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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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