on a pocket table), or causing it to jump off the table or lodge on the
cushion, or for missing altogether, nothing is forfeited other than
the stroke. The only penalty is that the ball so offending shall be
spotted upon the white-ball spot at the foot of the table, or if that
be occupied then on the nearest spot thereto unoccupied.
When the pins are arranged, the rotation of the players is determined
in like manner as in Fifteen-Ball Pool, after which each player
receives from the marker a little numbered ball which is placed in the
player’s cup on the pool board, and the number of which is not known to
any of his opponents.
The object of the player is to knock down as many pins as will count
exactly thirty-one when the number on the small ball held by him is
added to their aggregate; thus, if the small ball is No. 9, the player
will have to gain twenty-two points on the pins before calling game,
and whoever first gets exactly thirty-one points in this manner wins
the pool.
A white ball is spotted five inches from the lower end of the table, on
a line drawn down the centre; and the red ball placed upon its own spot
at the foot of the table.
Player No. 1 must play with the remaining white ball from any point
within the string-line at the head of the table at either the red or
white ball, or place his own on the string spot. Player No. 2 may play
with any ball on the table--red or white. After the first stroke has
been played, the players, in their order, may play with or at any ball
upon the board.
Unless the player has played on some ball upon the board before
knocking down a pin, the stroke under all circumstances goes for
nothing, and the pin or pins must be replaced and the player’s ball
put upon the white-ball spot at the foot of the table or if that be
occupied, on the nearest unoccupied spot thereto. But should two balls
be in contact the player can play with either of them, direct at the
pins, and any count so made is good.
If a player, with one stroke, knocks down the four outside pins and
leaves the black one standing on its spot, it is called a Natural, or
_=Ranche=_, and under any and all circumstances it wins the game.
When a player gets more than 31, he is _=burst=_, and he may either
play again immediately with the same ball he has in the pool rack,
starting at nothing of course, or he may take a new ball. If he takes a
new ball he may either keep it or keep his old one, but he cannot play
again until it comes to his turn.
THE LITTLE CORPORAL.
This game is the regular Three-Ball Carrom Game with a small pin
added, like those used in Pin Pool, which is set up in the centre of
the table. The carroms and forfeits count as in the regular Three-Ball
Game, but the knocking down of the pin scores five points for the
striker, who plays until he fails to effect a carrom or knock down the
pin. A ball must be hit by the cue-ball before the pin can be scored;
playing at the pin direct is not allowed. The pin must be set up where
it falls; but in case it goes off the table or lodges on the top of the
cushion it must be placed upon the centre spot. The pin leaning against
the cushion must be scored as down, and when the pin lodges in the
corner of the table, so that it cannot be hit with the ball, it is to
be set up on the centre spot. One hundred points generally constitute a
game, but any number of points may be agreed upon.
THE SPANISH GAME OF BILLIARDS.
This game is played in the South, California, and in Mexico and Cuba,
and is played with two white and one red ball, and five pins placed
similar to those in Pin Pool. The red ball is placed on the red-ball
spot, and the first player strikes at it from within the baulk
semicircle. The game is scored by winning and losing hazards, carroms,
and by knocking over the pins. It is usually played thirty points up.
The player who knocks down a pin after striking a ball gains _two_
points, if he knocks down two pins he gains _four_ points, and so on,
scoring two points for each pin knocked down. If he knock down the
middle pin alone he gains _five_ points. The player who pockets the red
ball gains _three_ points and two for each pin knocked down by the same
stroke. The player who pockets the white ball gains two points, and
two for each pin knocked over with the same stroke. Each carrom counts
two. The player who knocks down a pin or pins with his own ball before
striking another ball loses two for every pin so knocked down. The
player who pockets his own ball without hitting another ball forfeits
three points; for missing altogether he forfeits one point. The striker
who forces his own ball off the table without hitting another ball
forfeits _three_ points, and if he does so after making a carrom or
pocket he loses as many points as he would otherwise have gained. The
rules of the American Carrom Game, except where they conflict with the
foregoing rules, govern this game also.
BOTTLE POOL.
