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A
Guide to
Table Games Supervision
"The
intelligent way to start the day; by tying a noose around your neck"
By Dale
S. Yeazel
Chapter
15
Casino
math Part one
Part
Two
Explaining
the book “Practical Casino Math.”
In
the pursuit of knowledge that is usually only possessed by the elite of
the gaming industry, the seeker’s usual choice is to enroll in the much
acclaimed casino management course at UNLV. These acolytes are usually
motivated by the ambition to become casino managers themselves.
There
are some of us though, that have a reluctance to commit the time and money
to attend college and whose only ambition is to be over-qualified for
whatever position we might find ourselves filling.
“Practical
Casino Math” by Robert C. Hannum and Anthony N. Cabot is one of
those books that are highly
recommended by people of both the aforementioned groups. There is
certainly good reason for all of this fanfare. This book contains the
distilled knowledge of what is needed to understand the mathematics of
(and thus the mechanics) of casino games as well as the math behind such
topics as accounting and the comp system.
As
for myself; I always excelled in math in public school until I encountered
“Intermediate Algebra.” Perhaps this will give you some indication of
my math skills and why perhaps the authors under-estimated how many
idiots, like myself, would want to pursue this education.
In
any case, whether you are preparing to attend UNLV or are like myself and
want to read the books most coveted by the powers that be; I can’t
imagine you not benefiting from reading this article. I would not have
written it otherwise.
An
excellent resource for people like myself that managed to forget what
mathematics we did learn in school is: Basic Math and Pre-Algebra by
the real CliffNotes people and can be downloaded for $14.99 by clicking
here. This book has been referenced often in this work.
Chapter
1
Brief
History of Gambling
This
chapter manages to tell us two things that even the most retarded of us
already know: First, gambling has been around for thousands of years.
Second, the gaming industry makes a lot of money.
Chapter
2
Mathematical
Theory for Casino Games
I
have never read or attempted to read a book on probability that didn’t
start out easy and then at some point jump into Einstein-ion equations
without the least bit of warning. As a self-professed lump, I am reluctant
to blame the authors. I will attempt to explain the substance of this
chapter using additional emphasis where I deem necessary.
This
chapter asserts that the understanding of the mathematics of the games is
essential for table game managers. If this is the case, I imagine there
are and have been thousands of managers that were unqualified for their
positions. This chapter also states that the foundation of the required
knowledge is an understanding of probability, odds and percentages.
Probability
I
can’t imagine a more succinct explanation of probability than the one in
this book: “Outcomes in games of chance are, by definition, uncertain.
The branch of mathematics used to describe and calculate uncertainty is
called probability. Not surprisingly, the roots of modern
probability theory came from gambling problems posed by
seventeenth-century gamblers to mathematicians.”
The
“American Heritage Dictionary” defines probability as: “A number
expressing the likelihood of occurrence of a specific event, such as the
ratio of the number of experimental results that would produce the event
to the total number of events considered possible.” The concept that you
need to immediately embrace is: that the number representing probability
can be expressed as a fraction, decimal or percentage.
Fractions,
decimals and percentages
Let’s
start with the same moronic example that this and every other book use at
this point: the famed coin-flipping experiment of death. When citing this
and every other example in any book, the author feels compelled to bring
up the “fair” clause. That the coin, roulette wheel, dice or deck of
cards is “un-biased.” I think that most readers assumed that since
they were studying theoretical probabilities but I suppose that this is a
good opportunity to remind you that keeping our games random is a
responsibility not to be trivialized. 
The
chances of correctly predicting the outcome of a tossed coin is most often
expressed as a fraction:
Which
is often spoken as; “one half.” A description that is more revealing
is;” one divided by two.” The number one the left is the
“numerator” and the number on the right is the “denominator.” You
also see the fraction written as:
Since
the denominator is a larger number than the numerator, the number is less
than one (<1) but still greater than zero (>0). It is this type of
number we deal with in probability, as a value of “1” is considered a
certainty and a value of zero is considered an impossibility.
But
I regress from our example of the coin toss experiment. There are two
possible outcomes of a coin toss: heads or tails. The denominator
represents those two outcomes. In other words: the total number of
possible outcomes. The numerator represents the total number of
“desired” outcomes. In this case, heads or tails, whichever one
you called.
As
you may remember, I stated that a probability could be represented by a
fraction, a decimal or a percentage. How might we convert this fraction of
1/2 into a decimal? We might merely carry out the equation that the
fraction suggests and divide 1 by 2. 1/2 = .5 please note that the
number “. 5” is halfway between “1” and “0.”
The
easiest method of converting a decimal to a fraction is to: “read it,
write it and then reduce it.” You would read .5 as “five tens” and
then write it 5/10. By dividing both the numerator and denominator by the
largest number that can be evenly divided into both (5) we reduce it to
1/2.
To
convert from decimal to percentage we just multiply the number by 100 or
move the decimal point two spaces to the right (adding zeros as
necessary). So .5 becomes 50%. To convert a percentage to a fraction you
would first convert it to a decimal by dividing the percentage by 100
(move the decimal two spaces to the left) and then convert the decimal to
a fraction. 50% = .5 = 5/10 = 1/2.
Stating
the probability of an event
Armed
with this information, you should be able to describe the probability of
these events as a fraction, decimal or percentage:
Cutting
an ace from a full deck: 4/52 = 1/13 =
.076923076923076923076923076923077 ≈ .077 = 7.7%. There are four
aces in a deck that contains a total of fifty-two cards. 4/52 can be
reduced to 1/13. 1 divided by 13 = .076923076923076923076923076923077.
