Download 1 Probability Basics 1.1 Meaning of Probability It`s hand 229 of day 7

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Probability Basics
1.1 Meaning of Probability
It’s hand 229 of day 7 of the 2006 World Series of Poker
(WSOP) Main Event, the biggest poker tournament ever
played, and after 8,770 entrants have been eliminated, it is
now down to the final three players. The winner gets a cash
prize of $12 million, second place just over $6.1 million,
and third place gets about $4.1 million. Jamie Gold is the
chip leader with $60 million in chips, Paul Wasicka has $18
million, and Michael Binger has $11 million. The blinds are
$200,000 and $400,000, with $50,000 in antes. (Please note:
for readers unfamiliar with Texas Hold’em, consult the
Appendices A and B for a brief explanation of the game and
related terminology used in this book.) Gold calls with 4♠
3♣, Wasicka calls with 8♠ 7♠, Binger raises with A♥ 10♥,
and Gold and Wasicka call. The flop is 10♣ 6♠ 5♠. Wasicka
checks, perhaps expecting to raise, but by the time it gets
back to him, Binger has bet $3.5 million and Gold has
raised all-in. What would you do, if you were Paul Wasicka
in this situation?
There are, of course, so many issues to consider. One
relatively simple probability question that may arise is this:
given all the players’ cards and the flop, if Wasicka calls,
what is the probability that he will make a flush or a straight?
This is the type of calculation we will address in this book,
but before we get to the calculation, it is worth examining the
question a bit. What does it mean to say that the probability of
some event, like Wasicka’s making a flush or a straight, is,
say, 55%? It may surprise some readers to learn that
considerable disagreement continues among probabilists and
statisticians about the definition of the term probability. There
are two main schools of thought.
The frequentist definition is that if one were to record
observations under the exact same conditions over and over,
with each observation independent of the rest, then the event
in question would ultimately occur 55% of the time. In other
words, to say that the probability of Wasicka making a flush
or a straight is 55% means that if we were to imagine
repeatedly observing situations just like this one, or if we
were to imagine dealing the turn and river cards repeatedly,
each time reshuffling the remaining 43 cards before dealing,
then Wasicka would make a flush or a straight 55% of the
time.
The Bayesian definition is that the quantity 55% reflects
one’s subjective feeling about how likely the event is to
occur: in this case, because the number is 55%, the feeling is
that the event is slightly more likely to occur than it is not to
occur.
The two definitions suggest very different scientific
questions, statistical procedures, and interpretations of results.
For instance, a Bayesian may discuss the probability that life
exists on Mars, while a frequentist may argue that such a
construct does not even make sense. Frequentists and
Bayesians have had spirited debates for decades on the
meaning of probability, and consensus appears to be nowhere
in sight.
While experts may differ about the meaning of probability,
they have achieved unanimous agreement about the
mathematics of probability. Both frequentists and Bayesians
agree on the basic rules of probability—known as the Axioms
of Probability. These three axioms are discussed in Section
1.3, and the methods for calculating probabilities follow from
these three simple rules. There is also agreement about the
proper notation for probabilities: we write P(A) = 55%, for
instance, to connote that the probability of event A is 55%.
1.2 Basic Terminology
Before we get to the axioms of probability, a few terms
must be clarified. One is the word or. In the italicized
question about probability in Section 1.1 above, it is unclear
whether we mean the probability that Wasicka makes either a
flush or a straight but not both or the probability that Wasicka
makes a flush or a straight or both. English is ambiguous
regarding the use of or. Mathematicians prefer clarity over
ambiguity, and have agreed on the convention that A or B
always means A or B or both. If one means A or B but not
both, one must explicitly include the phrase but not both.
Of course, in some situations, A and B cannot both occur.
For instance, if we were to ask what the probability is that
Wasicka makes a flush or three 8s on this hand, one can see
that they both cannot happen: if the turn and river are both
8s, then neither the turn nor the river can be a spade, since
Wasicka already has the 8 of spades. If two events A and B
cannot both occur, i.e., if P(A and B) = 0, then we say that
the events are mutually exclusive. We usually use the
notation AB to denote the event A and B, so the condition for
mutual exclusivity can be written simply P(AB) = 0.
