192 THE TURF 



will now be T 4 ^ ; but ^ is equal to -f ; there- 

 fore, it will be 



4x1 = 1%; and 10-2 = 8: 

 hence the odds are 8 to 2 = 4 to i. 



Proof by Hedging. B begins to hedge by 

 betting an even one pound on the first 

 event, which, A winning, he wins. On the 

 subsequent event, B takes the odds, 3 to 2, 

 which, A winning, he also wins. Thus he 

 receives four pounds, which pays the 4 to i 

 he betted on A losing both events. 



Upon two several events, even betting on 

 the one, and 7 to 4 in favour of A on the 

 other, what odds may B lay against A 

 winning both? The one, as before, is J, 

 and the other is represented by T 7 T : 

 Then Jx T 7 T = ^; and 22-7 = 15: 



thus 15 to 7 is the odds. 



Proof by Hedging. The sum against 

 which B laid his odds is 7 ; therefore he 

 begins by laying seven pounds on the first 

 event ; which, as A wins, he wins. On the 

 next event he lays 14 to 8, or twice 7 to 

 twice 4, as per terms of question, which he 

 also wins; making together 7 and 8=15, 

 the odds he had laid with, and lost to A. 