That gay little symbol “≈” means, “is approximately equal to.”
For instance, a gram of blow bought from an man standing on a corner of a
street named after a letter of the alphabet, could be represented
mathematically as (score ≈ 1 gram) or more realistically (score <
1 gram).
I
approximated the number by “rounding.” When rounding to a number: if
the number to the right of the digit is 4 or less, leave the number the
same. If the number to the right is 5 or higher, then round up.
Rolling
an eleven with a pair of dice:
There
is a total 6 combinations (outcomes or events) that can be rolled with one
die
There
are a total of 36 combinations that can be rolled with two dice. There are
six sides that can occur on the first die, times the six sides of the
second die.
On
a pair of dice, two of the combinations are eleven. They are 6-5 and 5-6.
If you have trouble envisioning the two combinations you can imagine one
green die and one red die. You can then roll an eleven by a green die six
and a red die five or by a green die five and a red die six.
So
we may express the chances (probability) of rolling an eleven as 2/36,
since are two combinations of eleven out of a total of 36. 2/36 = 1/18
≈ .0556 = 5.56%
Remember
that since there is only one combination of all the pairs, you would still
require green six and a red six to roll a twelve. 1/36 ≈ .0278 =
2.78%
BTW,
there are 216 combinations that can be rolled with three dice. 6 x 6 x 6 =
216
Winning
a straight-up bet in roulette.
There
are 38 pockets on an American roulette wheel: the numbers 1 –36 (18 red
& 18 black), “0” (green) and “00” (green). This means the
chances of predicting the number is:
1/38
≈ .0263 = 2.63%
Since
there are only 18 red pockets and 18 black pockets, instead of the
necessary 19 needed to make a bet on red or black a true odds payoff, the
chances of predicting red or black are:
18/38
= 9/19 ≈ .4737 = 47.37%
Of
course us gaming people are more comfortable speaking in terms of the
“odds against” something occurring instead of the probability or
chances of something occurring. A crap dealer will usually describe the
chances of throwing a twelve as “35 to 1 against” instead of “1
chance in 36.” For now, all you need to understand is that these two
expressions mean the same thing. I will further explain this phenomenon,
as well as how to compute house percentage (HP), later in this chapter.
Basic
Probability Rules
In
the preceding examples, all events had an “equal chance of occurring.”
Each of the 52 cards, thirty-six dice combinations and 38 pockets of the
roulette wheel all had an equal chance of occurring.
In
order to calculate the probability of more complicated events, some
additional rules need to be utilized. The understanding of these rules is
essential in order to know how to construct the correct solutions to most
problems.
Probability
Rule 1
“The
probability of an event will always be between 0 and 1.” This was
touched on briefly earlier. If the probability is “0” an event is
impossible. If the probability is “1” it is certain. The closer the
probability is to “1” the more likely it is to occur. The probability
of a coin toss being heads is .5.
Probability
Rule 2
(Complement
Rule)
“The
probability of an event occurring plus the probability of the event not
occurring equals one.” A simple example is to again compute the
chances of rolling an eleven. The chances of doing it are 2/36 and the
chances of not doing it is: 34/36:
2/36
+ 34/36 = 36/36 = 1 = 100%.
You
may not appreciate the importance of this rule until you start to work out
problems on your own. Remember this: it is often easier to compute the
probability of an event not occurring and then subtract that number
from 1, in order to determine the probability of it occurring. Say I want
to bet someone that he can’t throw one heads in three flips of a coin.
The easiest way to compute the probability of that event is to compute the
chances of not doing it and then subtracting that probability from zero.
1/2 x 1/2 x 1/2 = (1/2)3 = 1/8 is the probability of
throwing three tails in a row, so the probability of throwing at least one
head in three tosses is 7/8 = .875 = 87.5%.
A
couple of points to prevent confusion: first, if I lost you on the “1/2
x 1/2 x 1/2” used to compute the probability of tails appearing three
consecutive tosses, that will be covered in Rule 4A. Second, the “3”
in the “(1/2)3” is called an “exponent” and the number
is read “one half to the third power” or “one half cubed.” And in
case you have forgotten, indicates how many times the number is used as a
factor of itself. The number will always be multiplied by itself one less
time than the number of the exponent.
Probability
Rule 3
(Addition
Rule)
“For
mutually exclusive events, the probability of at least one of these
events occurring equals the sum of their individual
probabilities.” The key phrase is “at least one.” In other
words, not all events must occur, only one of them. Remember the word
“sum” indicates addition, so the probabilities are added
together to compute the probability of at least one of them occurring.
Remember that we are talking about mutually exclusive events, where the
occurrence of one event precludes the occurrence of the other(s).
The
probability of cutting a red card or the ace of spades is:
26/52
+ 1/52 = 27/52 ≈ .5192 = 51.92%
Probability
Rule 4A
(Multiplication
Rule-Independent Events)
“For
independent events, the probability of all of them occurring equals
the product of their individual probabilities.”
Unlike
rule 3 where we only needed one event to occur, we now need all
of the events to occur. When we only needed one of the events to occur we
were allowed the luxury of adding the probabilities. Now that we need all
of them to occur, we must multiply the individual
probabilities to compute the product that represents the probability of
all events occurring. When multiplying fractions: all numerators are
multiplied and the product becomes the numerator of the answer and all
denominators are multiplied and the product becomes the denominator of the
answer.
Remember
the bet I mentioned about throwing heads at least once in three tosses? It
begins by reasoning that in order not to throw heads at least once, you
must throw tails three times in tow. That is the only combination of the
eight possible outcomes of three coins tosses (whether they are tossed one
a time or all three at once) that doesn’t include at least one heads:
1/2
x 1/2 x 1/2 = (1/2)3 = 1/8 = .125 = 12.5%. Thusly, we remember
rule #2 and reason that the probability of throwing at least one head is:
P
= (1 - .125) = .875 = 87.5% = 7/8.
