22 ON PROBABILITY. 
greater than ¢.(b — 1) times the parcel immediately preceding it, and very 
much greater than i times the sum of all the parcels which precede it. 
Exactly the same reasoning will show that if p,,, can be made greater than 
i.(a — 1) times p,,, _ ,y that is, iin > i.(a—1) the parcel of the second 
limit, will, with the same value of am, be more than i times greater than all 
the parcels which follow it. 
35. Let a be that one of the quantities « and 5, which is not the least, 
A value of m, which will make p™“ greater than i. (¢ — 1) times p,,,_.., and 
than .(@— 1) times Pp orm Will evidently satisfy both the above-mentioned 
conditions. 
Pua. (ma+1).. (ma m) GC) = (mab +b)..(mab-4 mb) 
Pmagm (mb—{m—11%1)..mb\a (mab— {m-1}%}a).. mab 
b 
The last pair of factors here is meh mh or el and any other 
3 mab a 
pair, such as the #*, is é 
 mab-rb sx so @1 
br TD 
if mab4+rb>@-+1)(@mb—m 47), 
that is, if b< a+ 1, 
which by supposition itis. Therefore 
na a + 1 nm 
iY 
Pmatm a 
for there are m factors in this continued product. 
Pos _ (mb+1)..(mb +m) NE paid auncidn } 
Pna-m (Mma—{m—1%})..ma\b (mab—.{m—1%}b)...mb 
Any factor, as the +, here 
: _ mab+ra wa a+ 1 
aU AT which is > £2 en 
if admdb+t+n>=@+1).b(ma—{m-r}), 
that is, if ra(@—0b)+ (m—1r).0>=0, 
which it must always be, for neither @ — 4, nor m — #, can ever become 
negative, 
{ Pua a - by 3 i: + 1 m ; 
oo — —or oe Pie AS —) = 3% 
log. ¢ (a — 
or mit one 1 
log.(@ + 1) — log. a 
and with this value of m both the necessary conditions are satisfied. 
36. Therefore the probability 
_ 1st parcel + 2d parcel + a, 
1st parcel + 2d parcel + a,, + sum of other parcels’ 
or since lst parcel +4 2d parcel > 4. (sum of other parcels) 
= ¢ (sum of other parcels) -} %,