The game of Bottle Pool is played on a pool table with one white ball,
the 1 and 2 ball, and pool-bottle. The 1 and 2 balls must be spotted,
respectively, at the foot of the table, at the left and right diamond
nearest each pocket, and the pool-bottle is placed standing on its neck
on the spot in the centre of the table, and when it falls it must be
set up, if possible, where it rests.
Carrom on the two object-balls counts 1 point; Pocketing the 1 ball
counts 1 point; Pocketing the 2 ball counts 2 points; Carrom from ball
and upsetting bottle counts 5 points. The game consists of 31 points.
The player having the least number of points at the finish of the game
shall be adjudged the loser.
Any number of persons can play, and the rotation of the players is
decided as in ordinary pool. Player No. 1 must play with the white
ball from any point within the string at the head of the table, at
either the 1 or 2 ball at his option. The player who leads must play
at and strike one of the object-balls before he can score a carrom on
the pool-bottle. If a player carrom on the bottle from either of the
object-balls, in such a way as to seat the bottle on its base, he wins
the game, without further play.
Should the 1 or 2 ball in any way, during the stroke, touch the bottle
and the bottle is in the same play knocked over or stood on its base
by the cue-ball, the stroke does not count. If the player forces the
bottle off the table or into a pocket, the bottle must be spotted on
its proper spot in the centre of the table, the player loses his shot
and forfeits one point, and the next player plays.
A player who makes more than 31 points is burst, and must start his
string anew; all that he makes in excess of 31 points count on his new
string, and the next player plays.
BILLIARD TEXT BOOKS.
_=American Game=_:--
Modern Billiards, Brunswick-Balke-Collender Co.
Billiard Laws, ” ” ” ”
Manuel du Biliard, by Vignaux.
Garnier’s Practice Shots.
_=English Game=_:--
Billiards Simplified, Burroughs and Watts.
Billiards, by W. Cook.
Billiards, by Joseph Bennet.
Billiards, by Maj.-Gen. Drayson.
Practical Billiards, by W. Dufton.
The Spot Stroke, by Joseph Bennet.
CHANCE AND PROBABILITY.
In calculating the probability of any event, the difficulty is not,
as many persons imagine, in the process, but in the statement of the
proposition, and the great trouble with many of those who dispute on
questions of chance is that they are unable to think clearly.
The chance is either for or against the event; the probability is
always for it. The chances are expressed by the fraction of this
probability, the denominator being the total number of events possible,
and the numerator the number of events favourable. For instance:
The probability of throwing an ace with one cast of a single die is
expressed by the fraction ⅙; because six different numbers may be
thrown, and they are all equally probable, but only one of them would
be an ace. Odds are found by deducting the favourable events from the
total, or the numerator from the denominator. In the example, the odds
against throwing an ace are therefore 5 to 1. The greater the odds
against any event the more _=improbable=_ it is said to be, and the
more _=hazardous=_ it is to risk anything upon it.
When an event happens which is very improbable, the person to whom it
happens is considered _=lucky=_, and the greater the improbability, the
greater his luck. If two men play a game, the winner is not considered
particularly lucky; but if one wanted only two points to go out and the
other wanted a hundred, the latter would be a very lucky man if he won.
It is a remarkable fact that luck is the only subject in the world on
which we have no recognised authority, although it is a topic of the
most universal interest. Strictly speaking, to be lucky simply means to
be successful, the word being a derivative of _gelingen_, to succeed.
There are a few general principles connected with luck which should be
understood by every person who is interested in games of chance. In the
first place, luck attaches to persons and not to things. It is useless
for an unlucky man to change the seats or the cards, for no matter
which he chooses the personal equation of good or bad luck adhering to
him for the time being cannot be shaken off. In the second place, all
men are lucky in some things, and not in others; and they are lucky or
unlucky in those things at certain times and for certain seasons. This
element of luck seems to come and go like the swell of the ocean. In
the lives of some men the tide of fortune appears to be a long steady
flood, without a ripple on the surface. In others it rises and falls
in waves of greater or lesser length; while in others it is irregular
in the extreme; splashing choppy seas to-day; a storm to-morrow that
smashes everything; and then calm enough to make ducks and drakes with
the pebbles on the shore. In the lives of all the tide of fortune is
uncertain; for the man has never lived who could be sure of the weather
a week ahead. In the nature of things this must be so, for if there
were no ups and downs in life, there would be no such things as chance
and luck, and the laws of probability would not exist.