Now
don’t spaz out on me just because I chose to use the universal symbol
for probability “P” instead of the word “probability.”
It
is at this point that the authors choose to remind us about independent
trials, where the outcome of a trial has no effect on any other
trial. Do you see where this is going? They are preparing us for
the “gambler’s fallacy” lecture. In a nutshell, the gambler’s
fallacy is when a bettor believes that, fore instance, if black numbers
have appeared on a roulette wheel fifty-seven thousand times in a row, it
is only logical to bet “red” because it must be “due.”
Lets
go back to my beloved coin-flipping bet. Suppose I am watching a friend
flipping a coin and he throws tails in the first two tosses He says;
“Hey math guy. You just saw me flip tails twice in a row. I know a
sophisticated gentleman like you knows that the chance of throwing tails
three times in a row are only one in eight or, 7 to 1 against. I’ll be a
sport and bet one dollar against your five dollars, that the next toss
will be tails (5 to 1 payoff).” I would say; “Not so quick there
buddy! Are you trying to tell me you have a goddamn magic coin that knows
it has flipped tails twice in a row? The chances of that coin showing
tails again is the same as any other time in it’s existence, one in
two.”
Probability
Rule 4B
(Multiplication
Rule – Dependent events)
“For
non-independent events, the probability of all of them occurring equals
the product of the conditional probabilities, where the conditional
probability of one event is affected by the event(s) that came before
it.”
Although
this rule is often used to compute probabilities in blackjack because the
odds in the game change as cards are removed from the deck: I find that
poker is an easier game to illustrate this rule.
What
is the probability of being dealt a club flush in five cards?
P
= 13/52 x 12/51 x 11/50 x 10/49 x 9/48 = 154440/311875200 = 15444/31187520
≈ .0004951981 = .04951981%. The probability of being dealt a club on
the first card is 13/52 because there are 13 clubs out of a total of 52.
If you are dealt a club on the first card, there are only 12 clubs
remaining out of 51 cards. And thus the progression continues with there
being one less club and one less total cards, after each card is dealt. In
a desperate attempt to avoid computing the largest common multiple of that
huge fraction so I could reduce it (I did take an unnecessary zero on the
end of both number by moving the decimal one to the left or dividing both
the numerator and denominator by 10) I chose to divide the fraction and
convert it to decimal. Of course the resulting number is not completely
accurate since I had to round of but I deemed it close enough for this
illustration.
What
is the probability of being dealt a flush in five cards?
P
= 52/52 x 12/51 x 11/50 x 10/49 x 9/48 = 11880/5997600 =
1188/599760 ≈ .001981 = .1981%. The first fraction represents the
chances of drawing a card of any suit out of 52 cards. Since the numerator
is the same as the denominator, I could “cancel” each of them out.
Just imagine I multiplied all the fractions and then divided both the
numerator and denominator by 52. You can cancel matching numerators and
denominators even if they belong to different fractions:
20/23
x 9/20 = 1/23 x 9/1 = 1/23 x 9 = 9/23
What
is the probability of being dealt a royal flush in five cards?
P
= (20/52) (4/51) (3/50 (2/49) (1/48) = 480//311875200 ≈ .00053978 =
.053978%.
I
remember that I once tried to write out this problem and came up with:
P
= (5/52) (4/51) (3/50) (2/49) (1/48)
This
friends, is the chances of being dealt a royal flush in a specific suit!
But my blunder can serve you well if it reminds you to think carefully
about how you decide to frame your problem.
Yes,
I decided to go with the parentheses this time instead of the “x” sign
for multiplication. The dividends of the fractions inside the parentheses
are multiplied by each other to compute the answer.
At
this point (assuming you have read this far) I imagine your reaction to be
one out of possible three:
I’m
sorry; I have tried to work you up as slowly as possible without skipping
any steps. You will thank me later because a certain amount of math
knowledge is required in order to study or even understand
probability.
If
you truly find this article an exercise in belaboring the obvious: your
time would be better spent by just reading “Practical Casino Math” and
skipping this prep course.
We
share this journey together brothers and sisters! I will teach you
everything in “Practical Casino Math” just as soon as I figure it out
myself.
Pascal
and Fermat
In
the beginning, before probability theory, Man lived like the
animals.
He
created game rules through intuition rather than reasoning.
Blaise
Pascal and Pierre de Fermat are generally considered to be the fathers of
probability. In the 17th century they worked out gambling problems for a
French nobleman and gentleman gambler by the name of the Chevalier de
Mere.
In
the year 1654, the Chevalier de Mere made considerable money betting that
he could throw at least one six in four throws. Since he incorrectly
reasoned that the probability of winning that bet was 4/6 (1/6 x 4
throws), he assumed that the chance of throwing a pair of sixes with two
dice must be six times as much, or 24. After losing money at this new
wager, he contacted Pascal and asked him to solve the mystery. Pascal
corresponded with de Fermat and together they worked out the problem and
the science of probability was born.
Lets
analyze these problems:
The
probability of throwing one six in four rolls of a single die:
We
are going to use rule 4 to determine the probability of not throwing a six
in four rolls and then subtracting that number from one (rule 2) to
determine the answer.
P
(no sixes) = (5/6) (5/6) (5/6) (5/6) = 625/1296 = .4823 = 48.23%.
