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					    CSci 2011
     Discrete
   Mathematics
Lecture 27 – ch 5.4, 5.5

         Fall 2008
       Yongdae Kim

         CSci 2011 Fall 2008
                          Admin
Assignment
   Assignment 11 will be posted.
E-mail
   CC TA for all of your e-mails.
   Put [2011] in front.
Quiz 5: Nov 21th, Covering Ch 4.1 ~ 5.2 (not 5.3)

New office hour
   M: 12:30 ~ 1:30 pm (before 11:00 ~ 12:00)
   T: 3:00 ~ 4:00 pm
   W, F: Same as before




                                                CSci 2011 Fall
                               Recap
 If there are n1 ways to do task 1, and n2 ways to do task 2
    Then there are n1n2 ways to do both tasks in sequence
    If these tasks can be done at the same time, then there are n1+n2 ways to
     do one of the two tasks
 The inclusion-exclusion principle: |A1UA2| = |A1|+|A2|-|A1∩A2|
 Pigeon-hole principle: If N objects are placed into k boxes, then there
  is at least one box containing N/k objects
 A permutation is an ordered arrangement of the elements of some set
  S: P(n, r) = n! / (n-r)!.
 When order does not matter, it is called combination: C(n, r) = n! /
  (r! (n-r)!)
 Let n and r be integers with 0 ≤ r ≤ n. Then C(n,r)=C(n,n-r).
 (x+y)n = i=0n nCi xi yn-i
 Let n and k be positive integers with n ≥ k. Then n+1Ck= nCk+nCk-1.




                                                            CSci 2011 Fall
               Pascal’s Identity
By Pascal’s identity: 7C5= 6C5+6C4 or 21=15+6

Let n and k be positive integers with n ≥ k.
Then n+1Ck= nCk+nCk-1

We will prove this via two ways:
   Combinatorial proof
   Using the formula for nCk




                                          CSci 2011 Fall
      Combinatorial proof of Pascal’s
                 identity
Prove C(n+1,k) = C(n,k-1) + C(n,k)
Consider a set T of n+1 elements
    We want to choose a subset of k elements
    We will count the number of subsets of k elements via 2 methods
Method 1: There are C(n+1,k) ways to choose such a
 subset
Method 2: Let a be an element of set T
Two cases
    a is in such a subset
        There are C(n,k-1) ways to choose such a subset
    a is not in such a subset
        There are C(n,k) ways to choose such a subset
Thus, there are C(n,k-1) + C(n,k) ways to choose a
 subset of k elements
Therefore, C(n+1,k) = C(n,k-1) + C(n,k)




                                                           CSci 2011 Fall
    Algebraic proof of Pascal’s identity
   (n + 1)!               n!               n!
               =                      +
k!(n + 1 - k )! (k - 1)!(n - (k - 1))! k!(n - k )!
          (n + 1)n!                         n!                       n!
                               =                            +
k (k - 1)!(n + 1 - k )(n - k )! (k - 1)!(n - k + 1)(n - k )! k (k - 1)!(n - k )!
   (n + 1)          1        1                                        Substitutions:
              =            +
k (n + 1 - k ) (n - k + 1) k
                                                            (n + 1 - k )!= (n + 1 - k ) * (n - k )!
   (n + 1)           k         (n - k + 1)
              =             +                                   (n + 1)!= (n + 1)n!
k (n + 1 - k ) k (n - k + 1) k (n - k + 1)
                                                            (n - k + 1) = (n - k + 1)(n - k )!
n +1 = k + n - k +1
n +1 = n +1




                                                                          CSci 2011 Fall
       Pascal’s triangle
n= 0                  sum = 1
                                 = 2n
   1                        2

   2                        4

   3                        8

   4                       16

   5                       32

   6                       64

   7                       128

   8                       256




                           CSci 2011 Fall
     Proof practice: corollary 1
Let n be a non-negative integer. Then
 i=0n nCi = 2n

Algebraic proof
  Binomial equation: (x+y)n = i=0n nCi xiyn-i

  Put x=1, and y = 1
  2n = i=0n nCi




                                          CSci 2011 Fall
      Proof practice: corollary 1
Let n be a non-negative integer. Then i=0n nCi =2i

Combinatorial proof
   A set with n elements has 2n subsets
     By definition of power set
   Each subset has either 0 or 1 or 2 or … or n elements
     There are nC0 subsets with 0 elements, nC1 subsets with 1
      element, … and nCn subsets with n elements
     Thus, the total number of subsets is i=0n nCi

   Thus, i=0n nCi =2i.