The greatest fallacy in connection with luck is the belief that certain
men _are_ lucky, whereas the truth is simply that they _have been_
lucky up to that time. They have succeeded so far, but that is no
guarantee that they will succeed again in any matter of pure chance.
This is demonstrated by the laws governing _=the probability of
successive events=_.
Suppose two men sit down to play a game which is one of pure chance;
poker dice, for instance. You are backing Mr. Smith, and want to know
the probability of his winning the first game. There are only two
possible events, to win or lose, and both are equally probable, so 2 is
the denominator of our fraction. The number of favourable events is 1,
which is our numerator, and the fraction is therefore ½, which always
represents equality.
Now for the successive events. Your man wins the first game, and they
proceed to play another. What are the odds on Smith’s winning the
second game? It is evident that they are exactly the same as if the
first game had never been played, because there are still only two
possible events, and one of them will be favourable to him. Suppose he
wins that game, and the next, and the next, and so on until he has won
nine games in succession, what are the odds against his winning the
tenth also? Still exactly an even thing.
But, says a spectator, Smith’s luck must change; because it is very
improbable that he will win ten games in succession. The odds against
such a thing are 1023 to 1, and the more he wins the more probable it
is that he will lose the next game. This is what gamblers call the
_=maturity of the chances=_, and it is one of the greatest fallacies
ever entertained by intelligent men. Curiously enough, the men who
believe that luck must change in some circumstances, also believe in
betting on it to continue in others. When they are _=in the vein=_ they
will “follow their luck” in perfect confidence that it will continue.
The same men will not bet on another man’s luck, even if he is “in the
vein,” because “the maturity of the chances” tells them that it cannot
last!
GAMES. ODDS.
One 1 to 1
Two 3 to 1
Three 7 to 1
Four 15 to 1
Five 31 to 1
Six 63 to 1
Seven 127 to 1
Eight 255 to 1
Nine 511 to 1
Ten 1023 to 1
If Smith and his adversary had started with an agreement to play ten
games, the odds against either of them winning any number in succession
would be found by taking the first game as an even chance, expressed
by unity, or 1. The odds against the same player winning the second
game also would be twice 1 plus 1, or 3 to 1; and the odds against his
winning three games in succession would be twice 3 plus 1, or 7 to 1,
and so on, according to the figures shown in the margin.
GAMES.
1st 2nd
1 1
1 0
0 1
0 0
That this is so may easily be demonstrated by putting down on a sheet
of paper the total number of events that may happen if any agreed
number of games are played, expressing wins by a stroke, and losses by
a cipher. Take the case of two games only. There are four different
events which may happen to Smith, as shown in the margin. He may win
both games or lose both; or he may win one and lose the other, either
first. Only one of these four equally probable events being favourable
to his winning both games, and three being unfavourable, the odds are
3 to 1 that he does not win both; but these are the odds _=before he
begins to play=_. Having won the first game, there are only two events
possible, those which begin with a win, and he has an equal chance to
win again.
GAMES.
1st 2nd 3rd
1 1 1
1 1 0
1 0 1
1 0 0
0 0 0
0 0 1
0 1 0
0 1 1
If the agreement had been to play three games, there would have been
eight possible events, one of which must happen but all of which were
equally probable. These are shown in the margin. If Smith wins the
first game, there are only four possible events remaining; those in
which the first game was won. Of these, there are two in which he may
win the second game, and two in which he may lose it, showing that it
is still exactly an even thing that he will win the second game. If he
wins the second game, there are only two possible events, the first
two on the list in the margin, which begin with two wins for Smith.
Of these he has one chance to win the third game, and one to lose it.
No matter how far we continue a series of successive events it will
always be found that having won a certain number of games, it is still
exactly an even thing that he will win the next also. The odds of 1023
to 1 against his winning ten games in succession existed only before
he began to play. After he has won the first game, the odds against
his winning the remaining nine are only 511 to 1, and so on, until it
is an even thing that he wins the tenth, even if he has won the nine
preceding it.