The collection of all possible outcomes is sometimes called
the sample space, and an event is a subset of elements of the
sample space. For instance, in the WSOP example described
in the beginning of Section 1.1, if one were to consider the
possibilities that might come on the turn, one might consider
the sample space to be all 52 cards. Of course, if we know
what cards have been dealt to the three players and which
cards appeared on the flop, then these nine cards have zero
probability of appearing on the turn, and one may consider the
remaining 43 cards equally likely. The event that the turn is
the 7♦ is an event consisting of a single element of the sample
space, and the event that the turn is a diamond is an event
consisting of 13 elements.
Given an event A, we use the notation Ac to mean the
complement of A, or in other words, the event that A does not
occur. If A is the event that the turn is a diamond, for instance,
then Ac is the event that the turn is a club, heart, or spade. For
any event A, the events A and Ac are always mutually
exclusive and exhaustive, meaning that together they cover
the entire sample space.
1.3 Axioms of Probability
Three basic rules or axioms must govern all probabilities:
1: P(A) ≥ 0.
2: P(A) + P(Ac) = 1.
3: If A1, A2, A3,… are mutually exclusive events, then
P(A1 or A2 or … An ) = P(A1 ) + P(A2 ) + … + P(An ),
where n may be a positive integer or ∞.
Axiom 1 says that all probabilities must be at least 0, and
because of this and Axiom 2, no probability can be greater
than 1. According to Axiom 2, if the probability that A will
not occur, is, say, 45%, then we know that the probability of A
occurring is 55%. It is sometimes convenient to compute the
probability of A not occurring en route to computing the
probability of A.
Axiom 3 is sometimes called the addition rule for mutually
exclusive events. Note that it is usually rather obvious when
events are mutually exclusive. For example, if you play a
single hand of Texas Hold’em, A1 might be the event that you
are dealt pocket aces, and A2 the event you are dealt pocket
kings. They cannot both occur, so these two events are
mutually exclusive, and according to Axiom 3, the probability
of your being dealt either pocket aces or pocket kings on this
single hand is equal to the probability of your being dealt
pocket aces plus the probability of your being dealt pocket
kings. This may seem obvious, but it is worth noting that the
fact that we may add these two probabilities together to find
the probability that one will occur is not something that can
be proven; it is a fundamental principle we agree to assume
when discussing probabilities.
Note also that Axiom 3 does not apply directly to the
question discussed in Section 1.1, involving the probability of
Wasicka making a flush or a straight. It is possible (if, for
instance, the turn and river are the 9♣ and Q♠) for both to
occur, so the event that Wasicka makes a straight and the
event that he makes a flush are not mutually exclusive.
A final comment worth noting about the axioms of
probability is that they imply that if there are n equally likely
events, exactly one of which must occur, then the probability
associated with any one of the events is simply 1/n. This
may again seem obvious, but note that it can be proven
directly from Axiom 3 (see Exercise 1.3). This fact is
important for many questions involving Texas Hold’em,
because in many cases the outcomes are equally likely. The
subject of equally likely events is taken up further in Section
2.1.
1.4 Venn Diagrams
Most readers have likely seen Venn diagrams so they will be
reviewed only briefly here. One question that is sometimes
overlooked in discussing Venn diagrams is: why can one use
shapes such as those shown in Figure1.1 to address
probability questions?
The answer has everything to do with the Axioms of
probability discussed in Section 1.3. All the rules of
probability discussed in this book follow from these three
simple Axioms. Note that they apply not only to
probabilities, but also to other constructs including areas.
