Algorithms

Lecture 1: Discrete Probability [Sp’17]

The first lot fell to Jehoiarib, the second to Jedaiah, the third to Harim, the fourth to
Seorim, the fifth to Malkijah, the sixth to Mijamin, the seventh to Hakkoz, the eighthto Abijah, the ninth to Jeshua, the tenth to Shekaniah, the eleventh to Eliashib, thetwelfth to Jakim, the thirteenth to Huppah, the fourteenth to Jeshebeab, the fifteenth toBilgah, the sixteenth to Immer, the seventeenth to Hezir, the eighteenth to Happizzez,
the nineteenth to Pethahiah, the twentieth to Jehezkel, the twenty-first to Jakin, thetwenty-second to Gamul, the twenty-third to Delaiah, and the twenty-fourth to Maaziah.
This was their appointed order of ministering when they entered the temple of the
LORD, according to the regulations prescribed for them by their ancestor Aaron, as the
LORD, the God of Israel, had commanded him.
— 1 Chronicles 24:7–19 (New International Version)
The ring worm is not ringed, nor is it worm. It is a fungus.
The puff adder is not a puff, nor can it add. It is a snake.
The funny bone is not funny, nor is it a bone. It is a nerve.
The fishstick is not a fish, nor is it a stick. It is a fungus.
— Matt Groening, “Life in Hell” (1986)
When Pierre-Simon Laplace described probability theory as “good sense reduced to a
calculus,” he intended to disparage neither good sense nor probability theory thereby.
— Lorraine Daston, “Enlightenment Calculations” (1994)

1

Discrete Probability

Before I start discussing randomized algorithms at all, I need to give a quick formal overview of
the relatively small subset of probability theory that we will actually use. The first two sectionsof this note are deliberately written more as a review or reference than an introduction, although
they do include a few illustrative (and hopefully helpful) examples.

1.1

Discrete Probability Spaces

A discrete1 probability space (Ω, Pr) consists of a non-empty countable set Ω, called the sample
space, together with a probability mass function Pr: Ω → R such thatX
Pr[ω] ≥ 0 for all ω ∈ ΩandPr[ω] = 1.
ω∈ΩThe latter condition implies that Pr[ω] ≤ 1 for all ω ∈ Ω. I don’t know why the probability
function is written with brackets instead of parentheses, but that’s the standard;2 just go with it.
Here are a few simple examples:• A fair coin: Ω = {heads, tails} and Pr[heads] = Pr[tails] = 1/2.
• A fair six-sided die: Ω = {1, 2, 3, 4, 5, 6} and Pr[ω] = 1/6 for all ω ∈ Ω.
• A strangely loaded six-sided die: Ω = {1, 2, 3, 4, 5, 6} with Pr[ω] = ω/21 for all ω ∈ Ω.
(For example, Pr[4] = 4/21.)
1Correctly defining continuous (or otherwise uncountable) probability spaces and continuous random variables
requires considerably more care and subtlety than the discrete definitions given here. There is no well-definedprobability measure satisfying the discrete axioms when Ω is, for instance, an interval on the real line. This way lies
the Banach-Tarski paradox.
2Or more honestly: one of many standards© Copyright 2018 Jeff Erickson.
This work is licensed under a Creative Commons License (http://creativecommons.org/licenses/by-nc-sa/4.0/).
Free distribution is strongly encouraged; commercial distribution is expressly forbidden.
See http://jeffe.cs.illinois.edu/teaching/algorithms/ for the most recent revision.

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Algorithms

Lecture 1: Discrete Probability [Sp’17]• Bart’s rock-paper-scissors strategy: Ω = {rock, paper, scissors} and Pr[rock] = 1 andPr[paper] = Pr[scissors] = 0.
Other common examples of countable sample spaces include the 52 cards in a standard deck,the 52! permutations of the cards in a standard deck, the natural numbers, the integers, therationals, the set of all (finite) bit strings, the set of all (finite) rooted trees, the set of all (finite)
graphs, and the set of all (finite) execution traces of an algorithm.
The precise choice of probability space is rarely important; we can usually implicitly define Ω
to be the set of all possible tuples of values, one for each random variable under discussion.
1.1.1

Events and Probability

Subsets of Ω are usually called events, and individual elements of Ω are usually called sample
points or elementary events or atoms. However, it is often useful to think of the elements of Ωas possible states of a system or outcomes of an experiment, and subsets of Ω as conditions thatsome states/outcomes satisfy and others don’t.
The probability of an event A, denoted Pr[A], is defined as the sum of the probabilities of
its constituent sample points:
X
Pr[A] :=Pr[ω]ω∈A

In particular, we have Pr[∅] = 0 and Pr[Ω] = 1. Here we are extending (or overloading) the
function Pr: Ω → [0, 1] on atoms to a function Pr: 2Ω → [0, 1] on events.
For example, suppose we roll two fair dice, one red and the other blue. The underlying
probability space consists of the sample space Ω = {1, 2, 3, 4, 5, 6} × {1, 2, 3, 4, 5, 6} and theprobabilities Pr[ω] = 1/36 for all ω ∈ Ω.
• The probability of rolling two 5s is Pr[{(5, 5)}] = Pr[(5, 5)] = 1/36.
• The probability of rolling a total of 6 is
5Pr {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)} =.
36• The probability that the red die shows a 5 is
 1
Pr {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6)} = .
6• The probability that at least one die shows a 5 is
 11
Pr {(1, 5), (2, 5), (3, 5), (4, 5), (5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6), (6, 5)} =.
36• The probability that the red die shows a smaller number than the blue die isPr {(1, 2), (1, 3), (1, 4), (1, 5), (1, 6),(2, 3), (2, 4), (2, 5), (2, 6),(3, 4),5(3, 5), (3, 6), (4, 5),(4, 6), (5, 6)} =.
12

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1.1.2

Lecture 1: Discrete Probability [Sp’17]