So
now we subtract the probability of not throwing a six from one, to
determine the probability of throwing at least one:
P
(at least one six in four rolls) = (1 - .4823) = .5177 = 51.77%.
We’ll
work out what the HP (house percentage) was on this bet later. For now at
least we know he had a 51.77% chance of winning.
Now
you might be wondering; “If this Chevalier guy was such a gentleman
gambler, why didn’t he just bet that he could throw a six in three
rolls.
After
all: 1/6 + 1/6 + 1/6 = 3/6 = 1/2 = .5 = 50%.
If
you think this, you know just enough about probability to be someone’s
fish in a dice rolling contest during the renaissance period:
(5/6)
(5/6) (5/6) = 125/216 ≈ .5787 = 57.87%. This is the chance of not
throwing at six in three rolls.
(1
- .5787) = .4213 = 42.13%. You see, the Chevalier was a gentleman but not
a fool.
To
compute the probability of throwing one twelve in 24 rolls, we compute the
probability of not doing so:
(35/36)24
≈ .9722224 = 0.50856822447223562323575026806816 ≈
.50857 = 50.857%.
We
subtract that number from one to compute the chances of rolling twelve one
time in 24 rolls.
(1
- .50857) = 0.49143 = 49.143%.
The
first number indicates we are going to multiply 35/36 by itself 23 times.
To multiply the numerator by itself 23 times, multiply the denominator by
itself 23 times and then divide the numerator by the denominator, would be
the epitome of masochism. I divided 35 by 36 so I could convert the
number to a decimal and then multiply it times itself 23 times.
John
Scarne is his book, “Scarne’s Complete Guild to Gambling” tells a
tale from 1952, of two New York City gamblers by the names of “Fat the
Butch” and “The Brain.” The Brain offered to bet Fat the Butch $1000
that he couldn’t throw a twelve in 21 rolls. Fat the Butch figured that
since there are 36 dice combinations and only one of the twelve, on
average it should take 18 rolls to roll a twelve and that he was getting
the best of it. Twelve hours and $49,000 later, “Fat the Butch”
decided that there must be something wrong with his logic and gave up.
Which
just goes to show you that hustlers can use the oldest cons in the world,
as long as the sucker hasn’t heard of them.
Odds
“Odds
are another way of expressing probability. Whereas probability represents
the relationship between the desired event and all possible events, odds
describe the relationship between a desired event and all non-desired
events.”
Fortunately,
us casino folk are more accustomed to expressing probability as “odds
against” an event:
The
chances of winning a straight-up bet in roulette are 1/38 but it is
usually expressed as 37 to 1 against.
The
chances of rolling an eleven on the come-out roll (or any other time) is:
2/36
= 1/18 = 17 to 1
The
chances of the dealer having a ten under his ace and thus having a
blackjack is:
16/51
= 1/3.1875 = 2.1875 to 1.
There
are 16 face cards from a remaining 51 unknown cards, that fraction can be
reduced by dividing the denominator by the numerator (for a change). That
makes the numerator a value of 1 and makes the denominator the smallest
number possible.
And
in case you haven’t figured it out for your self yet, the formula for
converting probability to odds is:
If
the probability of an event is x/y, then the odds against the event are
y-x to x. Lets take an example from craps: there are six combinations of
seven out of thirty-six total. So the chances of rolling a seven are 6/36
= 1/6 = 5 to 1.
If
we reverse engineer this problem x = 1 and y = 6 so: 6 –1 to 1 = 5 to 1
The
difference between 5 to 1 and 5 for 1
In
craps, if a bet on “seven” paid 5 to 1, the bettor would bet a
dollar and if the bet won, he would collect a total of $6 and down.
But the bet is paid as it is listed on the layout as 5 for 1. The
“for” means “a total of and down” in other words, we’ll give you
five but we are going to keep the one. The number preceding “for” it
is what the bettor can expect to collect, for each dollar or check bet,
assuming he takes the bet down after it wins. So a payoff of 5 for 1 is
the same as 4 to 1.
When
training BJ dealer to be crap dealers, I would pose the question; “What
does a “3 to 2” snapper pay if we want to convert the “to” to a
“for?” I seldom get a correct response for this, for few can come up
with: 5 for 2. In other words, the bettor collects 5 and down, for every 2
bet.
Permutations
If you are anything like me, this is the
part where things will start to get difficult. If I have failed to reach
you thus far, you might consider re-reading the previous material,
although I promise to take this as slowly as we both may require.
This book defines “permutations” in
this way; “Permutations” refer to the number of different ways
in which objects can be arranged in order. In a permutation, each item can
appear only once and each order of the item’s arrangement constitutes a
separate permutation.
Again, I find that reading the dictionary
definition is also helpful: “An ordered arrangement of all or some of
the elements of a set.”
A clever observation I read in
“Probability Without Tears” by Derek Roundtree, asserts that a
“combination lock” is miss-named. If a locks’ combination was say:
22 – 10 – 32 and it was truly a combination lock, then dialing 10 –
22 – 32 should also open the lock. But the locks’ combination must be
dialed in a certain order. Since this is the case, the lock should be call
a “permutation lock” since the numbers must be permuted (chosen,
dialed) in a certain order. This book also gives a great example that
illustrates the difference between a combination and a permutation:
Any combination of things can be permuted
– re-ordered or re-arranged – in several different ways. For instance,
let’s select the first three letters from the alphabet. We could write
these letters down in six different ways:
ABC BCA CAB ACB BAC CBA
(i) How many combinations do we have here?
(ii) How may permutations?
(i) There is only one combination – the
first three letters of the alphabet are the same three letters, no matter
what order they are written in.
(ii) There are six permutations – each
different order is a new permutation.
