The San Francisco call., 1910-10-16
THE GAME DAILY AND FINDS THAT
THEY DO NOT GET AN EVEN BREAK,
THAT THE PERCENTAGE AGAINST THEM IS 23


the ticket for any amount from 8 cents
up Although great many play from
% cents to SIS on each ticket, the usual
amount per ticket is 10 cents. So we
will travel through Chinese lottery at
the 10 cent rate Remember, one can
only mark 10 numbers and there are
20 winning numbers on a ticket con-
tanning so numbers. The numbers are
sometimes referred to as spots. A
player may say he has caught six
spot, which would mean that of the
10 numbers he marked six proved to
be winning numbers I'm the drawing.
One must have at least five winning
numbers or his ticket is worthless. and
then he gets paid even money, that is
if he plays the ticket for 10 cents he
will get 90 cents for his ticket. The
rate of pay on 10 cent tickets is as
follows.
Five spot ...-........-...-........ $50
Seven spot -.-.---..-----. IBID
Eight spot $500
Nine spot -....--.---..-..-.-... leo.oo
Ten spot .................. 372.00
Now we will see what our chances
are to win. We will first find out how
many different ways one can mark a
ticket or, in other words, how many
different combinations of 10 numbers
each there are on ticket Of 80 num-
bers that will be the quotient obtained
by dividing the product Of the 10 high
est numbers in so by the product of the
1O<smallest. We multiply together So
70, rs, ~. 76, 75, "A, 73, 70 and II and
divide the product by that obtained
from multiplying together l. 2, 8, 4. 5.
G. Z, 8, and 10. We thus get 1885,
492, ,11O,120 as the total number of 10
spots on ticket of so numbers. We
will now see what our chances are of
catching a 10 spot. There are as many
10 spots possible in each drawing as
the total number of combinations of 10
spots in the 20 numbers that are drawn.
This result will be the quotient OBJ
tained by dividing the product of the
10 highest numbers in 90 by the product
OF the 10 smallest numbers. Thus we
have 2Ox1sx1Sx17x16x15x1
equals GO. 0.442,572, 500 and 1x2x3x4x5x6-
x7xSxsx1o equals 3,62S.S00 and $780,
442,572, 500 divided by s.62S.SOO equals
1876 as the total number of 10 spots
in one drawing Of 20 numbers.
Therefore, if were were to mark
l,646, 6,492, 120 tickets and mark them all
differently, as we have seen rs possible,
we would catch 1875 10 spots in any
one drawing. Therefore, our chance to
catch a 10 spot are as numerous as
1876 is contained times in 1,646,4s2,-
11o.I20, which is S,91I.711.176 plus
times, or one chance in about s,s11.711.
However, 1t is only necessary for us to
remember that out of 1,646, 492, ,110, ,120
trials we will get 1876 10 spots.
The number of nine spots we will get
greater. The total number OF COM
binations of nine spots in any one
drawing Of 20 numbers is the quotient
obtained by dividing the product of the
nine highest numbers IN 20 by the
product Of the nine smallest numbers
that is, the product Of 90. -. Is, II, 16,
15, II. 1s and II divided by the prod
uct of I, 2, 3, 4, 8, 6. z, and 8. We
thus get 167. $50 as the total number of
combinations of nine numbers each in
one urawing of 20 numbers, each Of
which can be taken as many times as
the remaining number marked can be
placed on the remaining numbers on
the ticket-that is, If we mark 10 num-
bers and only nine of them are winning
numbers. there is one remaining, which
can be any one OF the remaining. so
numbers on the ticket, which are not
winning numbers. Thus we have 167,-
$50 times 60. which is 1o.077,6 as the
number of nine spots we will catch In
1916, 6,4s2,110,l20 trials; or, If we were
to mark 1876. 92,11o.120 tickets no two
of which were alike we would have
10,077,600 tickets that contained nine
winning numbers each. or. in other
words, we would have caught 10,077,-
600 nine spots in any one drawing, no