Suppose you have a piece of paper of area 1, and that A1, A2,
A3, … represent shapes on the paper. Suppose also that P(A)
means the area occupied by shape A (instead of probability);
Ac is the portion of the paper not occupied by shape A; AB
denotes the intersection of A and B, i.e., the portion of the
paper occupied by both A and B; and A or B indicates the
union of A and B, i.e., the portion of the paper occupied by
shapes A or B or their intersection. Then the three axioms
would still be true: any imaginable shape on the paper would
necessarily have an area greater than or equal to zero; the
area covered by the shape plus the area not covered by the
shape would equal 1; and shapes would be mutually
exclusive if their area of intersection is 0, so if shapes A1, A2,
A3,… are mutually exclusive, then obviously the total area
covered by them is simply the area of A1 + the area of A2 +
the area of A3 + ….
Thus, the same axioms that govern probability also apply
to areas on a paper of total area 1. We will see the connection
between probability and area several times in this book, and it
is worth noting that in further studies of probability at the
graduate and research levels, the connections are made very
explicit between probability and measure theory, the latter of
which is essentially the study of principles related to areas,
volumes, and other means of measuring shapes or other
objects.
1.5 General Addition Rule
Axiom 3 in Section 1.3 is sometimes called the addition rule
for mutually exclusive events. How does one compute the
corresponding probability for events that may not be mutually
exclusive? In general, for any events A and B,
P(A or B) = P(A) + P(B) – P(AB).
The above may be called the general addition rule because it
applies to events A and B whether they are mutually exclusive
or not. Of course, when A and B are mutually exclusive, the
last term, P(AB), is equal to zero. The general addition rule
follows directly from the axioms of probability and may be
demonstrated quite easily by breaking up A or B into mutually
exclusive events. That is, observe that
A or B = ABc or AB or AcB,
and that these three events, ABc, AB, and AcB, are mutually
exclusive. Therefore,
P(A or B) = P(ABc or AB or AcB) = P(ABc) + P(AB) + P(AcB).
One can also form the partition A = ABc or AB, and since ABc
and AB are mutually exclusive, it follows from Axiom 3 that
P(A) = P(ABc) + P(AB). Similarly, B = AcB or AB, so P(B) =
P(AcB) + P(AB). Thus,
P(A) + P(B) - P(AB) = [P(ABc) + P(AB)] + [P(AcB) + P(AB)]
- P(AB)
c
= P(AB ) + P(AB) + P(AcB)
= P(A or B).
The following five examples refer to the WSOP hand
discussed previously. For each example, assume we have only
the information given in Section 1.1, i.e., that Wasicka had 8♠
7♠, Binger had A♥ 10♥, Gold had 4♠ 3♣, and the flop was 10♣
6♠ 5♠. Assume that the remaining 43 cards are all equally
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likely to appear as turn or river cards and suppose that
Wasicka had called.
Example 1.5.1 — What is the probability of Wasicka making
a straight flush on the turn?
Answer — Wasicka needs the 4♠ or 9♠ to make a straight
flush, but Gold has the 4♠, so the only remaining possibility is
the 9♠. Since there are 43 equally likely possibilities for the
turn, exactly one of which must occur, the probability
associated with each of them appearing on the turn must be
1/43, and therefore the probability of the turn being the 9♠ is
1/43.
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Example 1.5.2 — What is the probability of Wasicka making
a straight flush on the turn or river?
Answer — Note that, from the definition of the word or given
in Section 1.2, this question is really asking what the
probability is of Wasicka making a straight flush on the turn
or river or both. However, the only card that can give
Wasicka a straight flush is the 9♠, and it cannot appear both
on the turn and the river. Thus, if A is the event that the 9♠
comes on the turn and B is the event that the 9♠ comes on the
river, then A and B are mutually exclusive, and by Axiom 3,
the probability of A or B is P(A) + P(B), which is simply 1/43
+ 1/43 = 2/43.
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Note that P(B) in the answer above is 1/43, not 1/42. It might
help to think of P(B) as the frequency with which the river
will be the 9♠, if we were to repeat this scenario over and over
again. Clearly, each of the 43 remaining cards is equally
likely to appear on the river, so the probability associated with
the 9♠ is simply 1/43.
