THE TURF. 
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. 
Upon the same two events, what odds may B bet 
A that the latter does not lose both? Set down for 
the former i, for the latter -^ : 
1 4 4 
Then 2 X TT = 22 5 and 22 ~ 4 = 18: 
therefore 18 to 4 = 9 to 2 is the odds. 
Proof by Hedging. B bets first the sum to which 
he has laid his odds, namely, two pounds, which he 
wins ; and then taking 7 to 4 on the second event, he 
wins 2-f 7 = 9, which pays the nine pounds he lost to A; 
and had more favourable odds been offered, B must 
have been a winner without risk of losing. 
When three distinct events are pending, on the first 
of w r hich the betting is even ; on the second 3 to 2 in 
favour of A, and the third 5 to 4 ; what odds should B 
lay A that the latter does not name all the winners ? 
The first is expressed by -J, the second by , and the 
third by | : 
136 1 
Therefore, X + = (by cancelling) ; and 6 1 = 5 : 
hence the odds are 5 to 1. 
Proof by Hedging. B begins to hedge by betting 
an even two pounds that A wins the first event ; he 
then bets the odds on the next, viz. (3 to 2)-i-2 = lJ 
25G 