Combining Events

Because they are formally just sets, events can be combined using arbitrary set operations.
However, in keeping with the intuition that events are conditions, these operations are usually
written using Boolean logic notation ∧, ∨, ¬ and vocabulary (“and, or, not”) instead of theequivalent set notation ∩, ∪, · and vocabulary (“intersection, union, complement”). For example,
consider our earlier experiment rolling two fair six-sided dice, one red and the other blue.
Pr[red 5] = 1/6
Pr[two 5s] = Pr[red 5 ∧ blue 5] = 1/36
Pr[at least one 5] = Pr[red 5 ∨ blue 5] = 11/36
Pr[at most one 5] = Pr[¬(two 5s)] = 1 − Pr[two 5s] = 35/36
Pr[no 5s] = Pr[¬(at least one 5)]= 1 − Pr[at least one 5] = 25/36
Pr[exactly one 5] = Pr[at least one 5 ∧ at most one 5]= Pr[red 5 ⊕ blue 5] = 5/18
Pr[blue 5 ⇒ red 5] = Pr[¬(blue 5) ∨ red 5] = 31/36(As usual, p ⇒ q is just shorthand for ¬p ∨ q; implication does not indicate causality!)
For any two events A and B with Pr[B] > 0, the conditional probability of A given B isdefined as
Pr[A ∧ B].
Pr[A | B] :=Pr[B]For example, in our earlier red-blue dice experiment:
Pr[blue 5 | red 5] = Pr[two 5s | red 5] = 1/6
Pr[at most one 5 | red 5] = Pr[exactly one 5 | red 5]= Pr[¬(blue 5) | red 5] = 5/6
Pr[at least one 5 | at most one 5] = 2/7
Pr[at most one 5 | at least one 5] = 10/11
Pr[red 5 | at least one 5] = 6/11
Pr[red 5 | at most one 5] = 1/7
Pr[blue 5 | blue 5 ⇒ red 5] = 1/31
Pr[red 5 | blue 5 ⇒ red 5] = 6/31
Pr[blue 5 ⇒ red 5 | blue 5] = 1/6
Pr[blue 5 ⇒ red 5 | red 5] = 1
Pr[blue 5 ⇒ red 5 | red 5 ⇒ blue 5] = 26/31
Two events A and B are disjoint if they are disjoint as sets, meaning A ∩ B = ∅. For example,
in our two-dice experiment, the events “red 5” and “blue 5” and “total 5” are pairwise disjoint.
Note that it is possible for Pr[A ∧ B] = 0 even when the events A and B are not disjoint; consider
the events “Bart plays paper” and “Bart does not play rock”.
Two events A and B are independent if and only if Pr[A ∧ B] = Pr[A] · Pr[B]. For example,
in our two-dice experiment, the events “red 5” and “blue 5” are independent, but “red 5” and“total 5” are not.
3

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Lecture 1: Discrete Probability [Sp’17]

More generally,aQ
countable set of events {Ai | i ∈ I} is fully or mutually independent if andVonly if Pr[ i∈I Ai ] = i∈I Ai . A set of events is k-wise independent if every subset of k events
is fully independent, and pairwise independent if every pair of events in the set is independent.
For example, in our two-dice experiment, the events “red 5” and “blue 5” and “total 7” arepairwise independent, but not mutually independent.
1.1.3

Identities and Inequalities

Fix n arbitrary events A1 , A2 , . . . , An from some sample space Ω. The following observations
follow immediately from similar observations about sets.
• Union bound: For any events A1 , A2 , . . . An , the definition of probability implies that
 n 
n
X
_
Pr
Ai ≤
Pr[Ai ].
i=1

i=1

The expression on the right counts each atom in the union of the events exactly once; theleft summation counts each atom once for each event Ai that contains it.
• Disjoint union: If the events A1 , A2 , . . . An are pairwise disjoint, meaning Ai ∩ A j = ∅ forall i 6= j, the union bound becomes an equation:
 n 
n
X
_
Pr
Ai =
Pr[Ai ].
i=1

i=1

• The principle of inclusion-exclusion describes a simple relationship between probabilities
of unions (disjunctions) and intersections (conjunctions) of arbitrary events:
Pr[A ∨ B] = Pr[A] + Pr[B] − Pr[A ∧ B]
This principle follows directly from elementary boolean algebra and the disjoint union
bound:
Pr[A ∨ B] + Pr[A ∧ B] = Pr[(A ∧ B) ∨ (A ∧ B) ∨ (A ∧ B)] + Pr[A ∧ B]= Pr[A ∧ B] + Pr[A ∧ B] + Pr[A ∧ B] + Pr[A ∧ B]
= Pr[A ∧ B] + Pr[A ∧ B] + Pr[A ∧ B] + Pr[A ∧ B]
= Pr[A] + Pr[B]Inclusion-exclusion generalizes inductively any finite number of events as follows:
 n 


X
_
^
|I|
Pr
Ai = 1 −(−1) Pr
Ai
i=1

I⊆[1 .. n]i∈I• Independent union: For any pair A and B of independent events, we have
Pr[A ∨ B] = Pr[A] + Pr[B] − Pr[A ∧ B]
= Pr[A] + Pr[B] − Pr[A] Pr[B][inclusion-exclusion][independence]= 1 − (1 − Pr[A])(1 − Pr[B]).
More generally, if the events A1 , A2 , . . . An are mutually independent, then
 n 
n
Y
_
Pr
Ai = 1 −(1 − Pr[Ai ]) .
i=1

i=1

4

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Lecture 1: Discrete Probability [Sp’17]• Bayes’ Theorem: If events A and B both have non-zero probability, the definition of
conditional probability immediately implies
Pr[B | A]
Pr[A | B]= Pr[A ∧ B] =
.
Pr[A]
Pr[B]
and therefore

1.2

Pr[A | B] · Pr[B] = Pr[B | A] · Pr[A].

Random Variables

Formally, a random variable X is a function from a sample space Ω (with an associated probability
measure) to some other value set. For example, the identity function on Ω is a random variable,
as is the function that maps everything in Ω to Queen Elizabeth II, or any function mapping Ω to
the real numbers. Random variables are almost universally denoted by upper-case letters.

A random variable is not random, nor is it a variable.
The value space of a random variable is commonly described either by an adjective preceding
the phrase “random variable” or a noun replacing the word “variable”. For example:• A function from Ω to Z is called an integer random variable or a random integer.
• A function from Ω to R is called an real random variable or a random real number.
• A function from Ω to {0, 1} is called an indicator random variable or a random bit.
Since every integer is a real number, every integer random variable is also a real random variable;similarly, every random bit is also a random real number. Not all random variables are numerical;
for example:• A random graph is function from some sample space Ω to the set of all graphs.
• A random point in the plane is a function from some sample space Ω to R2 .
• The function mapping every element of Ω to Queen Elizabeth II could be called a random
Queen of England, or a random monarch, or (if certain unlikely conspiracy theories aretaken seriously) a random shape-shifting alien reptile.
1.2.1

It is a fungus.