The most important thing to remember is
that in a combination, the order isn’t important but in a permutation it
is. Also, there will always be more permutations than combinations.
Creating a “probability tree” is often
the easiest way to frame a solution to a problem. Imagine there were three
ping pong balls, labeled “a” “b” and “c.” If the first ball
drawn is an “a” then the second on can be a “b” or a “c.”
After the second ball is drawn, the identity of the third ball is obvious,
as there is only one ball remaining. You can see from this tree that there
is 6 permutations of the three letters and since they all have an equal
chance of appearing, there is a:
1/6 ≈ .1667 = 16.67%, chance of any
of them appearing.
At this point you might well imagine that
since there were three possible outcomes of the first draw and two
possible outcomes of the second; the mathematical shortcut we are looking
for is:
3 * 2 = 6.
It is at this point in “Practical Casino
Math” the authors illustrate the permutation formula by using an example
of picking the first two horses to cross the finish line in a five horse
race. I found that the way they described this problem was more confusing
than helpful. I was fortunate enough to find myself sitting next to a math
professor, while I was passing time in the waiting room of a car
dealership by reading “Practical Casino Math.” With the help he gave
me and the insights I gained from the book “The Mathematics of Games and
Gambling” by Edward Packel, I will illustrate this problem in a way I
think makes it easier to understand.
But first lets work our way up to this by
studying some simpler problems since you must learn to understand simple
problems and their solutions before you can even aspire to achieve oneness
with the esoteric beauty of mathematical law. So here is the best of the
problems from TMOGAG that are designed to make you understand the forces
you hope to control.
But really I think Dr. Packel said it best;
“A crucial concept in what follows will be the distinction between a permutation
and combination of objects (people, horses, cards, letters, or
other entities). A permutation is a selections of objects in which order
is taken into account. A combination is a selection in which order is
unimportant.”
Question 1.
In how many different ways can a bridge
player arrange a particular (13 card) bridge hand?
Solution:
Remember the example of the first three
letters of the alphabet being the first three letters and thus the same
combination of three letters no matter how you arrange them? Well these 13
cards are the same 13-card combination, no matter how you arrange them.
This means this is a permutation problem.
If we think of ordering the given hand from
positions 1 through 13, there are 13 choices for the first position, then
12 choices for the second, then 11 for the third, and so on down to 2
choices for the twelfth position with the 1 remaining card going in the
last position.
Thus there are: 13*12*11* ... *3*2*1 =
6,227,020,800 arrangements of a given bridge hand.
Notice that I choose to use the asterisk to
denote multiplication and this “…” thing is called an “ellipsis”
and is the way most writers choose to avoid writing out the entire
equation but not using “factorials” that the reader hasn’t been
introduced to yet.
If you want to see that equation written
out fully it would be:
13*12*11*10*9*8*7*6*5*4*3*2*1 =
6,227,020,800
Why do we multiply rather than add? Because
we are following probability rule 4B which calls for multiplication of
conditional events.
Question 2.
How many different finishes among the first
3 places can occur in an 8 horse race (excluding the possibility of ties)?
Solution:
There are 8 choices for the winner, then 7
for the places horse (2nd place), then 6 for the show horse (3rd place).
Accordingly there are 8*7*6 = 336 different orders of finish as far as the
payoff windows are concerned (ignoring ties).
Notice that the question isn’t asking how
many combinations of three horses will be the first to cross the finish
line, it makes the distinction of order, so this is a permutation problem
and rule 4B was used again.
Question 3.
How many foursomes can be formed from among
7 golfers?
Solution:
Consider a score card sighed by the
foursome. There are 7 possibilities for the first name on the card, six
possibilities for the 2nd name, then 5 possibilities for the 3rd and 4
possibilities for the last name. This gives 7*6*5*4 = 840 possibilities,
where different orderings of the same 4 players have been counted as
different foursomes. But the wording of the question indicates that order
is unimportant, so we realize that the result of 840 contains considerable
duplication. In fact a given foursome can be ordered in 4*3*2*1 = 24
different ways, so our total of 840 has counted each foursome 24 times. We
see then that there are 840/24 = 35 possible foursomes that can be chosen.
What we are supposed to learn from this is:
that in order to compute the number of combinations, we must compute the
number of permutations and then divide that number by the number of
possible permutations that can be formed by the number of objects in the
selection.
Question 4.
How many different 13-card bridge hands are
there?
Solution:
Again order is unimportant here. We wish to
know how many different hands of 13 cards can be chosen from a 52-card
deck. If order were important there would be
possibilities. Since order is unimportant
we must divide by:
13*12*11*10*9*8*7*6*5*4*3*2*1
We thus obtain the smaller but still
astronomical count of
different bridge hands.
Question 5.
How many different baseball-batting orders
are possible with a 25-player roster in which any player can bat in any
position?
Solution:
How the wording of the question demands
that order be taken into account, so we do not factor out duplications.
The answer is:
a rather large number of batting orders.
Two more points from TMOGAG for summary and
we will be finished with our discussion on permutations. First, all
examples involve choosing without replacement some subset from a set of
objects. Thus, as each choice is made (position is filled) the number of
choices for the next position decreases by one. This accounts for the
repeated appearance of consecutive integers* (decreasing by
one) and motivates the notation developed in the next section.
Second, you use only one calculation to
compute the number of permutation but to determine the number of
combinations requires an additional calculation,
*Integer: Any member of the set of
positive whole numbers (1, 2, 3…), negative whole numbers (-1, -2,
-3…), and zero.
