound & hare. Hare is 60 hound leaps ahead. Hound makes 63 leaps to hare's 100, but 61 hound leaps equal 140 hare leaps. Says Pacioli does it wrong, but I don't see that this is the same as Pacioli's problem?? [The phrasing of the problem is awkward. H&S 72 gives the Latin and a mistranslation of it.]

Section 12, f. CC.iii.r (p. 139) (iii is misprinted ii). O-(5, 3; 20, 0).

Section 13, ff. CC.iii.r - CC.iii.v (p. 139) (iii is misprinted ii). 1 + 2 + 4 + 8 + 16 + ... + 3 + 4 + 6 + 9 + 13 + ... = 330. He interpolates between the 7th and 8th days to get 7 95/137 = 7.69343, while the correct(?) answer is 7.83659. Says Pacioli could not do this.

Section 14, ff. CC.iii.v - CC.iiii.r (p. 139) (iii is misprinted ii). Birds flying around the world. 1 + 2 + 3 + 4 + ... + 1 + 8 + 27 + 64 + ... = 44310. Seems to get 20½ days, but 20 is the answer. (H&S 74 gives Italian and says similar problems are in Pacioli and Stifel.)

Section 15, ff. CC.iiii.r - CC.iiii.v (pp. 139-140).

1 + 2 + 4 + 8 + ... = 2 + 4 + 6 + 8 + ....

Section 66, ff. FF.ii.r - FF.ii.v (p. 155). (66 is not printed in the Opera Omnia). O (3, 5; 30, 0).

Section 154, ff. MM.iiii.r - MM.iiii.v (p. ??). An overtaking problem, but with times for the entire route given which can be considered as M-(10, -7) with delay of 2, giving (n+2)/10 = n/7. Described in H&S 73.

Buteo. Logistica. 1559.

Prob. 21, p. 219. A simple hound and hare problem, O (6b/5, b), D = 1.

Prob. 28, pp. 229-230. States O-(1, 1; 22, 0), but does O (1, 1; 18, 0).

Prob. 29, pp. 230-231. O-(1, 1; 20, 0).

Prob. 30, p. 231. O-(2, 2; 15, 0).

Prob. 31, pp. 231-233. O-(24, -1; 10, 2). Finds common distance of 199⅜ by linear interpolation on 11th day.

Prob. 32, pp. 233-234. M(3, 2) with D = 200 given.

Baker. Well Spring of Sciences. 1562? ??check if this in the Graves copy of the 1562/1568 ed.)

Prob. 3, 1580?: ff. 36v-37r; 1646: pp. 62-63; 1670: pp. 76-77. O-(1, 1; 8, 0).

Prob. 4, 1580?: ff. 37r-37v; 1646: pp. 63-65; 1670: pp. 77-78. MR-(1, 1; 2, 2; 200). He notes that they meet when the first has done 200/3 = 66 2/3, which is 2/3 on the 12th day. He assumes they travel at constant speed on each day, so they meet after 11 1/18 = 11.0555.... [18 is misprinted 28 at one point.] But if the velocity is increasing linearly, so that the distance covered to time t is t(t+1)/2, then the problem leads to a quadratic with solution t = 11.0578....

Gori. Libro di arimetricha. 1571.

F. 72r (p. 78). Deer is 60 ahead of a dog, which goes 7 for every 4 of the deer, i.e. O-(4, 7), D = 60.

F. 72v (p. 78). Fox is 80 ahead of a dog, which goes 7 for every 5 of the fox, i.e. O (5, 7), D = 80.

F. 81v (p. 79). Couriers between Paris and Siena meeting -- M(20, 24) with D = 800, but not used.

F. 82r (p. 79). Couriers meeting -- MR(40, 50, 1000).

F. 71r (p. 79). O-(1, 1; 30, 0).

F. 81v (p. 79). O-(1, 1; 40, 0).

van Halle. 1568. ??NYS -- cited by the editor of King, 1795, p. 155. O (11, 0; 1, 1).

Io. Baptiste Benedicti (= Giambattista Benedetti). Diversarum Speculationum Mathematicarum, & Physicarum Liber. Turin, (1580), Nicolai Bevilaquæ, Turin, 1585; (Venice, 1599). [Rara 364. Graves 141.f.16.] This has a number of overtaking and meeting problems, but he makes diagrams showing the sums of the arithmetic progressions involved.

Theorema CVI, pp. 68-69 (misprinted 70-71). O-(4, 1; 1, 1) with diagram showing 1 + 2 + ... + 7 = 4 x 7.

Theoremae CVII & CVIII, pp. 69 (misprinted 71) - 70. O-(k, 0; 1, 2) with diagrams showing 1 + 3 + ... + 15 = 8 x 8 and 1 + 3 + ... + 13 = 7 x 7.

Theoremae CIX & CX, pp. 70-71. O-(k, 0; 2, 2) with diagrams showing 2 + 4 + ... + 14 = 7 x 8 and 2 + 4 + ... + 16 = 8 x 9.

Theorema CXI, p. 71. O-(k, 0; 3, 3) with diagram showing 3 + 6 + ... + 21 = 7 x 12.

Theorema CXII, pp. 71-73. O-(11, 0; 3, 3), which does not have an integral solution. He draws a diagram of 3 + 6 + ... + 21 and tries to make it into a rectangle of side 11, but sees this does not work. He then sees the answer is between 6 and 7 days, so he does linear interpolation over this period, getting 6.3 days instead of the correct 6 1/3. He then uses a simple triangle to show this interpolation process.

Theorema CXIII, pp. 73-74. O-(a, b) with first having headstart of time T done in general. Does O-(20, 25), T = 8. Gives a diagram of a x n+T = b x n. There is an Appendix to this Theorema on pp. 74-75 which applies it to Jupiter overtaking Saturn but I can't make it out.