Although random variables are formally not variables at all, we typically describe and manipulatethem as if they were variables representing unknown elements of their value sets, without referring
to any particular sample space.
In particular, we can apply arbitrary functions to random variables by composition. Fir any
random variable X : Ω → V and any function f : V → V 0 , the function f (X ) := f ◦ X is a randomvariable over the value set V 0 . In particular, if X is a real random variable and α is any real
number, then X + α and α · X are also real random variables. More generally, if X and X 0 are
random variables with value sets V and V 0 , then for any function f : V × V 0 → V 00 , the functionf (X , X 0 ) is a random variable over V 00 , formally defined asf (X , X 0 )(ω) := f (X (ω), X 0 (ω)).
5

Algorithms

Lecture 1: Discrete Probability [Sp’17]

These definitions extend in the obvious way to functions with an arbitrary number of arguments.
If φ is a boolean function or predicate over the value set of X , we implicitly identify the
random variable φ(X ) with the event {ω ∈ Ω | φ(X (ω))}. For example, if X is an integer randomvariable, thenPr[X = x] := Pr {ω | X (ω) = x}andPr[X ≤ x] := Pr {ω | X (ω) ≤ x}

and



Pr[X is prime] := Pr {ω | X (ω) is prime} .

In particular, we typically identify the boolean values True and False with the events Ω and ∅,respectively. Predicates with more than one random variable are handled similarly; for example,

Pr[X = Y ] := Pr {ω | X (ω) = Y (ω)} .
1.2.2

Expectation

For any real (or complex or vector) random variable X , the expectation of X is defined as
X
E[X ] :=
x · Pr[X = x].
xThis sum is always well-defined, because the set {x | Pr[X = x] 6= 0} ⊆ Ω is countable. For integer
random variables, the following definition is equivalent:
XX
E[X ] =Pr[X ≥ x] −Pr[X ≤ x]
=

x≥0

x≤0

X


Pr[X ≥ x] − Pr[X ≤ −x] .

x≥0

If moreover A is an arbitrary event with non-zero probability, then the conditional expectationof X given A is defined as
X
X x · Pr[X = x ∧ A]
E[X | A] :=x · Pr[X = x | A] =
Pr[A]
x
x
For any event A with 0 < Pr[A] < 1, we immediately have
E[X ] = E[X | A] · Pr[A] + E[X | ¬A] · Pr[¬A].
In particular, for any random variables X and Y , we haveX
E[X ] =
E[X | Y = y] · Pr[Y = y].
y

Two random variables X and Y are independent if, for all x and y, the events X = x and Y = y
are independent. If X and Y are independent real random variables, then E[X · Y ] = E[X ] · E[Y ].
(However, this equation does not imply that X and Y are independent.) We can extend thenotions of full, k-wise, and pairwise independence from events to random variables in similar
fashion. In particular, if X 1 , X 2 , . . . , x n are fully independent real random variables, then
 n

n
Y
Y
E
Xi =
E[X i ].
i=1

i=1

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Algorithms

Lecture 1: Discrete Probability [Sp’17]

Linearity of expectation refers to the following important fact: The expectation of any
weighted sum of random variables is equal to the weighted sum of the expectations of thosevariables. More formally, for any real random variables X 1 , X 2 , . . . , X n and any real coefficients
α1 , α2 , . . . , αn ,
 n

n
X
X

E
(αi · X i ) =αi · E[X i ] .
i=1

i=1

Linearity of expectation does not require the variables to be independent.
1.2.3

ExamplesConsider once again our experiment with two standard fair six-sided dice, one red and the otherblue. We define several random variables:• R is the value (on the top face) of the red die.
• B is the value (on the top face) of the blue die.
• S = R + B is the total value (on the top faces) of both dice.
= 7 − R is the value on the bottom face of the red die.

The variables R and B are independent, as are the variables
variables is independent.

R

R

•

and B, but no other pair of these

1+2+3+4+5+6 7=
62
E[R + B] = E[R] + E[B] = 7
E[R] = E[B] = E[ ] =RE[R + ] = 7[linearity]R[trivial distribution]

E[R + B + ] = E[R] + E[B] + E[ ] =

21
2

R

R

49
4
1 · 6 + 2 · 5 + 3 · 4 28
E[R · ] =
=
3
3
2
2
2
2
2
91
1
+
2
+
3
+
4
+
5
+ 62
E[R2 ] ==6
6
329
E[(R + B)2 ] = E[R2 ] + 2 E[RB] + E[B 2 ] =6
E[R + B | R = 6] = E[R | R = 6] + E[B | R = 6]
E[R · B] = E[R] · E[B] =[linearity][independence]R= 6 + E[B] = 19/2

[linearity][linearity][independence]

1+2+3+4+5=3
5
12 + 22 + 32 + 42 + 52
E[R2 | R + B = 6] == 11
5
(1 + 6) + (2 + 3)
E[R + B | R · B = 6] ==6
2
(1 · 5) + (2 · 4) + (3 · 3) + (4 · 2) + (5 · 1)
E[R · B | R + B = 6] ==7
5
E[R | R + B = 6] =7

Algorithms

1.3

Lecture 1: Discrete Probability [Sp’17]

Common Probability Distributions

A probability distribution assigns a probability to each possible value of a random variable.
More formally, X : Ω → V is a random variable over some probability space (Ω, Pr), the probability
distribution of X is the function P : V → [0, 1] such thatX
P(x) = Pr[X = x] ={Pr(ω) | X (ω) = x}.
The support of a probability distribution is the set of values with non-zero probability; this is asubset of the value set V . The following table summarizes several of the most useful discrete
probability distributions.
nameintuition

parameterssupportPr[X = x]E[X ]trivialGood ol’ Rock,nothing beats that!

—singleton set {a}1uniformfair die roll—finite set S 6= ∅

1
|S|

a
P

¨
Bernoullibiased coin flip0≤p≤1

{0, 1}

binomialn biased coin flips

0≤p≤1
n≥0

[0 .. n]geometric#tails before first head

0<p≤1

N

negativebinomial#tails before nth head

0<p≤1
n≥0

N

if x = 1p

1 − p if x = 0
 
n x
p (1 − p)n−x
x



S

|S|
p
np

(1 − p) x p

1−p
p


n+ x −1(1 − p) x p nxn(1 − p)p

Figure 1.1. Common discrete probability distributions.

