Sources page probability recreations

Pp. 198-199, no. 117. Hare is 50 springs ahead of a hound. Their springs are of equal length, but the hound makes 27 while the hare makes 25. How many springs does the hare make before being overtaken? Answer is misprinted 675 instead of 625.

P. 360, no. 47. = Vyse, prob. 12.

P. 360, no. 48. See in 7.P.6.

John Radford Young (1799-1885). Simple Arithmetic, Algebra, and the Elements of Euclid. IN: The Mathematical Sciences, Orr's Circle of the Sciences series, Orr & Co., London, 1854. [An apparently earlier version is described in 7.H.]

No. 5, p. 177. O-(9, 5), T = 10.

No. 5, p. 216. A and B travel 90 miles. A goes 1 mile per hour faster than B and arrives 1 hour before him. What were their speeds?

No. 8, p. 216. Same as last with data 150, 3, 8⅓.

No. 4, p. 228. MR-(1, 1; 20, -2; 165).

Vinot. 1860. Art. LXIII: Problème du Renard et du Lévrier. Fox is 72 (fox-)leaps in front of a greyhound. Fox makes 9 leaps to hound's 5, but fox-leaps are 3/7 the size of hound-leaps.

Edward Brooks. The Normal Mental Arithmetic A Thorough and Complete Course by Analysis and Induction. Sower, Potts & Co., Philadelphia, (1858), revised, 1863. Further revised as: The New Normal Mental Arithmetic: A Thorough and Complete Course by Analysis and Induction; Sower, Potts & Co., Philadelphia, 1873. Many examples. I mention just one example.

1863 -- pp. 132-13, no. 19; 1873 -- p. 161, no. 18. "E takes 60 steps before he is overtaken by D; how many steps does D take to catch E, provided E takes 4 steps while D takes 3, and 5 of D's equal 8 of E's, and how far ahead was E when they started?"

Todhunter. Algebra, 5th ed. 1870. Several examples, including the following.

Examples X, no. 26, pp. 86 & 577. Hare is 80 hare leaps ahead of a greyhound who starts after her. Hare does three leaps for every two of the hound, but hare leaps are half the size of hound leaps. How many leaps does the hare make before it is caught?

Examples XXX, no. 53, pp. 268 & 589. O-(5, 0; 3, ½) with first having a headstart of time 4½.

Examples XXXIII, no. 7, pp. 285 & 590. O-(1, 1; 12, 0) with first having headstart of time 5. [He says the second overtakes the first after the first travels 36, i.e. after time 8, but the first then overtakes the second after time 15!]

Miscellaneous Examples, no. 77, pp. 550 & 605. A and B start on a circular walking race. After 30 min, A has done three circuits and B has done 4½. When do they meet again? This is just O-(6, 9) with the first having a headstart of distance ½.

Mittenzwey. 1880. Prob. 78, pp. 15 & 66; 1895?: 85, pp. 19 & 69; 1917: 85, pp. 18 & 65. Hare is 50 leaps ahead and makes 6 leaps while the hound makes 5, but 7 houndleaps are as long as 9 hare leaps. How leaps does the hare make before being caught?

William J. Milne. The Inductive Algebra .... 1881. Op. cit. in 7.E. No. 4, pp. 166 & 334. General problem: ax = b (n - x). Solves for n   x.

Clark. Mental Nuts. 1897, no. 46; 1904, no. 62. The fox and the hound. Fox is 60 (fox )leaps in front of a hound. Hound takes 6 leaps to fox's 9, but the hound leaps are 7/3 times as long as the fox's. (= Lauremberger)

Dudeney. Weekly Dispatch (17 May 1903) 13 & (14 Jun 1903) 16.

Perelman. 1937. MCBF. At the cycle track, prob. 150, pp. 255-256. Circular track of circumference 170m. When cyclists are going in opposite directions, they meet every 10 sec; when going the same direction, the faster passes the slower every 170 sec.

Haldeman-Julius. 1937. No. 109: Hare and hounds problem, pp. 13 & 26. Gives a verse version -- the only one I have seen -- and says the problem 'is about 150 years old.'

As I was walking o'er my forest grounds

Up jumped a hare before my two greyhounds.

The distance that she started up before

Was fourscore rods, just, and no more.

My dogs did fairly run

Unto her 16 rods just 21.

Now I would have you unto me declare

How far they ran before they caught the hare.

This is O-(a, [21/16]a) with headstart D = 80.

