April 1, 1890.] 



KNOWLEDGE. 



121 



holds the two of spades, and consequently knows that A's 

 three (trick 1) was not a penultimate, i.e. that A has only 

 three more spades. Therefore, imless Y's lead at trick 5 

 was fi'om three clubs only, A must hold the other trump. 



Trick 8. — B nevertheless draws the trump. If Y holds 

 the major tenace in spades, the game is hopeless, honours 

 being declared against AB ; but if he holds king, knave, 

 against A's ace, queen (the only other possible case), the 

 game may be saved by throwing the lead into liis hand at 

 the right moment, and compelling him to lead spades up 

 to A. Continuing the clubs at this point would be fatal 

 if the other trump should turn out to be with Y. 



Y's discard is the best under the circumstances (see B's 

 inference fi'om trick 8 in the analysis given below). 







O 

 O O 



\. 



B 



Tricks— AB, n ; YZ, 4. 



^ m 



Tnrks—AB, 6 ; YZ, 4. 



Notes. — Triik 9. — Z probably has either queen, knave, 

 of clubs, or knave guarded; and, if B now continues with 

 king and another club, Z, after getting in, will force him in 

 diamonds, obliging him, after all, to lead spades up to Y. 



Y again discards a club for the same reason as before. 

 It is, however, interesting to notice that, if he discards a 

 spade, B must not think to place the lead in his hand by 

 playing out kuig and another club ; for Y will throw his 

 queen on the king, leaving the command with Z, after 

 which there is no way for AB to save the game. 



Trich- 10. — It is, of course, necessary for B to dispose of 

 his king of clubs before leading the spades. 



Thick 11. 

 A 



n 



Tricks— AB, (! ; YZ, 5. TricKs—AB, 7 ; YZ, .5. 



Trick 13.- — A makes the ace of spades, and 



AB SCORE TWO BY C.UIDS AND WIN THE GAJIE. 

 A's Hand. B'b Hand. 



II.— 10, 8, 5. II.— Kg, 9, 7, 6, 3. 



S.— Ace, Qn, 1, :!. S.— 10, 5, 2. '' 



D.— Kg, 10, 4. D.— 8, 6. 



C— Ace, 8, 7. C— Kg, 9, 2. 



Y's Hand. Z's Hand. 



H.— Kn, 4, 2. H.— Ace, Qn. 



S.— Kg, Kn, i), H, (i. S.— 7. 



D.— 2. D.— Ace, Qn, Kn, !), 7, 5, 3. 



C— Qn, G, 4, 3. C— Kn, 10, 5. 



In place of our usual elementary explanation of the 

 play, we append an analysis of the play of B's hand and 

 the inferences drawn by him from each trick. 



Tricks. Play. 



1. Plays highest card, third 



Leads trumps, liolding Z has 



five to an honour. 

 Opens with penultimate 



to show number. 



3 & 4. Follows suit. 



Plays lowest card, second 

 in hand. 



Heads the trick. He 

 might perhaps hold up 

 the king, for Z cannot 

 have anything but the 

 ace. .and if the ace had 

 been with Y he would 

 almost certainly have 

 put it on the ten. But 

 this course might leave 

 A in doubt as to the 

 position of the king, 

 and nothing would be 

 gained by it, as B re- 

 tains the command with 

 the nine. 



Trumps adverse winning 

 card. 



8. Draws the losing trump, 

 its position being doubt- 

 ful. 

 If Y has it, and if B now 

 goes on with clubs, 

 either Y or Z will win 

 the third round, and B, 

 after being forced by a 

 club or a diamond, as 

 tbo case may bo, will 

 have eventually to lead 

 up to Y's tenace in 



9. 



Accordingly, B leads an- 

 other trump. The stu- 

 dent will find that 

 any other lino of play 

 loses the game, whe- 

 ther A finesses or not. 



Leads the best club. 

 Otherwise Y, after get- 

 ting in with a spade, 

 would dispossess him- 

 self of the lead by 

 continuing with the 

 queen of clubs. 



Loads a spade accordingly. 



Inferences. 



A has exactly four spades ; for, 

 since B himself holds the two, the 

 three cannot be a " penultiftate." 



Z has eight and nine of spades, or 

 no more. 



Either A has ace, rjueen, of spades 

 and Y has king, knave, or rice 

 versa. In any other case, one or 

 other of them would have played 

 an honour. 



no more tramps, or ace of 

 trumps and no more ; or he has 

 ace, knave, and has played a 

 false card. For, except in these 

 cases, he would not put in his 

 queen, second in hand, unless he 

 also held the king ; and the king 

 is in B's own hand. 



Z has led from ace, queen, knave, 

 of diamonds and at least two 

 others. 



Y being void, the remaining six 

 diamonds must lie between A and 

 Z ; and A cannot have more than 

 two, or he would have led dia- 

 monds originally instead of spades. 

 Therefore Z must have at least 

 four. 



Y may have four clubs, or he may 

 be leading from a short suit in 

 preference to returning the spadco 

 up to A. 



Since A wins the ten of clubs with 

 the ace, YZ must hold queen and 

 knave between them. 



The remaining tnimp may be with 

 either A or Y ; for A would in 

 either case return the ten, and Y, 

 holding_knave, eight, might prefer 

 to cover ten with knave. YZ are 

 out by honours unless AB can 

 forestall them by making two by 

 cards. 



A has no more diamonds, and Z has 

 the remaining four. Y's discard 

 shows that Z has no more spades 

 (see inference from trick 1), and 

 therefore Z has two clubs. 



Y is evidently in a difijculty, as- 

 .suming that the tenace in spades 

 is against him. If he discards 

 another spade, B will be able to 

 lead spades, to which A will play 

 his queen as the only chance of 

 making game ; Y, after winning 

 with the king, must lead a club ; 

 B, winning with king in his turn, 

 will lead another spade : and .A. 

 will make his :h-o, and afterwards 

 the small one. 



It is now certain that Y docs not 

 hold the major tenace in spades ; 

 with ace, queen, he would secure 

 the game by discarding from that 

 suit. 



He has clearly determined to leave 

 the protection of the club suit 

 to Z. 



Z is left with the best club. There- 

 fore the only way of vrinning the 

 game is to lead a .spado, trusting 

 to -\ to pass it and remain with 

 tenace over Y. 