• The trivial distribution describes the outcome of a “random” experiment that always
has the same result. A trivially distributed random variable takes some fixed value withprobability 1. Yes, this is still randomness.
• The uniform distribution assigns the same probability to every element of some finite
non-empty set S. For example, if the random variable X is uniformly distributed overthe integer range [1 .. n], then Pr[X = x] = 1/n for each integer x between 1 and n, andE[X ] = (n+1)/2. This distribution models (idealized) fair coin flips, die rolls, and lotteries;
consequently, this is what many people incorrectly think of as the definition of “random”.
• The Bernoulli distribution models a random experiment (called a Bernoulli trial) withtwo possible outcomes: success and failure. The probability of success, usually denoted p,
is a parameter of the distribution; the failure probability is often denoted q = 1 − p.
Success and failure are usually represented by the values 1 and 0. Thus, every indicatorrandom variable has a Bernoulli distribution, and its expected value is equal to its successprobability:
X
E[X ] =
x · Pr[X = x] = 0 · Pr[X = 0] + 1 · Pr[X = 1] = Pr[X = 1] = p.
x

The special case p = 1/2 is a uniform distribution with two values: a fair coin flip. The
special cases p = 0 and p = 1 are trivial distributions.
8

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Lecture 1: Discrete Probability [Sp’17]• The geometric distribution describes the number of independent Bernoulli trials (all withthe same success probability p) before the first success. If X is a geometrically distributed
random variable, then X = x if and only if the first x Bernoulli trials fail and the (x + 1)thtrial succeeds:
 x
YPr[X = x] =Pr[ith trial fails] · Pr[(x + 1)th trial succeeds] = (1 − p) x p.
i=1

• The binomial distribution is the sum of n independent Bernoulli distributions, all with thesame probability p of success. If X is a binomially distributed random variable, then X = x
if and only if x of the n trials succeed and n − x fail: 
n xPr[X = x] =p (1 − p)n−x .
x
If n = 1, this is just the Bernoulli distribution.
• The negative binomial distribution describes the number of independent Bernoulli trials(all with the same success probability p) that fail before the nth successful trial. If X is anegative-binomially distributed random variable, then X = x if and only if exactly x of thefirst n + x − 1 trials are failures, and the (n + x)th trial is a success:n+ x −1
Pr[X = x] =(1 − p) x p n
x
If n = 1, this is just the geometric distribution.

1.4

Coin Flips

Suppose you are given a coin and you are asked generate a uniform random bit. We distinguishbetween two types of coins: A coin is fair if the probability of heads (1) and the probability oftails (0) are both exactly 1/2. A coin where one side is more likely than the other is said tobe biased. Actual physical coins are reasonable approximations of abstract fair coins for most
purposes, at least if they’re flipped high into the air and allowed to bounce.3 Physical coins canbe biased by bending them.
1.4.1

Removing Unknown Bias

In 1951, John von Neumann discovered the following simple technique to simulate fair coin flips
using an arbitrarily biased coin, even without knowing the bias. Flip the biased coin twice. If thetwo flips yield different results, return the first result; otherwise, repeat the experiment fromscratch.
VonNeumannCoin( ):x ← BiasedCoin()y ← BiasedCoin()
if x 6= yreturn x
else
return VonNeumannCoin( )3Persi Diaconis, Susan Holmes, and Richard Montgomery published a thorough analysis of physical coin-flipping in
2007, which concluded (among other things) that coins that are flipped vigorously and then caught come up in thesame state they started about 51% of the time. The small amount of bias arises because flipping coins tend to precess
as they rotate. Letting the coin bounce instead of catching them appears to remove the bias from precession.

9

Algorithms

Lecture 1: Discrete Probability [Sp’17]

This is weird sort of algorithm, isn’t it? There is no upper bound on the worst-case running
time; in principle, the algorithm could run forever, because the biased coin always just happens
to flip heads. Nevertheless, I claim that this is a useful algorithm for generating fair random bits.
We need two technical assumptions:(1) The biased coin always flips heads with the same fixed (but unknown!) probability p. To
simplify notation, we let q = 1 − p denote the probability of flipping tails. For example, afair coin would have p = q = 1/2.
(2) All flips of the biased coin are mutually independent.
First, I claim that if the algorithm halts, then it returns a uniformly distributed random bit.
Because the two biased coin flips are independent, we havePr[x = 0 ∧ y = 1] = Pr[x = 1 ∧ y = 0] = pq,
and therefore (assuming pq > 0)Pr[x = 0 ∧ y = 1 | x 6= y] = Pr[x = 1 ∧ y = 0 | x 6= y] =

pq
1
= .
2pq
2

Because the biased coin flips are mutually independent, the same analysis applies without
modification to every recursive call to VonNeumannCoin. Thus, if any recursive call returns abit, that bit is uniformly distributed.
Now let T denote the actual running time of this algorithm; T is a random variable that
depends on the biased coin flips.⁴ We can compute the expected running time E[T ] by considering
the conditional expectations in two cases: The first two flips are either different or equal.
E[T ] = E[T | x 6= y] · Pr[x 6= y] + E[T | x = y] · Pr[x = y]
Because the two biased coin flips are independent, we havePr[x 6= y] = Pr[x = 0 ∧ y = 1] + Pr[x = 1 ∧ y = 0] = 2pq
and therefore Pr[x = y] = 1 − 2pq. If the first two coin flips are different, the algorithm ends
after two flips; thus, E[T | x 6= y] = 2. Finally, if the first two coin flips are the same, theexperiment starts over from scratch after the first two flips, so E[T | x = y] = 2 + E[T ]. Putting
all the pieces together, we have
E[T ] = 2 · 2pq + (2 + E[T ]) · (1 − 2pq).
Solving this equation for E[T ] yields the solution E[T ] = 1/pq. For example, if p = 1/3, theexpected number of coin flips is 9/2.
Alternatively, we can think of VonNeumannCoin as performing an experiment with adifferent biased coin, which returns a result (“heads”) with probability 2pq. Thus, the expected
number of unsuccessful iterations (“tails” before the first “head”) is a geometric random variablewith expectation 1/2pq − 1, and thus the expected number of iterations is 1/2pq. Because each
iteration flips two coins, the expected number of coin flips is 1/pq.
⁴Normally we use T (n) to denote the worst-case running time of an algorithm as a function of some input parametern, but this algorithm has no input parameters!

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1.4.2

Lecture 1: Discrete Probability [Sp’17]

Removing Known Bias

But what if we know that p = 1/3? In that case, the following algorithm simulates a fair coinwith fewer biased flips (on average):
FairCoin( ):x ← BiasedCoin(1/3)
y ← BiasedCoin(1/3)
if x 6= y〈〈probability 4/9〉〉
return 0
else if x = y = 1 〈〈probability 4/9〉〉
return 1
else〈〈probability 1/9〉〉
return FairCoin( )The algorithm returns a fair coin because
Pr[x 6= y] =4= Pr[x = y = 0].
9The expected number of flips satisfies the equation
E[T ] = 2 +1
E[T ],
9which implies that E[T ] = 9/4, a factor of 2 better than von Neumann’s algorithm.

