des; it is evident that the present value or worth
of each of their expectations will be L25, and the probabilities 25/50
or 1/2. For, if they had agreed to divide the prize between them,
according as the bets should be at the time of their starting, they
would each of them be entitled to L25; but if A had been thought so much
superior to B that the bets had been 3 to 2 in his favour, then the real
value of A's expectation would have been L30, and that of B's only L20,
and their several probabilities 30/50 and 20/50.

Example II. Let us suppose three horses to start for a sweepstake,
namely, A, B, and C, and that the odds are 8 to 6 A against B, and 6 to
4 B against C--what are the odds--A against C, and the field against A?
Answer:--2 to 1 A against C, and 10 to 8, or 5 to 4 the field against A.
For

A's expectation is 8
B's expectation is 6
C's expectation is 4
----
18

But if the bets had been 7 to 4 A against B; and even money B against
C, then the odds would have been 8 to 7 the field against A, as shown in
the following scheme:--

7 A
4 B
4 C
----
15

But as this is the basis upon which all the rest depends, another
example or two may be required to make it as plain as possible.

Example III. Suppose the same three as before, and the common bets 7 to
4 A against B; 21 to 20 (or 'gold to silver') B against C; we must state
it thus:--7 guineas to 4 A against B; and 4 guineas to L4, B against C;
which being reduced into shillings, the scheme will stand as follows:--

147 A's expectation. 81 B's expectation.
80 C's expectation.----311

By which it will be 164 to 147 the field against A, (something more
than 39 to 35). Now, if we compare this with the last example, we may
conclude it to be right; for if it had been 40 to 35, then it would have
been 8 to 7, exactly as in the last example. But, as some persons may
be at a loss to know why the numbers 39 and 35 are selected, it is
requisite to show the same by means of the Sliding Rule. Set 164 upon
the line A to 147 upon the slider B, and then look along till you see
two whole numbers which stand exactly one against the other (or as near
as you can come), which, in this case, you find to be 39 on A, standing
against 35 on the slider B (very nearly). But as 164/311 and 147/311 are
in the lowest terms, there are no less numbers, in the same proportion,
as 164 to 147,--39 and 35 being the nearest,