C. Dudley Langford. Note 1558: A graphical method of solving problems on "Rate of Work" and similar problems. MG 25 (No. 267) (Dec 1941) 304-307. + Note 2110: Addition to Note 1558: "Rate of Work" problems. MG 34 (No. 307) (Feb 1950) 44. Uses a graph to show (a, b) cistern problems as meeting problems. Also solves problems (A, x) in B and (a, -b), the latter appearing as an overtaking problem. The Addition gives a clearer way of viewing (a,  b) problems as overtaking problems

Gamow & Stern. 1958. Pp. 9 10, 59 63. Elevator problem.
10.A.1. CIRCLING AN ARMY
The linear form has an army of length L moving with velocity v. A rider goes at velocity V from the rear to the front and then back to the rear, reaching the rear when it has advanced d. How far, D, does he go? Since one version gives an erroneous answer, I will give the basic equations. Let t be the time for the rider to reach the front of the army and T be the time to get from the front back to the rear. We then have: Vt = vt + L; v(t+T) = d; d = V(t-T); D = V(t+T). Here may be insufficient equations to determine all the values (as also occurs in 10.A.3), but the value of D can be found as D  =  L + (L² + d²). Note that D/d = V/v. Other versions of the problem can be solved, sometimes more simply, e.g. L = (V² - v²)d/2vV; if V = rv, then L = (r²-1)d/2r.

If the army has a width W, and the rider goes across the moving army at each end, the situation is more complex -- see the first example and Loyd.

J. Gale, proposer; Joseph Edwards, Jr. & Mr. Coultherd, solvers. Question III. A Companion to the Gentleman's Diary; ... for the year 1798, pp. 59-60 & The Gentleman's Mathematical Companion, for the year 1799, pp. 16-17. Wagoner walking around his wagon and team while it is travelling. L = 20 yd, W = 4 yd. He can walk V = 4 mph and he walks 74⅔ yd in his circumambulation (hence taking T = 7/11 min). A complication arises as to how he goes crossways. The proposer says he turns at right angles and passes 2 yd clear at the front and at the back. (He also says the walker passes 2 yd clear on each side, but he never gives a width, taking the distance between the two side paths as 4, which I have accounted for by taking W = 4. In some cases, we can account for the 2 yd at each end by taking L = 24.) How fast, v, is the team going? Both solvers get v = 2 mph, but I find neither is viewing the problem correctly and neither has correctly formulated the problem he has described!

Edwards starts in the middle, 2 in front of the horses, and says the the walker must go 2 to the left, 24 back and 2 to the right to get the the middle in back. He then says the time required is 28/(V+v). But this assumes the crossways motion is at the same relative speed as the lengthwise relative motion and seems definitely incorrect to me. Edwards then says that a similar argument gives the time in the other direction as 28/(V-v). Setting the sum of these equal to T does give v = 2.

Coultherd starts in the middle, 2 behind the wagon, but I will rephrase it to parallel Edwards' solution. It takes the man 2/V to go 2 to the left. During this time, the wagon moves ahead 2v/V, so he is now only 2 - 2v/V in front. When he gets to the back, he only needs to be 2 - 2v/V behind the wagon when he turns to cross, so that he will be exactly 2 behind at the middle. So he must make a relative motion of 24   4v/V at the relative velocity V + v. Similarly, for the forward trip, he makes relative motion 24 + 4v/V at relative velocity V - v. Adding the times for these to 8/V for the crossways trips and setting equal to T, I get v = 2.056 mph, which seems to be the correct answer. Coultherd forgets to account for the 2 - 2v/V at the end of his first lengthwise trip and confuses distance travelled at V and at V-v, which simplifies his algebra to the same final equation as Edwards.

I am rather surprised at the basic errors in both solutions.

Clark. Mental Nuts. 1897, no. 88; 1904, no. 98; 1916, no. 99. A West Pointer. Column is 25 miles long. Courier goes from the rear to the front and returns to the rear and sees that he is now where the front of the column was when he started. I.e. his trip takes the same time as the time the column moves 25 miles. How far did he go? Answer is 60 miles, 1876 feet, which is correct. Here L = 25 = d, so D = 60.36.

Loyd. The courier problem. Cyclopedia, 1914, pp. 315 & 382. (= MPSL2, prob. 146, pp. 103 & 167 168, with solution method provided by Gardner -- Loyd only gives the values "following the rule for solving puzzles of this kind".) Army 50 miles long. Rider goes from back to front to back of the army in the time it moves forward 50 miles. Here L = 50 = d, so D = 120.71 -- Loyd says "a little over 120 miles". Note this is the double of Clark's problem.

