Dec. 30, 1881.] 



KNOVV^LEDGE ♦ 



191 



hi- is not very strong in that suit, in which ease, by leading through 



him, you put hira at a disadvantage. Many pUvyers seem to think 



tliat the cxcellont general rule, lead through strength (that is, lead 



■■nit in which your left-hand adversary has high cards), is a 



■ to be universally followed when you have no good suit 



your own, and do not know which is your partner's 



-t suit. Bnt if yonr left-hand adversary leads from 



suit both strong anil long, and you, making first trick, 



ul through him in that suit, you are simply playing his game. 



the other two suits (outside trumps), you select that which you 



:: lead with least chance of aiding the adversaries, and, as a rule, 



I play the kest of the suit. It is an even chance that your partner 

 strong in it. 



I I you are second player, and take the first trick, you can 

 inlly go wrong. Leading the suit your right-hand adversary 



a led would be doubly disadvantageous: you would be pro- 

 lily leading up to strength, and certainly helping to establish 

 - suit. Of the other two suits, outside trumps, you select the 

 -; . and play the card most likely to help your partner. If you 

 ■.!■ thus led from a suit in which your left-hand adversary is 

 I ing, you at any rate lead through his strength. If your high 

 r-.l makes, and yon then play a low one, youi- partner knows you 

 ■■•• led from a short suit (or that you have made a forced lead), 

 I infers that either you have four trumps, and no other four-card 

 i: . or that yonr only four-card suit is very weak. His own hand 

 will help to show which of the two explanations is the more 

 [iroljable. . 



With a five-card suit, however smaU the individual cards, it is 



■ rircely ever wrong to lead from tha long suit. Cavendish, in an 



I ising story in his " Card Essays," gives it as his opinion that 



long suit should be led from, even with such a hand as this : — 



, King, Queen of Spades; Eight of Clubs; Ace, King, Queen, 



1 Three of Diamonds (trumps) ; Nine, Eight, Six, Four, Three of 



A YARBOROUGH HAND AT WHIST. 

 \T the close of the letter to which we referred last week, 

 H. P. H." suggests a plan far calculating the odds that there will 

 r. t be a Yarborough at a given deal. He says : " We must find the 

 number of ways in which the pack must be dealt so as not to 

 include a Yarborough. Suppose, for instance, one hand contains 

 one of the high cards, another three of them, the third six, and the 

 fourth ten. This arrangement may happen in 



20 . 19 ■ 18 . 17 . 16 . 15 .'14 . 13 . 12 . 11 ^ , 

 2.3.6.5.4.3.2 

 By taking all such arrangements as these, and adding the number 

 of ways, we get the number of different arrangements of the whole 

 pack which will not include a Yarborough ; the ratio of this 

 number to the number of different arrangements of the pack, 



namely, - — is the chance against a Yarborough happening. 



This would give a long piece of work, but, perhaps, some of your 

 leaders may find a short method."' H. P. H. 



• The reasoning here is unsound. We leave the problem as an 

 exercise for our readers (not proposing, however, to publish all 

 eolations which may be sent to us). 



(9uv iHatftfmatiral Column. 



MATHEMATICAL QUERIES. 

 [8} — Are there any exact solutions of the equations : — 



l+v/l-:c 



= v/l- 



and 



1 -H v/1 + J 



Vl-.-^ 



1 ■*■ v^l ■^ X 1 + v^l 



[The equations are really the same, so far as any difficulty in their 

 solution is concerned, for they differ only in the sign of x, so that 

 whatever root we find for one, the same quantity, taken negatively, 

 is a root of the other. But it will be found on trial that is the 

 only value of x which satisfies either equation. The solution may 



run thus (taking first equa tion) : — 



1 + v/i^==V'l-j'H- (l + .T)yi-x 



l = yi-x(v/l-fx--t.T (A) 



l = (v/l + x-^x)(^l-^-.l.■-a,^ (B) 

 v/l- i=v/l 4-x-x (C) 



2-2v/l-i:'=x' 

 l-.r'-2v^-x= + l = 



V'l-.t' = l 

 x = 0. 



