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1 Great Theoretical Ideas In Computer Science Victor Adamchi CS 5-5 Carnegie Mellon University Probability - I The Story The theory of probability was originally developed by a French mathematician Blaise Pascal. Pascal had a friend, fond of gambling, who ased him for strategies. Probability was invented to analyze gambling. This is not Great Theoretical Ideas in Gambling, but in Computer Science. Plan Sample and vents Conditional Probability Bayes Law Law of Total Probability Use of Generating Functions ample Mary flips a fair coin. If it s a head, she rolls a -sided die. If it s a tail, she rolls a -sided die. What is the probability die roll is or higher? Draw a probability tree. Prob: -die A Probability Tree H coin T -die (H,) (H,) (H,) (T,) (T,) (T,) (T,) / / / /8 /8 /8 /8 Label the leaves with outcomes Under each, write its probability: multiply along the path Outcome: A leaf in the probability tree. I.e., a possible sequence of values of all calls to generators in an eecution. Sample Space: The set of all outcomes..g., { (H,), (H,), (H,), (T,), (T,), (T,), (T,) } Probability: ach outcome has a nonnegative probability. Sum of all outcomes probabilities always.

2 ample Mary flips a fair coin. If it s heads, she rolls a -sided die. If it s tails, she rolls a -sided die. What is the probability die roll is or higher? vent: A subset of outcomes. In our eample, = { (H,), (T,), (T,) }. = sum of the probabilities of the outcomes in. coin H T die die (H,) (H,) (H,) (T,) (T,) (T,) (T,) Prob: / / / /8 /8 /8 /8 = roll is or higher = / + /8 + /8 = 5/ We will consider eperiments with a finite number of possible outcomes w, w,..., w n The sample space of the eperiment is the set of all possible outcomes. The roll of a die: = {,,,,5,} ach subset of a sample space is defined to be an event. The event: = {,,} Probability of a event Let X be a random variable which denotes the value of the outcome of a certain eperiment. We will assign probabilities to the possible outcomes of an eperiment. We do this by assigning to each outcome w j a nonnegative number p(w j ) in such a way that p(w ) + + p(w n )= The function p(w j ) is called the probability distribution function of the random variable X, Pr[X = w ] Probabilities For any subset of, we define the probability of to be the number given by event w p(w ) probabilities of outcomes From Random Variables to vents For any random variable X and value a, we can define the event that X = a = Pr[X=a] = Pr[{t X(t)=a}]

3 From Random Variables to vents Random Variable It s a function on the sample space ->[0, ) It s a variable with a probability distribution on its values A coin is tossed ten times. The random variable X is the number of tails that are noted. X can only tae the values 0,,, 0, so X is a discrete random variable. You should be comfortable with both views Theorem The probabilities satisfy the following properties: Pr[ ] = 0 Pr[A B] = Pr[A] + Pr[B] Pr[A B] Pr[A c ] = - Pr[A] Pr[A] = Pr[A B] + Pr[A B c ] B A Uniform Distribution When a coin is tossed and the die is rolled, we assign an equal probability to each outcome. The uniform distribution on a sample space containing n elements is the function p defined by p(w) = /n for every w These should mae sense if we thin of events as sets. ample ample Consider the eperiment that consists of rolling a pair of standard dice. What is the probability of getting a sum of 7 or a sum of? S = { (,), (,), (,), (,), (,5), (,), (,), (,), (,), (,), (,5), (,), (,), (,), (,), (,), (,5), (,), (,), (,), (,), (,), (,5), (,), (5,), (5,), (5,), (5,), (5,5), (5,), (,), (,), (,), (,), (,5), (,) } F We assume that each of outcomes is equally liely. P()= * / P(F)= * / P( F)= 8/

4 A fair coin is tossed 00 times in a row What is the probability that we get eactly half heads? The sample space is the set of all outcomes (sequences) {H,T} 00 ach sequence in is equally liely, and hence has probability / =/ 00 Binomial Distribution = all sequences of 00 tosses Visually t = HHTTT TH Set of all 00 sequences {H,T} 00 vent = Set of sequences with 50 H s and 50 T s P(t) = / S Probability of event = proportion of in S / 00 Birthday Parado How many people do we need to have in a room to mae it a favorable bet (probability of success greater than /) that two people in the room will have the same birthday? Sample space Visually = 5 vent = { w two numbers not the same } ( )( 5 We must find sequences that have no duplication of birthdays. )( 5 ) 5

