model Which does the God use?
Consider the experiment when two dice are tossed. If the dice are identiable, the sample space has 36 sample points. If the dice are not-identiable, the sample space has 21 points. If each outcome is equally likely, what is the probability that the outcome is (1, 1)?
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January 13, 2011
1 / 17
Which model does the God use?
Assuming identiability: the answer is 1/36 = 2.8%; Assuming non-identiability: the answer is 1/21 = 4.8%;
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January 13, 2011
2 / 17
Which model does the God use?
Assuming identiability: the answer is 1/36 = 2.8%; Assuming non-identiability: the answer is 1/21 = 4.8%; We may reveal this secrete by experimentation.
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January 13, 2011
2 / 17
Which model does the God use?
Help to toss the pair of dice 20 times, and keep a record of how many times the outcome is (1, 1); and how many times the sum is 4. Give it to student behind you to repeat it again. My calculation indicates that we need about 1000 tosses to be able to tell a dierence of 1%. Let us work hard.
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January 13, 2011
3 / 17
Dancing party problem
Ten boys brought their girlfriends to have a party. When the music starts, each boy randomly invited a girl to dance. What is the probability that none of them were dancing with his own girlfriend?
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January 13, 2011
4 / 17
Dancing party problem
By randomly inviting, we assumed that each girl is equally likely to dance with each of the ten boys. Assigning ten girls to ten boys, there are 10! = 362880 possibilities. That is, #(S) = 362880. It is apparent that identifying and then enumerate all sample points in which case no boy with his own girlfriend is not practical.
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January 13, 2011
5 / 17
Dancing party problem
Yet we can try to work on something less complex. Let A1 be the event that the rst boy was dancing with his own girlfriend. Computing the probability of A1 is not dicult.
P (A1 ) =
9! 10!
=
1 10 .
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January 13, 2011
6 / 17
Dancing party problem
Computing the probability of A1 A2 is not so hard either.
P (A1 A2 ) =
8! 10! .
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January 13, 2011
7 / 17
Dancing party problem
Computing the probability of A1 A2 Ak is simple.
P (A1 Ak ) =
(10k ) ! 10! .
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January 13, 2011
8 / 17
Dancing party problem
What is the probability that some boy danced with his own girlfriend?
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January 13, 2011
9 / 17
Dancing party problem
What is the probability that some boy danced with his own girlfriend? The answer is: P (A1 A2 A10 )
=
P ( Ai ) P ( Ai Aj ) + P ( Ai Aj Ak ) +
+(1)10+1 P (A1 A10 )
10 (10 k )! 10! k k =1 1 1 1 . = 1 + 2! 3! 10!
=
( 1) k + 1
10
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January 13, 2011
9 / 17
Dancing party problem
What is the probability that none of them danced with his own girlfriend?
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January 13, 2011
10 / 17
Dancing party problem
What is the probability that none of them danced with his own girlfriend? The answer is: 1 P (A1 A2 A10 ) = 1 1 1 1 1 + ++ . 1! 2! 10!
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January 3! 13, 2011
10 / 17
Dancing party problem
What is the probability that none of them danced with his own girlfriend? The answer is: 1 P (A1 A2 A10 ) = 1 1 1 1 1 + ++ . 1! 2! 3! 10!
What is the limit when 10 is replaced by n = ?
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January 13, 2011
10 / 17
Dancing party problem
What is the probability that none of them danced with his own girlfriend? The answer is: 1 P (A1 A2 A10 ) = 1 1 1 1 1 + ++ . 1! 2! 3! 10!
What is the limit when 10 is replaced by n = ? Challenging question: what is the probability that exactly 3 boys danced with their own girlfriends?
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January 13, 2011
10 / 17
Counting the ballot
In an election in which 200 ballots were casted. There were two candidates, A and B. In the end, A won 110 votes and B won 90 votes. Suppose the ballots are counted in random order. What is the probability that candidate A was strictly in lead all the time?
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January 13, 2011
11 / 17
Counting the ballot
The random experiment is very clear. What are the possible outcomes in this problem? The following is a possible ballot: YY Y NN N
110 90
where Y means a vote for A, and N means a vote for B. We may regard the outcome the experiment is to place these 110 Y ballots among 200. When Y votes are not distinguished, there are 200 110 such arrangements.
() January 13, 2011 12 / 17
Counting the ballot
Let E be the event that Candidate A always strictly led B during the ballot counting. We need to nd the size of E to compute its probability. This task is not so straightforward.
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January 13, 2011
13 / 17
Counting the ballot
It is easy to see that any such sample point in E must be led by Y . There are 199 109 such paths. Yet not every sample point led by Y is an sample point in E . We nd it simpler to count the number of sample points led by Y but not in E .
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January 13, 2011
14 / 17
Counting the ballot
We will draw a sample path on blackboard. Let Xn = number of dierence in Y and N after n ballots counted. We may regard (n, Xn ) as a point on a two-dim space. Plot (n, Xn ); n = 0, 1, 2, . . . , 200 and connect them together gives us sample path from (0, 0) to (200, 20).
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January 13, 2011
15 / 17
Counting the ballot
Among these that goes through (1, 1), how many of them will touch x-axis before they arrive at (200, 20)? The number is the same as the number of paths from (1, 1) to (200, 20) by reection principle (to be explained). The answer is 199 110
b ecause the sample path moves up 110 out of 199 total moves. Hence, the number of sample paths from (1, 1) to (200, 20) without touching x-axis is 199 199 109 110
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January 13, 2011
16 / 17
Counting the ballot
Recall E = Candidate A was in strict lead all the time. Final answer is: P (E ) = The general answer is D T where D is the dierence in votes, T is the total number of ballots. P (E ) =
(199) (199) 20 110 109 = . 200 200 (110)
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January 13, 2011
17 / 17