Example 1.5.3 — What is the probability that the turn card
does not give Wasicka a straight flush and the river card also
does not give Wasicka a straight flush.
Answer — This is the complement of the event in Example
1.5.2, so by Axiom 2, the answer is 1 – 2/43 = 41/43.
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Example 1.5.4 — What is the probability of Wasicka making
a flush on the turn?
Answer — Eight spades are left in the deck: A♠, K♠, Q♠, J♠,
10♠, 9♠, 3♠, and 2♠. As noted in Example 1.5.1, each of the 43
remaining cards has a probability of 1/43 of appearing on the
turn. Let A1 be the event that the turn is the A♠, A2 be the event
that the turn is the K♠, etc. Note that A1, A2, …, A8 are mutually
exclusive, since it is impossible for more than one of these cards
to appear on the turn. Thus by Axiom 3,
P(A1 or A2 or … or A8) = P(A1) + P(A2) + … + P(A8)
= 1/43 + 1/43 + … + 1/43
= 8/43
~ 18.6%.
Two simple approximations used frequently by poker players
are known as the Rule of Two and the Rule of Four. If you are
behind your opponent presently, we call an out a card that will
give you the lead. Let x be the number of outs left in the deck.
The Rule of Two says that your probability of an out coming
on the next card (e.g. the turn) is approximately 2x%. The
Rule of Four says that your probability of an out coming on
the turn or river (or both) is approximately 4x%. Of course,
the term out can be replaced by an element of any collection
of x cards. The idea behind the Rule of Two is that, if one is
interested in the probability of one of x possible cards coming
on the turn, given only one's hole cards and the 3 flop cards,
then there are 47 cards left in the deck, each equally likely to
appear, so the probability of one of these x cards coming is
x/47 ~ x/50 = 2x/100. For the Rule of Four, the idea is that,
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again given only one's hole cards and the 3 flop cards,
P(out on turn or river) = P(out on turn) + P(out on river) –
P(both). If x is small, then this last term, the probability of an
out coming on both the turn and river, will be small and is
therefore neglected, and the other terms are each equal to x/47
~ 2x/100. Note that 2x/100 is a slight underestimate of x/47,
but this difference is offset by the exclusion of the term
P(both), resulting in a quite accurate estimate for small x. The
approximation becomes poor, however, when x is large, as
discussed further in Chapter 2.4.
In Example 1.5.4, x = 8, so the Rule of Two yields the
approximation 16%, whereas the actual answer was 18.6%.
However, in Example 1.5.4 we also assumed the cards of
Binger and Gold were known; otherwise, the correct answer
would be 8/47 ~ 17.0%.
Example 1.5.5 — What is the probability of Wasicka making
a flush on the turn or the river?
Answer — This one is not so simple. If A is the event that the
turn is a spade and B is the event that the river is a spade, then
A and B are not mutually exclusive. We will return to the
correct solution to this problem after discussing permutations
and combinations, in Example 2.4.3.
For now, we may apply the Rule of Four to obtain an
approximate answer. 8 spades remain, so x = 8, and the Rule
of Four yields 32% as the approximate solution.
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Example 1.5.6 — Mark Newhouse, who became one of the
WSOP main event's final 9 players (called the “November 9”)
in 2013, made the November 9 again in 2014. Of the 6683
entrants in the 2014 main event, if each player uses the same
strategy, then what is the probability that any one given player
would become part of the November 9?
Answer — The given player would be equally likely to wind
up in 1st place, 2nd place, …, or 6683rd place, so the
probability of being in the top 9 is 9/6683 ~ 0.135%.
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Example 1.5.7 — On day 5 in the 2015 WSOP Main Event,
with Andrew Moreno went all in with A♦K♦ and Dax Greene
called with A♠A♥. The flop came 5♦ K♠ K♥, giving Moreno 3
of a kind. Suddenly Greene needed the A♣ to come (and the
K♣ not to come) to win the hand. Given the players' cards and
the 3 cards on the flop, what is the probability of the A♣
coming on the turn or river?