Preparing to learn factorials
If I dreaded actually thinking about and (gasp)
remembering the probability rules so I could construct solutions to
probability problems, I cringed at the thought of conquering factorials so
I could construct (or at least understand someone else’s) solutions to
life’s more complex (and thus more meaningful) questions. Questions
like; “Well I don’t know too many games where everyone agrees to
play five-card showdown. What are my chances of bettering my pair of jacks
or better if I draw three cards?” “What if I keep a kicker?” Or;
“What are my chances of hitting of hitting a three-spot in keno? I
really need to talk to that man standing on a corner of a street named
after a letter of the alphabet.”
The source that I found easiest to
understand when it came to factorials was a website called
“Purplemath” by Elizabeth Stapel. It can be found at
http://www.purplemath.com/index.htm
The section dealing with factorials can be
found at http://www.purplemath.com/modules/factorial.htm.
The following is excerpts from that web
page along with my own commentary and analysis.
“Factorials are very simple things.
They’re just products, indicated by an exclamation mark. For instance,
“four factorial” is written as “4!” and means 1*2*3*4 = 24. In
general, n! (“enn factorial”) means the product of all the whole
numbers from 1 to n
n! = 1*2*3*...*n
For various reasons, 0! is defined to be
equal to 1, not 0. Memorize this now.”
I find this to be a wonderful explanation
except that the lists the integers in ascending order (1*2*3*4 = 24). I am
more accustomed to seeing writers express that in descending order
(4*3*2*1 = 24). Which of course would make the last equation read: n! = n*
... *3*2*1
I would think the easiest possible
explanation would be: a factorial is the number in the factorial,
multiplied by every whole number that is less than it. You often see
factorials referred to as “products.” This is certainly true as the
only way to “simplify” a factorial is by multiplication, either by
paper and pencil or by calculator.
Evaluate 6!
“Evaluate,” means the same as “write
it out.”
6*5*4*3*2*1
Simplify 6!
“Simplify” means to compute the final
product.
6*5*4*3*2*1 = 720
I discovered an excellent online calculator
(Java) that you can use to simplify factorials as well as use the
permutation and combination formulas that are about to be explained.
http://www.wiley.com/college/mat/gilbert139343/java/java05_s.html
Another online calculator that can be used
to for these functions and in addition it can be used to divide factorials
can be found at.
http://www.hutchon.net/Factorials.htm
Simplify 6!/4!
You can do this on the calculator or you
can do it with paper and pencil.
The numbers that have the red lines through
them have been “cancelled.” Canceling identical numerators and
denominators is much like reducing a fraction by dividing the numerator
and denominator by the same number. Since the entire denominator was
cancelled, it has a value of 1 and I didn’t even bother to show it.
Yes the denominator indicates that it is
the product of 14! * 3! Now you might be thinking the first step in
simplifying this fraction is to determine what the product of 14!3! is.
No, we can utilize those factorials to cancel.
In the second fraction of the top row, I
realized that both 17! and 14! contain the numbers
14*…*4*3*2*1. I could then cancel them
out and was left with 17*16*15 for a numerator and 3*2*1 for a
denominator. In the first fraction of the bottom row, I reduced the
fraction by dividing numbers in the numerator by numbers in the
denominator.
Another definition of
the letter “P.”
Up until now, the letter “P” has stood for
“probability.” But it can also stand for “the number of different
permutations.” When used for this purpose, you will see either nPr or
Pn,r. “n” represents the total number of objects and “r”
represents the number of objects to be selected. It may be spoken as “n
objects taken r at a time” or “n taken r.”
This method of expressing permutations can be used
to easily compute the number of permutations. If a horse race has five
horses and we wish to compute the number of exactas that are possible
(first two horse to cross the finish line and order is important) we can
express that problem as 5P2 or P5,2. As you will soon learn,
there is a permutation formula that we can use that involves dividing 5!
by the number of unselected items. The simpler method is to factor 5 (n)
by the number of times of the value of r or in this case 2.
So: 5P2 = 5*4 = 20.
Let’s do one more problem to make sure we
understand this process. You have twelve contestants in a beauty pageant
and there will be three selected: one for first place, one for second
place and one for third place. 12P3 = 12*11*10 = 1320 you see we factored
the total number of contestants by the number of selections.
If all of the objects are to be included in a
selection then the permutation can be represented as nPn. Since all the
objects are included, there is no unused portion of the objects to divide
into the total number of objects. Subsequently, nPn or Pn,n is
the same as n!.
I have one more concept that I need to explain
before I can go back to PCM.
What is the value of
?
Remember that the numerator of the first fraction
represents all objects of a group are selected so in the next fraction the
numerator is 4! The denominator of the first fraction can be simplified to
2!
The importance of this equation becomes clearer when
you consider that it proves:
which is the basis for the permutation formula that
we are finally ready to learn.
The dreaded permutation formula of death
I believe that I have finally learned enough about
math (and shared it with you) that I can now begin to solve and explain
the horseracing example in PCM.
This is the permutation formula that can be used to
compute the number of permutations in any example. Although, there is
sometimes an easier way to compute the number of permutations, this
formula will always work. It states that the total number of permutations
of n number of objects taken r at a time equals n factorial divided by the
total number of objects (n) minus the number of objects selected
factorial. The denominator is just a method of eliminating all of the
un-chosen objects. And remember, even if Pn,n and the denominator equals
0!, 0! equals 1, so the equation still works.
The example used in PCM is to compute the number of
exactas possible in a five horse race. Remember, since in an exacta the
order of finish is important, this is a permutation problem. So, n = 5
(the total number of objects) and r = 2 (the first two horses to finish).