                                                       CSci 2011 Fall
     Proof practice: corollary 2
Let n be a non-negative integer. Then
 i=0n (-1)nnCi = 0

Algebraic proof
  Binomial equation: (x+y)n = i=0n nCi xiyn-i

  Put x=-1, and y = 1
  0 = i=0n (-1)nnCi




                                          CSci 2011 Fall
         Vandermonde’s identity
Let m, n, and r be non-negative integers with r not
 exceeding either m or n. Then
 m+rCn = i=0 (mCr-i nCi)
             r

Consider two sets, one with m items and one
 with n items
   Then there are m+rCn ways to choose r items from the union
    of those two sets
Next, we’ll find that value via a different means
   Pick i elements from the set with n elements
   Pick the remaining r-i elements from the set with m
    elements
   Via the product rule, there are mCr-i nCi ways to do that for
    EACH value of i
   Lastly, consider this for all values of i.


                                                    CSci 2011 Fall
                         Examples
How many bit strings of length 10 contain exactly
 four 1’s?
   Find the positions of the four 1’s
   The order of those positions does not matter
     Positions 2, 3, 5, 7 is the same as positions 7, 5, 3, 2
   Thus, the answer is C(10,4) = 210


Generalization of this result:
   There are C(n,r) possibilities of bit strings of length n
    containing r ones




                                                       CSci 2011 Fall
      ch 6.1
Discrete Probability
       Fall 2008
     Yongdae Kim

      CSci 2011 Fall 2008
                Terminology
Experiment
  A repeatable procedure that yields one of a given
   set of outcomes
  Rolling a die, for example
Sample space
  The range of outcomes possible
  For a die, that would be values 1 to 6
Event
  One of the sample outcomes that occurred
  If you rolled a 4 on the die, the event is the 4




                                            CSci 2011 Fall
           Probability definition
The probability of an event occurring is:
 p(E) = |E| / |S|
  Where E is the set of desired events (outcomes)
  Where S is the set of all possible events
   (outcomes)
  Note that 0 ≤ |E| ≤ |S|
     Thus, the probability will always between 0 and 1
     An event that will never happen has probability 0
     An event that will always happen has probability 1




                                                CSci 2011 Fall
                  Dice probability
What is the probability of getting “snake-eyes” (two
 1’s) on two six-sided dice?
    Probability of getting a 1 on a 6-sided die is 1/6
    Via product rule, probability of getting two 1’s is the
     probability of getting a 1 AND the probability of getting a
     second 1
    Thus, it’s 1/6 * 1/6 = 1/36
What is the probability of getting a 7 by rolling two
 dice?
    There are six combinations that can yield 7: (1,6), (2,5),
     (3,4), (4,3), (5,2), (6,1)
    Thus, |E| = 6, |S| = 36, P(E) = 6/36 = 1/6




                                                    CSci 2011 Fall
Poker




CSci 2011 Fall 2008
Playing Card




               CSci 2011 Fall
                The game of poker
 You are given 5 cards (this is 5-card stud poker)
 The goal is to obtain the best hand you can
 The possible poker hands are (in increasing order):
    No pair
    One pair (two cards of the same face)
    Two pair (two sets of two cards of the same face)
    Three of a kind (three cards of the same face)
    Straight (all five cards sequentially – ace is either high or low)
    Flush (all five cards of the same suit)
    Full house (a three of a kind of one face and a pair of another
     face)
    Four of a kind (four cards of the same face)
    Straight flush (both a straight and a flush)
    Royal flush (a straight flush that is 10, J, K, Q, A)




                                                          CSci 2011 Fall
    Poker probability: royal flush
What is the chance of
 getting a royal flush?
   That’s the cards 10, J, Q, K,
    and A of the same suit

There are only 4 possible
 royal flushes

Possibilities for 5 cards: C(52,5) = 2,598,960

Probability = 4/2,598,960 = 0.0000015
   Or about 1 in 650,000




                                          CSci 2011 Fall
  Poker probability: four of a kind
What is the chance of getting 4 of a kind when dealt
 5 cards?
   Possibilities for 5 cards: C(52,5) = 2,598,960


Possible hands that have four of a kind:
   There are 13 possible four of a kind hands
   The fifth card can be any of the remaining 48 cards
   Thus, total possibilities is 13*48 = 624


Probability = 624/2,598,960 = 0.00024
   Or 1 in 4165




                                                     CSci 2011 Fall
         Poker probability: flush
What is the chance of getting a flush?
   That’s all 5 cards of the same suit
We must do ALL of the following:
   Pick the suit for the flush: C(4,1)
   Pick the 5 cards in that suit: C(13,5)
As we must do all of these, we multiply the values
 out (via the product rule)

This yields C(4,1) C(13,5) = 5148

Possibilities for 5 cards: C(52,5) = 2,598,960
Probability = 5148/2,598,960 = 0.00198
   Or about 1 in 505




