1 EQUALLY LIKELY OUTCOMES
Lecture 2
ORIE3500/5500 Summer2009 Chen
Equally Likely Outcomes and Combinatorics
1 Equally Likely Outcomes
In many applications, it turns out that the sample space is ﬁnite and all the
outcomes are “equally likely”,
ie
, there is no reason to believe that one of the
outcomes has any special preference. Example: tossing a fair coin, rolling a
fair die, choosing a card from a wellshuﬄed deck etc.
[Classical Deﬁnition of Probability]
Consider an experiment with a ﬁnite
number of equally likely outcomes. Then for any event
A
, the probability of
A
is deﬁned by
P
(
A
) =
#
A
#Ω
Example
Suppose that Brian and Charles are gambling. Both of them roll a die and the
person who has the higher number gets $10 from the other. No transaction
takes place in case of a tie. We can pose this problem as following: set Ω =
{
(
i,j
) : 1
≤
i,j
≤
6
}
and let
P
(
{
(
i,j
)
}
) = 1
/
36 for all (
i,j
)
∈
Ω(Note that
#Ω = 36). In every element of Ω the ﬁrst coordinate denotes the outcome
of Brian’s dice. If
B
is the event that Brian wins, then
B
=
{
(
i,j
) :
i > j
}
and
P
(
B
) = #
B/
36 = 5
/
12. Similarly the probability of Charles winning is
5
/
12 and there is 1
/
6 chance that there is a draw. One important comment
here is that this is not the only way to pose this problem. We can also set
this problem in the following way: take Ω =
{
i
:

5
≤
i
≤
5
}
, where
i
is
the score of Charles subtracted from that of Brian. In this formulation the
outcomes are surely not equally likely. Here the event that Brian wins will
be
B
=
{
1
,
2
,
3
,
4
,
5
}
. It is not as easy to ﬁnd the probability of
B
in this
setup. So the formulation can often change the ease with which you reach
the desired results.
Example
Five Cornell students Anna, Brian, Charles, Donna and Elena, appear for
an examination. They all try to come out with their best eﬀorts and per
form better than the others. An outcome is the rank of the students in the
1
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examination. For example (
C,B,A,E,D
) can be one order in which the stu
dents can be ranked and this will be one outcome. So the sample space, Ω,
will consist of all the permutations of
A,B,C,D
and
E
and hence contains
5
·
4
·
3
·
2
·
1 = 5! = 120 outcomes. For any subset
S
of Ω, we can compute
P
(
S
) by adding the probability of the occurrence of each outcome in
S
. For
example, the probability of the event that Brian gets rank 1, will be the
sum of the likelihoods of the 24 (All permutations of
B,C,D,E
) outcomes
in which Brian comes ﬁrst.
2 Combinatorics
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 Summer '08
 WEBER
 Conditional Probability, Probability theory, total number, Books of Kings

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