1.5

Pokémon Collecting

A distressingly large fraction of my daughters’ friends are obsessed with Pokémon—not the
cartoon or the mobile game, but the collectible card game. The Pokémon Company sells small
packets, each containing half a dozen cards, each describing a different Pokémon character. The
cards can be used to play a complex turn-based combat game; the more cards a player owns,
the more likely they are to win. So players are strongly motivated to collect as many cards, andin particular, as many different cards, as possible. Pokémon reinforces this message with their
oh-so-subtle theme song “Gotta Catch ‘Em All!” Unfortunately, the packets are opaque; the only
way to find out which cards are in a pack is to buy the pack and tear it open.⁵
Let’s consider the following oversimplified model of the Pokémon-collection process. In eachtrial, we purchase one Pokémon card, chosen independently and uniformly at random from theset of n possible card types. We repeat these trials until we have purchased at least one of eachtype of card. This problem was first considered by the French mathematician Abraham de Moivre
in his seminal 1712 treatise De Necessitate ut Caperent Omnium Eorum.⁶
1.5.1

After n Trials

How many different types of Pokémon do we actually own after we buy n cards? Obviously inthe worst case, we might just have n copies of one Pokémon,⁷ but that’s not particularly likely. To
⁵See also: cigarette cards, Dixie cups, baseball cards, Pez dispensers, Beanie Babies, Iwako puzzle erasers, Shopkins,
and Guys Under Your Supervision.
⁶The actual title was De Mensura Sortis seu; de Probabilitate Eventuum in Ludis a Casu Fortuito Pendentibus, whichmeans “On the measurement of chance, or on the probability of events in games depending on fortuitous chance”.
⁷“Dave Guy!”

11

Algorithms

Lecture 1: Discrete Probability [Sp’17]

analyze the expected number of types that we own, we introduce an incredibly useful techniquethat lets us exploit linearity of expectation: decomposing more complex random variables into
sums of indicator variables.
P
For each index i, define an indicator variable X i = [we own Pokémon i] so that X = i X i isthe number of cards we own. Linearity of expectation implies
E[X ] =n
X

E[X i ] =

i=1

n
X

Pr[X i = 1].

i=1The probability that we don’t own card i is (1 − 1/n)n ≈ 1/e, so
E[X ] =n
X

E[X i ] ≈i=1nX
(1 − 1/e) = (1 − 1/e)n ≈ 0.63212n.
i=1

In other words, after buying n cards, we expect to own a bit less than 2/3 of the Pokémon.
Similar calculations implies that we expect to own about 86% of the Pokémon after buying
2n cards, about 95% after buying 3n cards, about 98% after buying 4n cards, and so on.
1.5.2

Gotta Catch ‘em All

So how many Pokémon packs do we need to buy to catch ‘em all? Obviously in the worst case,we might never have a complete collection⁸, but assuming each type of card has some non-zeroprobability of being in each pack, the expected number of packs we need to buy is finite. Let
T (n) denote the number of packs (or “time”) required to collect all n Pokémon. For purposes ofanalysis, we partition the random purchasing algorithm into n phases, where the ith phase ends
just after we see the ith distinct card type. Let us writeX
T (n) =Ti (n)iwhere Ti (n) is the number of cards bought during the ith phase. Linearity of expectation implies
X
E[T (n)] =E[Ti (n)].
i

We can think of each card purchase as a biased coin flip, where “heads” means “got a new Pokémon”
and “tails” means “got a Pokémon we already own”. For each index i, the probability of heads (thatis, the probability of a single purchase being a new Pokémon) is exactly p = (n − i + 1)/n: each of
the n Pokémon is equally likely, and there are n− i +1 Pokémon that we don’t already own. By ourearlier analysis, the expected number of flips until the first head is E[Ti ] = 1/p = n/(n − i + 1).
We conclude that
E[T (n)] =

X
i

n
X

n

Xn
n
=
= nH n .
E[Ti (n)] =
n−i+1
j
i=1
j=1

Here Hn denotes the nth harmonic number, defined recursively as(0
if n = 0
Hn =1
H n−1 +
otherwisen⁸“Dave Guy!”

12

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Lecture 1: Discrete Probability [Sp’17]

Approximating the summation H n =Pn1i=1 i above and below by integrals implies the boundsln(n + 1) ≤ H n ≤ (ln n) + 1.
Thus, the expected number of cards we need to buy to get all n Pokémon is Θ(n log n).
In particular, to catch all 150 of the original Pokémon, we should expect to buy 150 · H150 ≈
838.67709 cards, and to own at least one copy of each of the 9184 Pokémon card types available
in 2013, we should expect to buy 9184 · H9184 ≈ 89107.65186 cards. (In practice, of course, thisestimate is far too low, because some cards are considerably more common than others.)

1.6

Random Permutations

Now suppose we are given a deck of n (Pokémon?) cards and are asked to shuffle them. Ideally,
we would like an algorithm that produces each of the n! possible permutations of the deck withthe same probability 1/n!.
There are many such algorithms, but the gold standard is ultimately based on the millenniaold tradition of drawing or casting lots. “Lots” are traditionally small pieces of wood, stone, or
paper, which were blindly drawn from an opaque container. The following algorithm takes a
set L of n distinct Lots (arbitrary objects) as input and returns an array R[1 .. n] containing aRandom permutation of those n lots.
DrawLots(L):n ← |L|for i ← 1 to n
remove a random lot x from L
R[i] ← x
return R[1 .. n]

There are exactly n! possible outcomes for this algorithm—exactly n choices for the first lot, thenexactly n − 1 choices for the second lot, and so on—each with exactly the same probability andeach leading to a different permutation. Thus, every permutation of lots is equally likely to be
output.
A modern formulation of lot-casting was described by (and is frequently misattributed to)statisticians Ronald Fisher and Frank Yates in 1938. Fisher and Yates formulated their algorithm
as a method for randomly reordering a List of numbers. In their original formulation, thealgorithm repeatedly chooses at random a number from the input list that has not been previouslychosen, adds the chosen number to the output list, and then strikes the chosen number from the
input list.
We can implement this formulation of the algorithm using a secondary boolean array
Chosen[1 .. n] indicating which items have already been chosen. The randomness is provided
by a subroutine Random(n), which returns an integer chosen independently and uniformly
at random from the set {1, 2, . . . , n} in O(1) time; in other words, Random(n) simulates a fairn-sided die.