Answer — Since the A♣ is Greene's only out, the Rule of
Four suggests an estimate of 4% for the A♣ to come on the
turn or river. Computed exactly,
P(A♣ on turn or river) = P(A♣ on turn) + P(A♣ on river)
– P(A♣ on both turn and river)
= P(A♣ on turn) + P(A♣ on river)
= 1/45 + 1/45 ~ 4.44%.
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(In the actual hand, the turn and river were the 7♠ and 8♣, so
Moreno won the pot.)
Exercises
1.1 Sometimes it matters whose hand is second best. With
the Gold, Wasicka, and Binger hands, for instance, if
both Wasicka and Binger had called all-in and Binger
had won the hand overall but Wasicka’s hand had beaten
Gold’s hand, then Wasicka would not have been
eliminated since he had more chips than Binger: Binger
would have ended up with 33 million chips, Wasicka
with 14 million chips, and Gold with 42 million chips. If
all 3 players were all-in and Gold had won the hand, of
course, the tournament would have been over and Gold
would have won because he had more chips than his
opponents at the start of the hand; by convention in such
situations, the player who started the hand with more
chips, which in this case was Wasicka, would get the
prize money for second place.
Suppose all 3 players went all-in on the flop, and that
the probability that Gold wins the hand is 17.17%, the
probability that Wasicka wins the hand is 53.82%, the
probability that Binger wins the hand and Wasicka’s
hand beats Gold’s hand is 24.03%, and the probability
that Binger wins the hand and Wasicka and Gold split
the side-pot is 1.00%. What is the probability that
Binger wins the hand and Gold’s hand beats Wasicka’s
hand?
1.2 Suppose that just before the hand described in the
beginning of Section 1.1, Gold’s probability of winning
the tournament was 60/89, and Binger’s probability of
winning the tournament was 11/89. What is the probability
that Wasicka would win the tournament? Which Axiom of
probability are you using?
1.3 Suppose that events A1, A2,…, An are equally likely
events and that exactly one of them must occur. Using
Axiom 3, prove carefully that P(A1 ) = 1/n.
1.4 Suppose you are at the final table of the WSOP main
event and you feel that, based on the number of chips
you have and your experience, you have a 30% chance
of winning the tournament. Is this a Bayesian or
frequentist type of probability?
1.5 Find the general addition rule for three events. That is,
find a general formula for P(A or B or C), for any events
A, B, and C, in terms of P(A), P(B), P(C), P(AB), P(AC),
P(BC), and P(ABC).
1.6 Using the Axioms of probability, prove Boole’s
inequality which states that P(A1 or A2 or …An ) ≤ P(A1 )
+ P(A2 ) + … + P(An ), for any events A1, A2,… An, where
n may be a positive integer or ∞.
1.7 In a huge hand from the 2015 WSOP Main Event, with
just 14 players left, Justin Schwartz called with 3♠3♣,
Alex Turyansky raised with Q♥Q♣, Joe Mckeehen called
with 6♠6♣, and Schwartz called. The flop came 2♦ 3♦ 6♥,
giving both Mckeehen and Schwartz three of a kind.
Schwartz checked, Turyansky bet 700,000, Mckeehen
called, Schwartz went all in, Turyansky folded, and
Mckeehen called. Schwartz would need the 3♥ (and no
6♦) to take the lead. Given these 3 players' cards and the 3
cards on the flop, what is the probability of the 3♥
coming on the turn or river? (In the actual hand, the turn
and river were 7♣ 5♣. Schwartz was eliminated in 14th
place, Mckeehen won the huge 22.32 million chip pot and
went on the win the tournament and the $7.7 million first
place prize.)
1.8 Compare the answer to the previous exercise involving
Schwartz, Mckeehen and Turyansky in the 2015 WSOP
Main Event to the approximation obtained using the Rule
of Four. Why does the Rule of Four yield an
underestimate in this case?
1.9
1.10
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