Bnt 



Whence 

 and 



It might seem that since, after obtaining (A) and (B), which give 

 (yi-t-x + x) (v/l + x-x) = v^l^» (s/lTe-n) 

 wo divide by v'l -H x -H x to get c, the equation should be satisfied by 

 the roots of 



which are 



v/l + x + x=0 

 1 + ^/5 



2 



but this is not the case. Neither root will satisfy the original 

 equation, whatever signs wc give the quantities ■^l—x and t/l-t x. 

 —Ed.] 



[9] — Arithmetical Pbobi.em. — Would any reader of Knowledge 

 favour me with a solution of the following problem : — If twelve 

 horses eat ten acres of grass in sixteen weeks, and eighteen horses 

 eat ten acres in eight weeks, how many horses will eat forty acres 

 in six weeks ? The grass is supposed to grow uniformly. — G. H. 

 Mapleto.v. 



[10] — On base J)C are triangles BAC, BPC\ having equal peri- 

 meters, AB being equal to AC. If ACBD intersect in 0, show that 

 A0> DO. M.iTHEMATiccs. — [Like most problems of the kind this is 

 best dealt with indirectly. Thus, take at point 6 in OA (produced 

 if necessary) such that 06 = OD, and from OB the greater, cut off 

 OF, equal to OC the less, and join FG. Then obviotisly 

 fG = DCand GC = FD 

 Hence FG + 6C = FD+DC 



But BF+'^ + GC=BD + DC = BA + AC 

 ^ <BP + FA + AC 



.-. FO*OC<FA + AC 

 or PG + GO<FA + AO 

 whence it follows that miiifc lie between and .-1. For if G were 

 at A, FG + GO would be the same as FA +A0 ; and if G were in OA 

 produced FG + GO would be greater than FA + AO. Therefore AO 

 is greater than GO, that is, than DO. — Ed.] 



[11] — Perimetee of Inscribed Triangles. — Show that the peri- 

 meter of an equilateral triangle inscribed in a circle is greater than 

 the perimeter of any other isosceles triangle inscribed in the same 

 circle. — Mathematicus. 



[Prove as follows : — Let ABC be the greatest triangle with a 

 given perimeter. Then, if it be not equilateral, there must be, at 

 least, two sides not equal to each other. Let AB and BC be unequal. 

 Through B draw KBL parallel to base AC, draw CM perpendicular 

 to KL, and produce CM to N, making MN=NC. Join AN, cutting 

 KL in D, and join DC, BN. Then BN = BG, and therefore AB + BC 

 = AB + BN> AN> AD + DC, so that there must be same point P in 

 D:V, such that if PC be joined AP + PC = AB + BC, or there is a 

 triangle greater than ABC, and having the same perimeter, contrary 

 to our supposition. Hence, no two sides of the greatest triangle 

 with the given perimeter can be unequal, or the triangle is equi- 

 angular. — Ed.] 



[12] — I wish to prove that 



2n (2n - 1) (2>t - 2) ■ • ■ (n + 1) 

 1 • 2 ■ 3 • ■ • ■ n. 

 The middle term in the expansion of 2^" is equal to 

 /n (n - 1)Y 



the sum of the squares of terms of the expansion of 2". 



Proposing to do this by mathematical induction, I find, calling 

 the first expression 2, that when in it I write n + 1 for n, it 

 becomes 



<i-orTi)0 



Can I, at this point, without more ado, conclude thai since 

 (by hypothesis) when n becomes n + 1 that 2 will become 



which would prove what I want ? — F. B. 



[Your demonstration is beyond us. If you write (»+l) iu thc- 

 first expression, calling it S before the change, it becomes — 

 S/ 2(27.-H) \ 

 V (n + 1) ) 

 The best way to solve your problem is this ; — 



(l + x)°=H-C, x+C„ x' + &c.+ C^ x"-'-hC, x°-'-hx" 

 (x-f l)'' = x°-H C, x'-' + Cj x"-2 + &c.-H Cj x'-i- Ci x-l-1 

 where C,, Cj, C^ are the well-known co-eflicients, in the expansion 

 of the triunial (l-fx)°. Multiplying — 



(l-hx)™ = l-)-(C, + 1) x-H(2 C„ + C.^) x-x&c. 



+ {\ + C;'+C^^+C^-+ +C.^- + C,' + l) c" 



■v&c. -t- (1-h Ci) x'"-' + .C"'' 