Loyd also extends to the case of a square army and Gardner gives a solution, assuming the courier goes across the ends on an angle with velocity V, so his crossways velocity is w = (V²-v²). The total time of his trip is then T  =  L/(V v) + L/(V+v) + 2W/w. Assuming that the army advances its own length in this time, i.e. vT = L, and setting x = V/v leads to L  =  L/(x 1) + L/(x+1) + 2W/(x²-1). Assuming W = L simplifies the expression, but it remains a fourth degree equation. Gardner says the only relevant solution is x = 4.18112+. The courier's distance is VT = xvT = xL = 209.056+.

Abraham. 1933. Prob. 67 -- The column of troops, pp. 33 & 44 (19 & 116). Rider circling army -- same as first part of Loyd.

Haldeman-Julius. 1937. No. 74: Train problem, pp. 10 & 24. Brakeman walks from rear of a train going at V = 27 mph, to the front, thereby passing a point in time T = 2½ minutes earlier than he would have. How long, L, is the train? Though this seems like a problem of this section, in fact it simply says the train takes 2½ minutes to pass a given point and L = VT.

Perelman. 1937. MCBF. Reconnaissance at sea, prob. 149 a & b, pp. 253-255. Squadron moving and a reconnaissance ship is sent ahead. Part a gives distance to go ahead and asks how long it will take. Part b gives time for reconnaissance and asks when the reconnaissance ship will turn back.

McKay. At Home Tonight. 1940. Prob. 17: The orderly's ride, pp. 65 & 79. Same as first part of Loyd with 1 mile army.

William R. Ransom. Op. cit. in 6.M. 1955. An army courier, p. 103. Same as first part of Loyd with 25 mile army, but says it takes a day and asks how fast the courier rides.

G. J. S. Ross & M. Westwood. Problems drive, 1960. Eureka 23 (Oct 1960) 23-25 & 26. Prob. G. v = 15, d + L = 23, V = 60, find L. At first I thought the result was wrong, but the phrasing is a bit different than usual as the back of the army is 23 from where the front finishes.

Nathan Altshiller Court. Mathematics in Fun and in Earnest. Op. cit. in 5.B. 1961. Prob. m, pp. 189 & 191-192. Courier going from back to front and then back again at three times the speed of the army. Where is he when he gets to the back again?

Philip Kaplan. More Posers. (Harper & Row, 1964); Macfadden-Bartell Books, 1965. Prob. 76, pp. 78 & 107. L = 100, V = 3v, find d.

Julius Sumner Miller. Millergrams. Ure Smith, Sydney, 1966. Prob. 25, pp. 25 & 70. Army is 3 mi long, officer starts at back, goes to front and returns, reaching the back when it has advanced 4 miles. How far did he go? L = 3, d = 4, giving D = 8 miles. He says: "you can get the right answer by erroneous logic", but he doesn't explain how to get the answer!

Birtwistle. Math. Puzzles & Perplexities. 1971. Procession, pp. 43, 167 & 189. = Birtwistle; Calculator Puzzle Book; 1978; prob. 55, pp. 38-39 & 99. Procession 1½ miles long going at 2 mph. Marshall starts at head, walks to the back and then forward, reaching his starting point when half the army has passed. What is his speed? He continues on the head and returns to the same point. Where is the end of the procession when he gets back?
10.A.2. NUMBER OF BUSES MET
New section -- I haven't done much on this yet. Kraitchik has several examples.
Mittenzwey. 1880. Prob. 75, pp. 14-15 & 66; 1895?: 82, pp. 19 & 69; 1917: 82, pp. 18 & 65. Daily trains crossing America in 7 days.

[Richard A. Proctor]. Letters received and short answers. Knowledge 3 (26 Oct 1883) 264. Answer to Harry. Recalls it being posed on by the captain's wife, Mrs Cargill, on the S. S. Australasia, but no date is clear. Trains going between New York and San Francisco taking 7 days. How many does one meet on such a trip? Says he gave the wrong answer!

[Richard A. Proctor]. Editorial gossip. Knowledge 3 (23 Nov 1883) 318. Gives a careful answer to the problem stated above.

L. Carroll. A Tangled Tale. (1885) = Dover, 1958.