13

Algorithms

Lecture 1: Discrete Probability [Sp’17]

FisherYates(L[1 .. n]):for i ← 1 down to n
Chosen[i] ← False
for i ← n down to 1
repeatr ← Random(n)until ¬Chosen[r]R[i] ← L[r]Chosen[r] ← True
return R[1 .. n]

The repeat-until loop chooses an index r uniformly at random from the set of previously unchosen
indices. Thus, this algorithm really is an implementation of DrawLots. But now choosing thenext random lot element may require several iterations of the repeat-until loop. How slow is thisalgorithm?
In fact, FisherYates is equivalent to our earlier Pokémon-collecting algorithm! Each callto Random is a purchase, and Chosen[i] indicates whether we’ve already purchased the ith
Pokémon. By our earlier analysis, the expected number of calls to Random before the algorithm
halts is exactly nH n . We conclude that this algorithm runs in Θ(n log n) expected time.⁹
A most efficient implementation of lot-casting, which permutes the input array in place, wasdescribed by Richard Durstenfeld in 1961. In full accordance with Stigler’s Law, this algorithm isalmost universally called “the Fisher-Yates shuffle”.1⁰
SelectionShuffle(A[1 .. n]):for i ← n down to 1
swap A[i] ↔ A[Random(i)]The algorithm clearly runs in O(n) time. Correctness follows from exactly the same argument as
DrawLots: There are n! equally likely possibilities from the n calls to Random—n for the first
call, n − 1 for the second, and so on—each leading to a different output permutation.
Although it may not appear so at first glance, SelectionShuffle is an implementation of
lot-casting. After each iteration of the main loop, the suffix A[i .. n] (on the Right) plays the role ofR, storing the previously chosen input elements, and the prefix A[1 .. i − 1] (on the Left) plays therole of L, containing all the unchosen input elements. Unlike the original FisherYates algorithm,
SelectionShuffle changes the order of the unchosen elements as it runs. Fortunately, andobviously, that order is utterly irrelevant; only the set of unchosen elements matters.
We can also uniformly shuffle by reversing the order of the loop in the previous algorithm.
Again, this algorithm is usually misattributed to Fisher and Yates.
InsertionShuffle(A[1 .. n]):for i ← 1 to nswap A[i] ↔ A[Random(i)]Again, correctness follows from the observation that there are n! equally likely output permutations. Alternatively, we can argue inductively that for every index i, the prefix A[1 .. i] is
⁹In later editions of Fisher and Yates’ monograph, they instead described a different algorithm due to C. Radhakrishna
Rao, dismissing their earlier method as “tiresome, since each [item] must be deleted from a list as it is selected and a
fresh count made for each further selection.”
1⁰However, some authors call this algorithm the “Knuth shuffle”, because Donald Knuth described it in his landmark
Art of Computer Programming, even though he attributed the algorithm to Durstenfeld in the first edition, and to
Fisher and Yates in the second. It’s actually rather shocking that Knuth did not attribute the algorithm to Aaron, citing
the First Chronicles.

14

Algorithms

Lecture 1: Discrete Probability [Sp’17]

uniformly shuffled after the ith iteration of the loop. Alternatively, we can observe that running
InsertionShuffle is the same as running SelectionShuffle backward in time—essentiallyputting the lots back into the bag—and that the inverse of a uniformly-distributed permutation is
also uniformly distributed.
(The names SelectionShuffle and InsertionShuffle for these two variants are nonstandard. SelectionShuffle randomly selects the next card and then adds to one end of therandom permutation, just as selection sort repeatedly selects the largest element of the unsorted
portion of the array. InsertionShuffle randomly inserts the first card in the untouched portionof the array into the random permutation, just as insertion sort repeatedly inserts the next itemin the unsorted portion of the input into the sorted portion.)

1.7

ÆÆÆ

Properties of Random Permutations• For any subsequence of indices: the set of values and the permutation of those values
are uniformly and independently distributed.

• For any subsequence of values: the set of indices and the permutation of those indices
are uniformly and independently distributed.

• For example: In a randomly shuffled deck, the expected number of hearts among thefirst 5 face cards is 5/4.

Exercises
Several of these problems refer to decks of playing cards. A standard (Anglo-American) deck of
52 playing cards contains 13 cards in each of four suits: « (spades), ª (hearts), © (diamonds),and ¨ (clubs). The 13 cards in each suit have distinct ranks: A (ace), 2 (deuce), 3 (trey), 4, 5, 6,
7, 8, 9, 10, J (jack), Q (queen), and K (king). Cards are normally named by writing their rankfollowed by their suit; for example, J« is the jack of spades, and 10ª is the ten of hearts. For
purposes of comparing ranks in the problems below, aces have rank 1, jacks have rank 11, queens
have rank 12, and kings have rank 13; for example, J« has higher rank than 8©, but lower rank
than Q¨.
1. On their long journey from Denmark to England, Rosencrantz and Guildenstern amuse
themselves by playing the following game with a fair coin. First Rosencrantz flips the coinover and over until it comes up tails. Then Guildenstern flips the coin over and over until
he gets as many heads in a row as Rosencrantz got on his turn. Here are three typicalgames:
Rosencrantz: H H T
Guildenstern: H T H H
Rosencrantz: T
Guildenstern: (no flips)
Rosencrantz: H H H T
Guildenstern: T H H T H H T H T T H H H(a) What is the expected number of flips in one of Rosencrantz’s turns?
(b) Suppose Rosencrantz happens to flip k heads in a row on his turn. What is the
expected number of flips in Guildenstern’s next turn?
15

Algorithms

Lecture 1: Discrete Probability [Sp’17](c) What is the expected total number of flips (by both Rosencrantz and Guildenstern) ina single game?
Prove that your answers are correct. If you have to appeal to “intuition” or “common sense”,
your answer is almost certainly wrong! Full credit requires exact answers, but a correct
asymptotic bound (as a function of k) in part (b) is worth significant partial credit.
2. After sending his loyal friends Rosencrantz and Guildenstern off to Norway, Hamlet decides
to amuse himself by repeatedly flipping a fair coin until the sequence of flips satisfies somecondition. For each of the following conditions, compute the exact expected number of flips
until that condition is met. Some conditions depend on a positive integer parameter n.
(a) Hamlet flips n times.
(b) Hamlet flips heads.
(c) Hamlet flips both heads and tails (in different flips, of course).
(d) Hamlet flips heads twice.
(e) Hamlet flips heads twice in a row.
(f) Hamlet flips heads followed immediately by tails.
(g) Hamlet flips the sequence heads, tails, heads, tails.
(h) Hamlet flips heads n times.
(i) Hamlet flips heads n times in a row.
(j) Hamlet flips more heads than tails.
(k) Hamlet flips the same positive number of heads and tails.
(l) Either Hamlet flips more heads than tails or he flips n times, whichever comes first.
For each condition, prove that your answer is correct. If you have to appeal to “intuition”
or “common sense”, your answer is almost certainly wrong! Correct asymptotic bounds forthe conditions that depend on n are worth significant partial credit.
3. Suppose you have access to a function FairCoin that returns a single random bit, chosen
uniformly and independently from the set {0, 1}, in O(1) time. Consider the following
randomized algorithm for generating biased random bits.
OneInThree:
if FairCoin = 0
return 0
else
return 1 − OneInThree(a) Prove that OneInThree returns 1 with probability 1/3.
(b) What is the exact expected number of times that this algorithm calls FairCoin?
(c) Now suppose instead of FairCoin you are given a subroutine BiasedCoin that returns
an independent random bit equal to 1 with some fixed but unknown probability p, inO(1) time. Describe an algorithm OneInThree that returns either 0 or 1 with equal
probability, using BiasedCoin as its only source of randomness.
16