In the statistics of 4000 coups at roulette at Monte Carlo it was found
that if one colour had come five times in succession, it was an exactly
even bet that it would come again; for in twenty runs of five times
there were ten which went on to six. In the author’s examination of
500 consecutive deals of faro, there were 815 cards that either won or
lost three times in succession, and of these 412 won or lost out. In
a gambling house in Little Rock a roulette wheel with three zeros on
it did not come up green for 115 rolls, and several gamblers lost all
they had betting on the eagle and O’s. When the game closed the banker
informed them that the green had come up more than twenty times earlier
in the evening. They thought the maturity of the chances would compel
the green to come; whereas the chances really were that it would not
come, as it had over-run its average so much earlier in the evening.
The pendulum swings as far one way as the other, but no method of
catching it on the turn has ever yet been discovered.
_=Compound Events.=_ In order to ascertain the probability of compound
or concurrent events, we must find the product of their separate
probability. For instance: The odds against your cutting an ace from a
pack of 52 cards are 48 to 4, or 12 to 1; because there are 52 cards
and only 4 of them are aces. The probability fraction is therefore
1/13. But the probabilities of drawing an ace from two separate packs
are 1/13 × 1/13 = 1/169, or 168 to 1 against it.
Suppose a person bets that you will not cut a court card, K Q or J,
from a pack of 52 cards, what are the odds against you? In this case
there are three favourable events, but only one can happen, and as
any of them will preclude the others, they are called _=conflicting
events=_, and the probability of one of them is the sum of the
probability of all of them. In this case the probability of any one
event separately is 1/13, and the sum of the three is therefore 1/13 +
1/13 + 1/13 = 3/13; or ten to 3 against it.
In order to prove any calculation of this kind all that is necessary is
to ascertain the number of remaining events, and if their sum, added to
that already found, equals unity, the calculation must be correct. For
instance: The probability of turning a black trump at whist is 13/52
+ 13/52 = 26/52; because there are two black suits of 13 cards each.
The only other event which can happen is a red trump, the probability
of which is also 26/52, and the sum of these two probabilities is
therefore 26/52 + 26/52 = 52/52, or unity.
Another fallacy in connection with the maturity of the chances is shown
in betting against two successive events, both improbable, one of which
has happened. The odds against drawing two aces in succession from a
pack of 52 cards are 220 to 1; but after an ace has been drawn the odds
against the second card being an ace also are only 16 to 1, although
some persons would be mad enough to bet 1000 to 1 against it, on the
principle that the first draw was a great piece of luck and the second
ace was practically impossible. While the four aces were in the pack
the probability of drawing one was 4/52. One ace having been drawn, 3
remain in 51 cards, so the probability of getting the second is 3/51,
or 1/17. Before a card was drawn, the probability of getting two aces
in succession was the product of these fractions; 1/13 × 1/17 = 1/221.
On the same principle the odds against two players cutting cards that
are a tie, such as two Fours, are not 220 to 1, unless it is specified
that the first card shall be a Four. The first player having cut, the
odds against the second cutting a card of equal value are only 16 to 1.
_=Dice.=_ In calculating the probabilities of throws with two or more
dice, we must multiply together the total number of throws possible
with each die separately, and then find the number of throws that will
give the result required. Suppose two dice are used. Six different
throws may be made with each, therefore 6 × 6 = 36 different throws
are possible with the two dice together. What are the odds against one
of these dice being an ace? A person unfamiliar with the science of
probabilities would say that as two numbers must come up, and there
are only six numbers altogether, the probability is 2/6, or exactly
2 to 1 against an ace being thrown. But this is not correct, as will
be immediately apparent if we write out all the 36 possible throws
with two dice; for we shall find that only 11 of the 36 contain an
ace, and 25 do not. The proper way to calculate this is to take the
chances against the ace on each die separately, and then to multiply
them together. There are five other numbers that might come up, and the
fraction of their probability is ⅚ × ⅚ = 25/36, or 25 to 11 in their
favour.
Take the case of three dice: As three numbers out of six must come up,
it might be supposed that it was an even thing that one would be an
ace. But the possible throws with three dice are 6 × 6 × 6 = 216; and
those that do not contain an ace are 5 × 5 × 5 = 125; so that the odds
against getting an ace in one throw with three dice, or three throws
with one die, are 125/216, or 125 to 91 against it.
