Sources page probability recreations

Prob. 148: De planetis, pp. 72, 164 165, 214. Though described as conjunction by Vogel, this is really just the discussion of the general overtaking problem on a circle. The text gives a general solution.

Benedetto da Firenze. c1465. P. 67. O-(1, 1; 30, 0); O-(1, 2; 30, 0); O-(3, 3; 60, 0).

The Treviso Arithmetic = Larte de Labbacho. Op. cit. in 7.H. 1478.

Ff. 54v 55v (= Swetz, pp. 158 160). Messengers, M-(7, 9) with D = 250. (English is also in Smith, Source Book I 12 and Isis 6 (1924) 330.)

Ff. 55v 56r (= Swetz, pp. 160 161). Hare & hound, rates 6 and 10, starting 150 apart. (English also in Isis 6 (1924) 330.)

Muscarello. 1478.

F. 73v, p. 189. Hare is 70 leaps in front of a hound. Each hound leap is 7/5 as big as a hare's, but in the same time. Says the hound catches the hare after 175 leaps.

F. 79v, p. 196. Couriers, M(70, 80).

Ff. 81r-81v, pp. 196-197. Couriers, O-(1, 2; 15, 0).

Ff. 82r-82v, pp. 198-199. Couriers, O-(2, 2; 15, 0).

della Francesca. Trattato. c1480. F. 40r (102). 25 + 25n = 1 + 2 + 4 + 8 + ... + 2^n-1 = 2ⁿ - 1. The exact solution is 7.785540889. He does linear interpolation on the seventh day, getting 7 73/103 = 7.7087378641. English in Jayawardene.

Ulrich Wagner. Das Bamberger Rechenbuch, op. cit. in 7.G.1. 1483.

Von Wandern, pp. 112 & 223, O-(1, 1; 6, 0).

Regel vom Hasen, pp. 113 & 223 224. Hare goes 12 for hound's 15 and starts 100 ahead.

Chuquet. 1484. See also 7.L for a problem with geometric progression.

Prob. 22. English in FHM 204. Meeting problem, M-(7, 9). He erroneously takes 7 and 9 as rates.

Prob. 98-108, FHM 220-221, are problems involving arithmetic progressions, usually overtaking problems, often with a fractional number of terms. 101 is a cask draining.

Prob. 98. O-(1, 1; 11, 0).

Prob. 99. O-(1, 1; 10, 0), T = -3.

Prob. 105, English in FHM 220-221. O-(3, 3; 14, 0). Chuquet says another author, Berthelemy de Romans, gets 8 4/13 instead of 8 1/3. FHM note that this assumes the speed is constant each day, while Chuquet assumes the speed increases through the day.

Prob. 117. x + (x+1) + (x+2) + ... + (x+5) = 30.

Prob. 127. M-(7, 8). See FHM 204 for comparison with prob. 22.

Borghi. Arithmetica. 1484.

Ff. 109v-109r (1509: f. 92r). Two ships going between Venice and Padua, MR (5, 8; 2400). (H&S 74 gives Italian.)

F. 109r (1509: ff. 92r-92v). O-(1, 1; 16, 0).

Calandri. Aritmetica. c1485.

F. 92r, p. 184. Ships meeting between Livorno and Marseilles -- M(7, 4).

F. 92v, p. 185. Hare 3000 in front of hound who goes 8 while the hare goes 5.

HB.XI.22. 1488. P. 52 (Rath 247). (Rath says it is similar to Alcuin, but with different numbers.)

Calandri. Arimethrica. 1491. F. 97v.

F. 65r. O-(3, 3; 60, 0).

F. 66r. Hare is 3000 hare leaps ahead of a hound, but the hounds step's are 8/5 as long as the hare's and take the same time.

F. 69r. Ships meeting between Pisa and Genoa. (3, 5). Same woodcut as used for cistern problems -- see 7.H.

Pacioli. Summa. 1494.

Ff. 39r-40r, prob. 1-6 = the six problems on Fibonacci 168-169.

F. 41r, prob. 13. Man travelling. 25+x + 2(25+x) + ... + 2^x 1(25+x) = 300, i.e. (2^x   1)(25 + x) = 300. Pacioli sets y = x - 3. After 3 days, he has gone 196 + 7y and he travels 224 + 8y on the 4th day. Pacioli interpolates both linearly to get 196 + 7y + y(224 + 8y) = 300, obtaining x = 3.44341.... I get x = 3.52567....

Ff. 41r-41v, prob. 14. 100 = 1 + 2 + 3 + ... + a = 1 + 3 + 5 + ... + 2b-1 =  2 + 4 + 6 + .. + 2c = 4 + 8 + 12 + ... + 4d. When should each start so they all arrive at once? (H&S 73 gives Italian and English and H&S 74 says he gives a problem involving flying about the earth, but I haven't found it.)

F. 41v, prob. 16-18. These are examples of the general form treated in BR, i.e. (a, b, T, n) gives the equation a(n+T)  =  bn.

Prob. 16. (30, 35, 5, n) = O-(30, 35), T = 5.

Prob. 17. (32, b, 6, 25).

Prob. 18. (a, 35, 4, 20).

F. 41v, prob. 19. O-(1, 1; 25, 0).

F. 41v-42r, prob. 20. 25n = 1 + 2 + ... + 2^n-2. He gets 8 73/103 = 8.70874... by interpolation on the 8-th day. I get 8.78554.... = della Francesca.

F. 42r, prob. 21. M-(4, 5) with D = 100.

Ff. 42v-43r, prob. 27. Hound and hare. Hare is 60 ahead and 5 hound leaps = 7 hare leaps. Says the problem is very unclear. Assumes the 60 are hare leaps and interprets the relation as saying that 7 hare leaps are the same as 5 hound leaps and leaps take the same time. Later, he supposes the 60 are hound leaps and converts this to 84 hare leaps. See Cardan, whose description implies both the lengths and times of the leaps differ.

Swetz, op. cit. as Treviso Arithmetic in 7.H, p. 244, says this is the first listing of variants of the problem -- but we've seen lots of examples before and many of Pacioli's are taken from Fibonacci. Swetz gives examples from Wagner, Calandri (1491) and Köbel (1514).

PART II.

F. 64, prob. 80. Overtaking on a circular island with speeds 1, 1/12. Equivalent to O (1, 1/12) with headstart of distance 1.

F. 64, prob. 81. Meeting on a circular island with speeds 1, 1/12. Same as M(1, 12).

Blasius. 1513. F. F.iii.v: Decimasexta regula. O-(10, 13), T = 9.

Köbel. 1514. ??NYS -- given in H&S 73. O-(10, 13), T = 9.

Tagliente. Libro de Abaco. (1515). 1541.

Prob. 111, ff. 55v-56r. Couriers meeting between Rome and Venice -- M-(17, 20).

Prob. 114, ff. 56v-57r. O-(1, 1; 30, 0).

Prob. 118, f. 58r. Ships meeting between Venice and Candia -- M (3, 5).

Prob. 122, ff. 59r-59v. Hound chasing goat. Goat is 50 leaps ahead and hound makes 7 to goat's 5.

Ghaligai. Practica D'Arithmetica. 1521.

Prob. 17, f. 64r. O-(1, 1; 20, 0).

Prob. 18, f. 64r. M-(4, 5) with D = 80.

Tonstall. De Arte Supputandi. 1522.