5 Birthday Problem Pr[in students, some pair share a bday ]= -( )( 5 )( 5 ) 5 Birthday Problem What if there are N possible birthdays? Pr[in m students, some pair share a bday ] -( )( N )( N m ) N For what value of m is this /? m N Cryptographic Hash Functions 99: Rivest publishes MD5. (=8) 99: NSA publishes SHA-0. (=0) 995: NSA publishes SHA-. (=0) SHA- now used in SSL, PGP, 00: NSA also introduces SHA- Birthday Attac Imagine trying to find a collision for SHA-: Tae a huge number of strings, hash them all, hope that two hash to the same 0 bits. This is lie the Birthday Problem with N = 0! So # tries before good chance of collision: Birthday Attac A crypto hash function is considered broen if you can beat the birthday attac. Prof. Xiaoyun Wang ( 王 小 云 ) Infinite Sample Spaces A coin is tossed until the first time that a head turns up. = {,,,, } 005: SHA- collisions in < 9 Later (w/ coauthors): in < SHA- = broen A distribution function: m(n) = -n. Pr w m(w) 5

6 Infinite Sample Spaces Let be the event that the first time a head turns up is after an even number of tosses. = {,,, } Infinite Sample Spaces Suppose your eperiment involves throwing a dart, which is equally liely to land anywhere in the interval [0,]. What is the probability that the dart lands at eactly 0.5? Probability of landing at any rational number is the same If that > 0, then the sum of all probabilities will be >. Thus, the probability that the dart lands at eactly 0.5 is zero. Conditional Probability Consider our voting eample: three candidates A, B, and C are running for office. We decided that A and B have an equal chance of winning and C is only / as liely to win as A. Suppose that before the election is held, A drops out of the race. What are new probabilities to the events B and C P(B A)=/ P(C A)=/ F Conditional Probability Definition. The probability of event F given event is written P(F ) and is defined by P(F ) P(F ) P() We narrow our sample space to. proportion of F to Suppose we roll a white die and blac die What is the probability that the white die is given that the total is 7? S = { (,), (,), (,), (,), (,5), (,), (,), (,), (,), (,), (,5), (,), (,), (,), (,), (,), (,5), (,), (,), (,), (,), (,), (,5), (,), (5,), (5,), (5,), (5,), (5,5), (5,), (,), (,), (,), (,), (,5), (,) } event A = {white die = } event B = {total = 7} Pr[A B] = Pr[A B] A B = = Pr[B] B event A = {white die = } event B = {total = 7}

7 Independence! A and B are independent events if P(A B ) = P(A) P(A B ) = P(A) P(B) P(B A ) = P(B) Silver and Gold One bag has two silver coins, another has two gold coins, and the third has one of each One bag is selected at random. One coin from it is selected at random. It turns out to be gold What is the probability that the other coin is gold? Let G be the event that the first coin is gold Pr[G ] = / Let G be the event that the second coin is gold Pr[G G ] = Pr[G G ]/ Pr[G ] = (/) / (/) = / Note: G and G are not independent Another Parado Consider a family with two children. Given that one of the children is a boy, what is the probability that both children are boys? / Tree Diagram Tree Diagram / / / / girl boy girl boy / / / / / / / girl boy girl boy / / / / boy girl boy / / / 7

8 Monty Hall Problem The host show hides a car behind one of doors at random. Behind the other two doors are goats. You select a door. Announcer opens one of others with no prize. You can decide to eep or switch. This is Tricy! We are inclined to thin: After one door is opened, others are equally liely What to do? To switch or not to switch? Monty Hall Problem Sample space = { car behind door, car behind door, car behind door } ach has probability / Staying we win if we choose the correct door Pr[ choosing correct door ] = / Switching we win if we choose the incorrect door Pr[choosing incorrect door ] = / Monty Hall Problem Let the doors be called X, Y and Z. Let C, Cy, Cz be the events that the car is behind door X and so on. Let H, Hy, Hz be the events that the host opens door X and so on. Supposing that you choose door X, the possibility that you win a car if you switch is Pr[Hy Cz] + P[Hz Cy] = Pr[Hy Cz] P[Cz] + Pr[Hz Cy] Pr[Cy] = / + / = / Bayes formula Sometimes we need to now Pr[F ], but all we now is the reverse direction Pr[ F]. Law of Total Probability Theorem. Let F, F,, F n partition of a sample space. Then Theorem. P[F ] Pr[ F] Pr[F] n Pr[ F ] n Pr[ F ] Pr[F ] Proof. Proof. Note F are mutually eclusive Pr[F ] Pr[F ] Pr[ F] Pr[ F]Pr[F] U n F 8

9 Problem A fair die is tossed and its outcome is denoted by X. After that, X independent fair coins are tossed and the number of heads obtained is denoted by Y. Compute: Pr[Y=] ercise By the law of total probability Pr[Y ] Pr[X ] is /. Compute Pr[Y= X ] Pr[Y X ] *Pr[X ] Pr[Y X ] / Use of Generating Functions Compute the probability of rolling a sum of 8 on standard dice. Problem Create a generating polynomial for the problem Compute the probability of rolling a sum of 8 on standard dice. Tae the coefficient by 8. Note, 0 0 Sample and vents Conditional Probability Bayes Law Law of Total Probability Thus, the coefficient by 8 is Here s What You Need to Know 9

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