And we can restate this problem as:
However the way PCM presented the solution to this
problem and the way it presented the permutation formula left me very
confused (I was tripping, bro) and was the motivation for me to write this
article.
Let us start with the bottom line which defines n!.
We know that n! will equal n multiplied by every whole number that is less
than n (except zero). So, n! equals n * (n-1) * (n-2) … and the reason
for the “x 2 x 1” at the end is because the authors didn’t know how
large of a number n would be but they knew that the value of the last two
pairs of parentheses will always equal 2 and 1. In the case of our example
(n – 3) would equal 2 and (n – 4) would equal 1.
The last pair of parentheses on the top line is what
really had me thrown for a loop until I remembered what I had learned
about 5P2 and how to simplify it. Rather than compute the value of n! and
divide it by (n – r)! they decided to merely take n and factor it by the
number of objects indicated by r. Or in the case of our example of the
horse race, the part of the equation following the last equal sign would
read: 5 x 4, as (n – r + 1) would equal 4. This (n-r+1) thing is merely
a way of expressing that shortcut.
Example:
Numbers Games
PCM gives this example of a numbers game where the
bettor must pick a three-digit number. Since the numbers must be in exact
order, one might think this is a permutation problem. But since a single
digit number can appear more than once (4-4-3) it isn’t. The chances of
picking the correct three-digit number are 10*10*10 or one in ten
thousand.
Combinations
Of course we remember that in a combination the
order of the selected objects isn’t important. So, there is a different
formula to use for computing the number of combinations. Earlier we
learned that you compute the number of combinations by computing the
number of permutations and then factoring out the duplications. Now we
will use a precise formula for accomplishing this.
n again represent the total number of objects and x
represents the number of chosen objects. The equation to the left of the
equal sign is read “n choose x.” You may also see it expressed as C n,x
or nCx. Lets take some examples and see how this formula works.
Compute the number of possible quinellas in a five
horse race (first two finishers, order not important).
 .
Compute the number of foursomes that can be created
from seven golfers.
Now this last example exemplified my initial and
profound shock upon thinking that there was more to computing the number
of combinations than by dividing the total number of permutations by the
number of unused objects. After all, isn’t that what we did in solving
this problem without the sacred combination formula? No, not exactly or
maybe, I really don’t know. But in the previous solution, we only took 7
by a factor of 2 to deduce the total number of permutations. In this
solution, we took a full factor of 7!, which meant that the denominator
needed to be the amount of x factorial times the amount of the unused
objects, factorial.
Now at this point your reaction may very well mirror
my own, which says something like; “This is a bunch of shit! You show me
a perfectly good way of doing something, then you have to come up with
some bullshit complicated stupid bullshit way that requires me to
understand this shit!” But there is a method to this madness. You see,
by having a single all encompassing equation to compute permutations and a
matching one for combinations, they have eliminated almost all the
mistakes you can make setting up a solution. The only major (and fatal)
mistake you can make is not to carefully consider whether your problem is
one where the order of the selection is important. Remember, in a
permutation the order is important and in a combination it isn’t.
We will finish this discussion on combinations and
finally rid ourselves of what I have considered to be the most arduous and
torturous segment of “Practical Casino Math” by demonstrating the two
combination examples shown in the book.
Example: Lottery
The winner must pick all six numbers out of a total
of forty-two balls, numbered 1 – 42 and order is unimportant.
C 42, 6 = 42! / 6! (42-6)! =
5245786.000000001 Ha, I used the online calculator for that one by
entering the total objects (42) the number of objects in the first set (6)
and the second set (unused) (36). This means the pathetic humanoid that
buys a ticket has only one chance in about 5.2 million.
Example: Poker
The probability of being dealt a royal flush in five
cards, (I probably screwed up my last attempt at this one).
C 52, 5 = 52! / 5! (52-5)! =
2598959.999999999 (Yes, Mr. Calculator at work again). And yes, this
is the chance of being dealt one of the four royal flushes (or any other
hand really) because that number is the number of possible five card
combinations with a 52-card deck. So to get the final answer we divide 4
by 2598959.999999999 and get 1.5390771693292701696063040600856e-6 on our
Microsoft calculator. That “e-6” crap on the end means we have to move
the decimal place six spaces to the left. That will make it
“approximately equal to” .00000153907772 or 1 in 649,740. We got that
calculation by dividing 1 by .00000153907772
In case you are wondering why we divided 4 by the
total five-card combination that could occur, it all goes back to
numerator being the total number of desired results and the denominator
being the total number of equally possible results.
Expectation (Expected
Value)
Expected value isn’t a concept that I immediately
embrace, as I am more comfortable in the realm of house percentage (HP).
But perhaps that’s why I’m a lump, so lets make sure we have a handle
(no pun intended) on this expected value thing.
“Expectation, or expected value, represents how
much money a player can expect to win or lose in the long run on a
particular wager. If the player’s expectation is negative, as is typical
for most bets made in a casino. The player can expect to lose money over
the long haul.”
The explanation I think all these books dance around
but not quite nail, is this: It is the “taxes” taken from each bet,
whether it wins, loses or pushes. Which brings me to another point that I
have only read in a couple of sources. That is that there are some
gambling problems that must be assigned dollar values, in order to be
computed. Expected value is one those problems.
EV = ∑ (Net payi
x Pi)
“EV” means expected value. Ahoy, we have another
gay little symbol sighting! “∑” means “the sum of everything
inside the parentheses.” “Net payi” equals the net payoff
and “Pi” is the probability of winning the payoff. Again,
this is one of the sacred formulas, which if properly used, will tell the
expected value of any bet (assuming you can compute the chances of the bet
winning).