Algorithms

Lecture 1: Discrete Probability [Sp’17](d) What is the exact expected number of times that your OneInThree algorithm calls
BiasedCoin?
4. Describe an algorithm that takes a real number 0 ≤ p ≤ 1 as input, and returns an
independent random bit that is equal to 1 with the given probability p, using independent
fair coin flips as the only source of randomness. What is the expected running time of youralgorithm (as a function of p)?
5. (a) Describe an algorithm that simulates a fair three-sided die, using independent faircoin flips as the only source of randomness. Your algorithm should return 1, 2, or 3,
each with probability 1/3. What is the expected number of coin flips used by youralgorithm?
(b) Describe an algorithm that simulates a fair coin flip, using independent rolls of a fair
three-sided die as the only source of randomness. What is the expected number ofdie rolls used by your algorithm?
? (c)Describe an algorithm that simulates a fair n-sided die, using independent rolls of afair k-sided die as the only source of randomness. What is the expected number ofdie rolls used by your algorithm (as a function of n and k)? You may assume n and k
are relatively prime.

6. Describe an algorithm FairDie that returns an integer chosen uniformly at random fromthe set {1, 2, 3, 4, 5, 6}, using an algorithm LoadedDie that returns an algorithm from thesame set with some fixed but unknown non-trivial probability distribution. What is the
expected number of times your FairDie algorithm calls LoadedDie? [Hint: 3! = 6.]
7. (a) Suppose you have access to a function FairCoin that returns a single random bit,chosen uniformly and independently from the set {0, 1}, in O(1) time. Describe
and analyze an algorithm Random(n) that returns an integer chosen uniformly andindependently at random from the set {1, 2, . . . , n}, given a non-negative integer n as
input, using FairCoin as its only source of randomness.
(b) Suppose you have access to a function FairCoins(k) that returns an integer chosen
uniformly and independently at random from the set {0, 1, . . . , 2k − 1} in O(1) time,
given any non-negative integer k as input. Describe and analyze an algorithm
Random(n) that returns an integer chosen uniformly and independently at random
from the set {1, 2, . . . , n}, given any non-negative integer n as input, using FairCoinsas its only source of randomness.
8. You are applying to participate in this year’s Trial of the Pyx, the annual ceremony where
samples of all British coinage are tested, to ensure that they conform as strictly as possible
to legal standards. As a test of your qualifications, your interviewer at the Worshipful
Company of Goldsmiths has given you a bag of n commemorative Alan Turing half-guinea
coins, exactly two of which are counterfeit. One counterfeit coin is very slightly lighter
than a genuine Turing; the other is very slightly heavier. Together, the two counterfeit
coins have exactly the same weight as two genuine coins. Your task is to identify the twocounterfeit coins.
17

Algorithms

Lecture 1: Discrete Probability [Sp’17]

The weight difference between the real and fake coins is too small to be detected byanything other than the Royal Pyx Coin Balance. You can place any two disjoint setsof coins in each of the Balance’s two pans; the Balance will then indicate which of thetwo subsets has larger total weight, or that the two subsets have the same total weight.
Unfortunately, each use of the Balance requires completing a complicated authorization
form (in triplicate), submitting a blood sample, and scheduling the Royal Bugle Corps, so
you really want to use the Balance as few times as possible.
(a) Suppose you randomly choose n/2 of your n coins to put on one pan of the Balance,and put the remaining n/2 coins on the other pan. What is the probability that thetwo subsets have equal weight?
(b) Describe and analyze a randomized algorithm to identify the two fake coins. What isthe expected number of times your algorithm uses the Balance?
9. Consider the following algorithm for shuffling a deck of n cards, initially numbered inorder from 1 on the top to n on the bottom. At each step, we remove the top card from thedeck an insert it randomly back into in the deck, choosing one of the n possible positions
uniformly at random. The algorithm ends immediately after we pick up card n − 1 andinsert it randomly into the deck.
(a) Prove that this algorithm uniformly shuffles the deck, so that each permutation of thedeck has equal probability. [Hint: Prove that at all times, the cards below card n − 1
are uniformly shuffled.](b) Prove that before the algorithm ends, the deck is not uniformly shuffled.
(c) What is the expected number of steps before this algorithm ends?
10. Suppose we want to write an efficient algorithm Shuffle(A[1 .. n]) that randomly permutes
the input array, so that each of the n! permutations is equally likely.
(a) Prove that the following algorithm is not correct. [Hint: Consider the case n = 3.]
NaiveShuffle(A[1 .. n]):for i ← 1 to nswap A[i] ↔ A[Random(n)](The only difference from InsertionShuffle is that the argument to Random is ninstead of i.)(b) The algorithm BufferShuffle, shown in Figure 1.2, takes an extra parameter b
describing the size of the internal buffer array B[1 .. b].
i. Prove that BufferShuffle is correct for all b ≥ n.
ii. What is the expected running time of BufferShuffle when b = n?
iii. What is the expected running time of BufferShuffle when b = 2n?
? (c)Prove that the algorithm RaoShuffle, shown in Figure 1.2, is correct and analyze its
expected running time. This is the algorithm of C. Radhakrishna Rao that Fisher andYates found less “tiresome” than their own.

18

Algorithms

Lecture 1: Discrete Probability [Sp’17]

RaoShuffle(A[1 .. n]):〈〈Initialize n lists〉〉
for j ← 1 to n`[ j] ← 0〈〈Add each A[i] to a random list〉〉
for i ← 1 to n`[ j] ← `[ j] + 1
L[ j][`[ j]] ← A[i]〈〈Recursively shuffle the lists〉〉
for j ← 1 to n
if `[ j] > 1
Shuffle(L[ j][1 .. `[ j]])〈〈Concatenate the lists〉〉i←0
for j ← 1 to n
for k ← 1 to `[ j]
A[i] ← L[ j][k]i ← i+1

BufferShuffle(n, b):〈〈Initialize buffer 〉〉for j ← 1 to 2n
B[ j] ← Null〈〈Copy items randomly into buffer 〉〉for i ← 1 to nj ← Random(b)while (B[ j] == Null)j ← Random(n)B[ j] ← A[i]〈〈Compress buffer into input array〉〉i←1
for j ← 1 to b
if B[ j] 6= Null
A[i] ← B[ j]i ← i+1

Figure 1.2. Left: Shuffling using a buffer array. Right: Rao’s less “tiresome” shuffling algorithm.