To find the probability of getting a given total on the faces of
two or three dice we must find the number of ways that the desired
number can come. In the 36 possible throws with two dice there are 6
which will show a total of seven pips. The probability of throwing
seven is therefore 6/36, or 5 to 1 against it. A complete list of the
combinations with two dice were given in connection with Craps.
_=Poker.=_ In calculating the probability of certain conflicting
events, both of which cannot occur, but either of which would be
favourable, we must make the denominator of our fraction equal in both
cases, which will, of course, necessitate a proportionate change in our
numerator. Suppose a poker player has three of a kind, and intends to
draw one card only, the odds against his getting a full hand are 1/16;
against getting four of a kind, 1/48. To find the total probability of
improvement, we must make the first fraction proportionate to the last,
which we can do by multiplying it by 3. The result will be 3/48 + 1/48
= 4/48; showing that the total chance of improvement is 1 in 12, or 11
to 1 against it.
_=Whist.=_ To calculate the probable positions of certain named cards
is rather a difficult matter, but the process may be understood from
a simple example. Suppose a suit so distributed that you have four to
the King, and each of the other players has three cards; what are the
probabilities that your partner has both Ace and Queen? The common
solution is to put down all the possible positions of the two named
cards, and finding only one out of nine to answer, to assume that
the odds are 8 to 1 against partner having both cards. This is not
correct, because the nine positions are not equally probable. We must
first find the number of possible positions for the Ace and Queen
separately, afterward multiplying them together, which will give us
the denominator; and then the number of positions that are favourable,
which will give us the numerator.
As there are nine unknown cards, and the Ace may be any one of them, it
is obvious that the Queen may be any one of the remaining eight, which
gives us 9 × 8 = 72 different ways for the two cards to lie. To find
how many of these 72 will give us both cards in partner’s hand we must
begin with the ace, which may be any one of his three cards. The Queen
may be either of the other two, which gives us the numerator, 3 × 2 =
6; and the fraction of probability, 6/72, = 1/12; or 11 to 1 against
both Ace and Queen.
If we wished to find the probability of his having the Ace, but not
the Queen, our denominator would remain the same; but the numerator
would be the three possible positions of the Ace, multiplied by the
six possible positions of the Queen among the six other unknown cards,
in the other hands, giving us the fraction 18/72. The same would be
true of the Queen but not the Ace. To prove both these, we must find
the probability that he has neither Ace nor Queen. There being six
cards apart from his three, the Ace may be any one of them, and the
Queen may be any one of the remaining five. This gives us 6 × 5 = 30,
and the fraction 30/72. If we now add these four numerators together,
we have:--for both cards in partner’s hand, 6; for Ace alone, 18; for
Queen alone, 18; and for neither, 30; a total of 72, or unity, proving
all the calculations correct.
In some of the problems connected with Whist, it is important to
know the probability of the suits being distributed in various ways
among the four players at the table; or, what is the same thing, the
probable distribution of the four suits in any one hand. The author is
indebted to Dr. Pole’s “Philosophy of Whist” for these calculations. As
an example of the use of this table, suppose it was required to find
the probability of any other player at the table holding four or more
trumps if you had six. Take all the combinations in which the figure 6
appears, and add together the number of times they will probably occur.
That will be your denominator, 166. The numerator will be the number of
times that the combinations occur which contain a figure larger than 3,
in addition to the 6. This will be found to be 74, and the probability
will therefore be 74/166.
DISTRIBUTIONS. TIMES IN
1000.
8 2 2 1 2
8 3 1 1 1½
8 3 2 0 1
8 4 1 0 ½
8 5 0 0 0
7 2 2 2 5
7 3 2 1 19
7 3 3 0 3
7 4 1 1 4
7 4 2 0 3
7 5 1 0 1
7 6 0 0 0
6 3 2 2 57
6 3 3 1 35
6 4 2 1 47
6 4 3 0 13
6 5 1 1 7
6 5 2 0 6
6 6 1 0 1
5 3 3 2 155
5 4 2 2 106
5 4 3 1 130
5 4 4 0 12
5 5 2 1 32
5 5 3 0 9
4 3 3 3 105
4 4 3 2 215
4 4 4 1 30