                                             CSci 2011 Fall
     Inclusion-exclusion principle
 The possible poker hands are (in increasing order):
    Nothing
    One pair cannot include two pair, three of a kind, four of a kind, or
     full house
    Two pair cannot include three of a kind, four of a kind, or full
     house
    Three of a kind cannot include four of a kind or full house
    Straight cannot include straight flush or royal flush
    Flush cannot include straight flush or royal flush
    Full house
    Four of a kind
    Straight flush cannot include royal flush
    Royal flush




                                                         CSci 2011 Fall
 Poker probability: three of a kind
What is the chance of getting a three of a kind?
    That’s three cards of one face
    Can’t include a full house or four of a kind
We must do ALL of the following:
    Pick the face for the three of a kind: C(13,1)
    Pick the 3 of the 4 cards to be used: C(4,3)
    Pick the two other cards’ face values: C(12,2)
        We can’t pick two cards of the same face!
    Pick the suits for the two other cards: C(4,1)*C(4,1)
As we must do all of these, we multiply the values out
 (via the product rule)

This yields C(13,1) C(4,3) C(12,2) C(4,1) C(4,1) = 54,912

Possibilities for 5 cards: C(52,5) = 2,598,960
Probability = 54,912/2,598,960 = 0.0211
    Or about 1 in 47


                                                        CSci 2011 Fall
              Poker hand odds
The possible poker hands are (in increasing order):
   Nothing               1,302,540   0.5012
   One pair              1,098,240   0.4226
   Two pair              123,552     0.0475
   Three of a kind 54,9120.0211
   Straight        10,2000.00392
   Flush                 5,140       0.00197
   Full house            3,744       0.00144
   Four of a kind        624         0.000240
   Straight flush        36          0.0000139
   Royal flush           4           0.00000154




                                               CSci 2011 Fall
            More on probabilities
Let E be an event in a sample space S. The
 probability of the complement of E is:

                       
                     p E = 1 - p( E )

Recall the probability for getting a royal flush is
 0.0000015
    The probability of not getting a royal flush is
     1-0.0000015 or 0.9999985
Recall the probability for getting a four of a kind is
 0.00024
    The probability of not getting a four of a kind is
     1-0.00024 or 0.99976



                                                       CSci 2011 Fall
Probability of the union of two events
Let E1 and E2 be events in sample space S

Then p(E1 U E2) = p(E1) + p(E2) – p(E1 ∩ E2)

Consider a Venn diagram dart-board




                                         CSci 2011 Fall
Probability of the union of two events

                   p(E1 U E2)

                                           S




              E1                E2




                                     CSci 2011 Fall
Probability of the union of two events
If you choose a number between 1 and 100,
 what is the probability that it is divisible by 2
 or 5 or both?
Let n be the number chosen
  p(2|n)   = 50/100 (all the even numbers)
  p(5|n)   = 20/100
  p(2|n)   and p(5|n) = p(10|n) = 10/100
  p(2|n)   or p(5|n) = p(2|n) + p(5|n) - p(10|n)
                     = 50/100 + 20/100 – 10/100
                     = 3/5




                                         CSci 2011 Fall
                Dealing cards
Consider a dealt hand of cards
  Assume they have not been seen yet
  What is the chance of drawing a flush?
  Does that chance change if I speak words after
   the experiment has completed?
  Does that chance change if I tell you more info
   about what’s in the deck?

No!
  Words spoken after an experiment has
   completed do not change the chance of an event
   happening by that experiment
    No matter what is said


                                         CSci 2011 Fall
Monty Hall Paradox




      CSci 2011 Fall 2008
   What’s behind door number three?
The Monty Hall problem paradox
  Consider a game show where a prize (a car) is behind
    one of three doors
  The other two doors do not have prizes (goats
    instead)
  After picking one of the doors, the host (Monty Hall)
    opens a different door to show you that the door he
    opened is not the prize
  Do you change your decision?
Your initial probability to win (i.e. pick the right door)
 is 1/3
What is your chance of winning if you change your
 choice after Monty opens a wrong door?
After Monty opens a wrong door, if you change your
 choice, your chance of winning is 2/3
  Thus, your chance of winning doubles if you change
  Huh?

                                              CSci 2011 Fall
What’s behind door number one hundred?
Consider 100 doors
    You choose one
    Monty opens 98 wrong doors
    Do you switch?
Your initial chance of being right is 1/100
Right before your switch, your chance of being right is
 still 1/100
    Just because you know more info about the other doors doesn’t
     change your chances
       You didn’t know this info beforehand!
Your final chance of being right is 99/100 if you switch
    You have two choices: your original door and the new door
    The original door still has 1/100 chance of being right
    Thus, the new door has 99/100 chance of being right
    The 98 doors that were opened were not chosen at random!
       Monty Hall knows which door the car is behind


Reference:
 http://en.wikipedia.org/wiki/Monty_Hall_problem
                                                        CSci 2011 Fall

				
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