Suppose I create my own casino game. You cut a card
from a 52-card deck and if you cut a spade I pay you one dollar. If you
don’t cut a spade, you lose one dollar. I think I’ll call it “Rape
and Pillage.” I can just see the first drop box on a game, stenciled
with “RP-1.” But any way, here is the equation in use, showing us how
far I’m bending you over.
EV = ∑ [(+$1) * (1/4)] +
[(-$1) * (3/4)] = -$.50
There you go. Your dollar becomes fifty cents the
second you put it in action on my game, win, lose or draw. If you were
thinking that the answer was going to be closer to -$.66; you didn’t
visualize the problem correctly.
You lose $1 three times for a sub-total of -$3. You
win $1 one time for a sub-total of $1. Hence you net balance after all
four possible occurrences magically appear, is -$2. -$2 divided by the
four trials is -$.50
On an American “00” roulette wheel, the payoff
for a bet on a single number pays 35 to 1. If a player makes a $5 bet the
expected value of his bet is:
EV = ∑ [(+$175) (1/38)
+ (-$5) (37/38)] = -$0.263
Which means the player, on average, can expect to
lose $.26 every time he makes the bet. Conversely, the casino can expect
to win $.26 every time that bet is made.
This also demonstrates that not all gambling writers
choose to frame their problems on the hypothetical $1 bet. If the reader
didn’t read that problem carefully, he might think the player is losing
$.26 on every dollar bet.
House
Advantage
“House advantage, or house edge, is the opposite
or negative of the player expectation, expressed as a percentage of the
wager. Sometimes house advantage is referred to as PC, for percentage,
although care must be taken when discussing PC since this term is
sometimes used for hold percentage. Hold is not the same as house
advantage and there is much confusion amount some casino personnel
regarding the meaning of “PC.” Hold and hold percentage are discussed
in Chapter 3 when common casino measurements are examined in more
detail.”
”House advantage” (HA) or as I am more accustomed to referring to it:
“house percentage” (HP) is the house’s advantage (or expectation) on
a bet, expressed as a percentile, instead of a dollar amount. Expected
value changes every time the bet does. Since HA is expressed as a
percentage, it remains the same, regardless of the amount of the bet.
The authors mention two issues that can cause
confusion regarding HA. The first and one of particular interest to me, is
whether ties are included in HA in baccarat and most notably, a roll of
twelve on the come-out roll for the don’t pass bettor. Writers such as
John Scarne, believe that the amount bet on the don’t pass that is
pushed on a roll of twelve should be included in the total amount put at
risk by the bettor. Michael Shackleford AKA “The Wizard of Odds”
classifies the push on twelve is a “non-event” and thus comes up with
a slightly lower HA for the don’t pass. Both methods will be covered in
their entirety in chapter 4. The second issue involves whether to merely
include the “base bet” in games like Caribbean Stud Poker, Let It Ride
and Three Card Poker, or if the average amount put in action should be
considered.
Computing House
Advantage Directly From Odds
The authors of “Practical Casino Math” certainly
gained my respect with their inclusion of this section. I will present a
method of computing HA without the need to use the expected value formula.
I have only seen this taught by John Scarne and the technique is unknown
to many people that are considered to be experts at gambling related
mathematics.
If the payoff for a bet is x to 1 and the true odds
against are y to 1, the house advantage is:
(y-x) / (y+1)
If the payoff doesn’t “end in one” then the
process becomes a bit more difficult to remember. If the payoff is x to z
and the true odds against are y to w, the HA is equal to:
(yz-xw) / [z(y+w)] And in case you don’t
know, when two values are together, like in the case of “yz” then that
means they are multiplied.
But who wants to remember all that shit, right? Let
me tell you there is an easy way to remember this and I learned it from my
hero, John Scarne.
“The house advantage is obtained by dividing
the difference between what a bet is paid and what it should be paid, by
what the bet would have been paid “and down” if the bet paid true
odds.”
Now if that sounds like a mouthful, you should not
only understand it but it should become your new mantra when looking at a
game of chance. It will become clearer as you see me work out problems
with it.
One number straight-up in roulette:
One dollar pays $36 (35 to 1) and down when it
should pay $38 (37 losers versus one winner). What is the difference? Two.
2 is now divided by the amount the bet should have paid and down ($38).
2/38 ≈ 0.05263 = 5.263%
Don’t forget to move the decimal point over two
spaces in order to convert it to percentage.
Place bet on the six or eight:
A six-dollar place bet on the six pays $7. Since the
bettor is betting on six to be rolled before a seven, the true odds payoff
(6 to 5) for a six-dollar bet should be $7.20
Difference: $.20
Total of true odds payoff and down: $13.20
.20/13.20 = 0.01515 = 1.515%
Big 6 or Big 8:
A $1 bet on the big 6 should pay $1.20 but it only
pays $1.
Difference: $.20
Total of true odds payoff and down: $2.20
.20/2.20 ≈ .0909 = 9.09%
So you’re thinking; “Wow, I’d be really stupid
to bet on the corner red when I could put that money to good use by
investing it on the hard 6.” Well, lets do a little research for the
prospectus.
Hard 6 or 8:
A bet on the hard six is betting that the shooter
will throw a hard six (3-3) before he throws and easy six (5-1 or 4-2) or
a seven. A $1 bet pays $10 and down when it should pay $11 and down (10
for 1 or 9 to 1 payoff). That is because there are ten losing combinations
(4 easy way & 6 sevens) and only one winning combination.
Difference: $1
Total of true odds payoff and down: $11
1/11 ≈ .0909 = 9.09%
Continued
©
2005 Dale S. Yeazel
www.CasinoDealers.net
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