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Lecture2

# Lecture2 - 1 EQUALLY LIKELY OUTCOMES Lecture 2...

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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 well-shuﬄed 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 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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2 COMBINATORICS 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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Lecture2 - 1 EQUALLY LIKELY OUTCOMES Lecture 2...

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