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Discrete Probability

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Discrete Probability

Hsin-Lung Wu

Assistant Professor

Advanced Algorithms 2008 Fall

- Sample space (set) S of elementary event
- eg. The 36 ways of 2 dices can fall

- An eventA is a subset of S
- eg. Rolling 7 with 2 dices

- A probability distributionPr{} is a map from events of S to R
- Probability Axiom:

- A random variable (r.v.) X is a function from S to R
- The event “X = x” is defined as {sS : X(s) = x}
- eg. Rolling 2 dices:
- |S|=36 possible outcomes
- Uniform distribution: Each element has the same probability
1/|S|=1/36

- Let X be the sum of dice
Pr{ X = 5 } = 4/36, {(1, 4), (2, 3), (3, 2), (4, 1)}

- Expected value:
- Linearity:
- X1: number on dice 1
- X2: number on dice 2
- X=X1+X2, E[X1]=E[X2]=1/6(1+2+3+4+5+6)=21/6

- Two random variables X and Y are independent if

- Indicator random variables
- Given a sample space S and an event A, the indicator random variable I{A} associated with event A is defined as:

- E.g.: Consider flipping a fair coin:
- Sample space S = { H,T }
- Define random variableY with Pr{ Y=H } = Pr{ Y=T }=1/2
- We can define an indicator r.v.XH associated with the coin coming up heads, i.e.Y=H

- The birthday paradox:
- How many people must there be in a room before there is a 50% chance that two of them born on the same day of the year?

- (1)
- Suppose there are k people and there are n days in a year,bi : i-th person’s birthday, i =1,…,k
- Pr{bi=r}=1/n, for i =1,…,k and r=1,2,…,n
- Pr{bi=r, bj=r}=Pr{bi=r}．Pr{bj=r} = 1/n2

- Define event Ai : Person i’s birthday is different from person j’s for j < i
- Pr{Bk}= Pr{Bk-1∩Ak}= Pr{Bk-1}Pr{Ak|Bk-1}where Pr{B1}= Pr{A1}=1

- (2) Analysis using indicator random variables
- For each pair (i, j) of the k people in the room, define the indicator r.v.Xij, for 1≤ i < j ≤ k, by

- When k(k-1)≥ 2n, the expected number of pairs of people with the same birthday is at least 1

- Balls and bins problem:
- Randomly toss identical balls into b bins, numbered 1,2,…,b. The probability that a tossed ball lands in any given bin is 1/b
- (a) How many balls fall in a given bin?
- If n balls are tossed, the expected number of balls that fall in the given bin is n/b

- (b) How many balls must one toss, on the average, until a given bin contains a ball?
- By geometric distribution with probability 1/b

- (c) (Coupon collector’s problem)How many balls must one toss until every bin contains at least one ball?
- Want to know the expected number n of tosses required to get b hits. The ith stage consists of the tosses after the (i-1)st hit until the ith hit.
- For each toss during the ith stage, there are i-1 bins that contain balls and b-i+1 empty bins
- Thus, for each toss in the ith stage, the probability of obtaining a hit is (b-i+1)/b
- Let ni be the number of tosses in the ith stage. Thus the number of tosses required to get b hits is n=∑bi=1 ni
- Each ni has a geometric distribution with probability of success (b-i+1)/b→ E[ni]=b/b-i+1

- Flip a fair coin n times, what is the longest streak of consecutive heads?Ans:θ(lg n)
- Let Aik be the event that a streak of heads of length at least k begins with the ith coin flip
- For j=0,1,2,…,n, let Lj be the event that the longest streak of heads has Length exactly j, and let L be the length of the longest streak.

- We look for streaks of length s by partitioning the n flips into approximately n/s groups of s flips each.

s

s

s

n

- The probability that a streak of heads of length does not begin in position i is

- Using indicator r.v. :

- If c is large, the expected number of streaks of length clgn is very small.
- Therefore, one streak of such a length is very likely to occur.

- To hire an assistant, an employment agency sends one candidate each day. After interviewing that person you decide to either hire that person or not. The process stops when a person is hired.
- What is the trade-off between minimizing the amount of interviewing and maximizing the quality of the candidate hired?

Pi

Pk-1

Pk-1

<?

P2

P1

Pk

….

What is the best k?

- Let M(j) = max 1ij{score(i)}.
- Let S be the event that the best-qualified applicant is chosen.
- Let Si be the event the best-qualified applicant chosen is the i-th one interviewed.
- Si are disjoint and we have Pr{S}= ni=1Pr{Si}.
- If the best-qualified applicant is one of the first k, we have that Pr{Si}=0 and thus
- Pr{S}= ni=k+1Pr{Si}.

- Let Bi be the event that the best-qualified applicant must be in position i.
- Let Oi denote the event that none of the applicants in position k+1 through i-1 are chosen
- If Si happens, then Bi and Oi must both happen.
- Bi and Oi are independent! Why?
- Pr{Si} = Pr{Bi Oi} = Pr{Bi} Pr{Oi}.
- Clearly, Pr{Bi} = 1/n.
- Pr{Oi} = k/(i-1). Why???
- Thus Pr{Si} = k/(n(i-1)).