11. (a) Prove that the following algorithm, modeled after quicksort, uniformly permutes its
input array,each of the n! possible output permutations is equally likely.
 meaningnn!
[Hint: k = k!(n−k)!
]
QuickShuffle(A[1 .. n]):
if n ≤ 1
returnj ← 1; k ← nwhile j ≤ kwith probability 1/2
swap A[ j] ↔ A[k]k ← k−1elsej ← j+1
QuickShuffle(A[1 .. k])
QuickShuffle(A[ j .. n])(b) Prove that QuickShuffle runs in O(n log n) expected time. [Hint: This will be mucheasier after reading the next chapter.](c) Prove that the algorithm MergeShuffle shown on the next page, which is modeled
after mergesort, does not uniformly permute its input array.
(d) Describe how to modify RandomMerge so that MergeShuffle does uniformly
permute its input array, and prove that your modification is correct. [Hint: Modify
the line in red.]19

Algorithms

Lecture 1: Discrete Probability [Sp’17]

MergeShuffle(A[1 .. n]):
if n > 1m ← bn/2c
MergeShuffle(A[1 .. m])
MergeShuffle(A[m + 1 .. n])
RandomMerge(A[1 .. n], m)RandomMerge(A[1 .. n], m):i ← 1; j ← m + 1
for k ← 1 to n
if j > nB[k] ← A[i]; i ← i + 1
else if i > m
B[k] ← A[ j]; j ← j + 1
else with probability 1/2
B[k] ← A[i]; i ← i + 1
else
B[k] ← A[ j]; j ← j + 1
for k ← 1 to n
A[k] ← B[k]12. Clock Solitaire is played with a standard deck of playing cards. To set up the game, deal
the cards face down into 13 piles of four cards each, one in each of the “hour” positions ofa clock and one in the center. Each pile corresponds to a particular rank—A through Q inclockwise order for the hour positions, and K for the center. To start the game, turn over acard in the center pile. Then repeatedly turn over a card in the pile corresponding to thevalue of the previous card. The game ends when you try to turn over a card from a pilewhose four cards are already face up. (This is always the center pile—why?) You win if
and only if every card is face up when the game ends.
What is the exact probability that you win a game of Clock Solitaire, assuming that thecards are permuted uniformly at random before they are dealt into their piles?
13. Professor Jay is about to perform a public demonstration with two decks of cards, onewith red backs (“the red deck”) and one with blue backs (“the blue deck”). Both decks lie
face-down on a table in front of the good Professor, shuffled so that every permutation ofeach deck is equally likely.
To begin the demonstration, Professor Jay turns over the top card from each deck. If
one of these two cards is the three of clubs (3¨), the demonstration ends immediately.
Otherwise, the good Professor repeatedly hurls the cards he just turned over into the thick,pachydermatous outer melon layer of a nearby watermelon, and then turns over the next
card from the top of each deck. The demonstration ends the first time a 3¨ is turned over.
Thus, if 3¨ is the last card in both decks, the demonstration ends with 102 cards embeddedin the watermelon, that most prodigious of household fruits.
(a) What is the exact expected number of cards that Professor Jay hurls into the watermelon?
(b) For each of the statements below, give the exact probability that the statement is trueof the first pair of cards Professor Jay turns over.
i. Both cards are threes.
ii. One card is a three, and the other card is a club.
iii. If (at least) one card is a heart, then (at least) one card is a diamond.
iv. The card from the red deck has higher rank than the card from the blue deck.
(c) For each of the statements below, give the exact probability that the statement is trueof the last pair of cards Professor Jay turns over.

20

Algorithms

Lecture 1: Discrete Probability [Sp’17]

i. Both cards are threes.
ii. One card is a three, and the other card is a club.
iii. If (at least) one card is a heart, then (at least) one card is a diamond.
iv. The card from the red deck has higher rank than the card from the blue deck.
14. Penn and Teller agree to play the following game. Penn shuffles a standard deck of playing
cards so that every permutation is equally likely. Then Teller draws cards from the deck,
one at a time without replacement, until he draws the three of clubs (3¨), at which point
the remaining undrawn cards instantly burst into flames.
The first time Teller draws a card from the deck, he gives it to Penn. From then on,
until the game ends, whenever Teller draws a card whose value is smaller than the lastcard he gave to Penn, he gives the new card to Penn.11 To make the rules unambiguous,
they agree beforehand that A = 1, J = 11, Q = 12, and K = 13.
(a) What is the expected number of cards that Teller draws?
(b) What is the expected maximum value among the cards Teller gives to Penn?
(c) What is the expected minimum value among the cards Teller gives to Penn?
(d) What is the expected number of cards that Teller gives to Penn? [Hint: Let 13 = n.]
15. Consider a random walk on a path with vertices numbered 1, 2, . . . , n from left to right. At
each step, we flip a coin to decide which direction to walk, moving one step left or one step
right with equal probability. The random walk ends when we fall off one end of the path,either by moving left from vertex 1 or by moving right from vertex n.
(a) Prove that the probability that the walk ends by falling off the right end of the path isexactly 1/(n + 1).
(b) Prove that if we start at vertex k, the probability that we fall off the right end of thepath is exactly k/(n + 1).
(c) Prove that if we start at vertex 1, the expected number of steps before the random
walk ends is exactly n.
(d) What is the exact expected length of the random walk if we start at vertex k, as afunction of n and k? Prove your result is correct. (For partial credit, give a tightΘ-bound for the case k = (n + 1)/2, assuming n is odd.)[Hint: Trust the recursion fairy.]16. Suppose n lights labeled 0, . . . , n − 1 are placed clockwise around a circle. Initially, every
light is off. Consider the following random process.
11Specifically, he hurls it directly into the back of Penn’s right hand.

21

Algorithms

Lecture 1: Discrete Probability [Sp’17]

LightTheCircle(n):k←0turn on light 0
while at least one light is off
with probability 1/2k ← (k + 1) mod nelsek ← (k − 1) mod n
if light k is off, turn it on(a) Let p(i, n) denote the probability that the last light turned on by LightTheCircle(n, 0)
is light i. For example, p(0, 2) = 0 and p(1, 2) = 1. Find an exact closed-form expression for p(i, n) in terms of n and i. Prove your answer is correct.
(b) Give the tightest upper bound you can on the expected running time of this algorithm.
[Hint: You may find the previous exercise helpful.]
17. Let T be an arbitrary, not necessarily balanced, binary tree T with n nodes.
(a) Consider a random walk downward from the root of T , as described by the following
algorithm.
RandomTreeDescent(T ):v ← T.root
while v 6= Nullwith probability 1/2v ← v.leftelsev ← v.right

Find a tight bound on the worst-case expected running time of this algorithm, as afunction of n, and prove your answer is correct. What is the worst-case tree?
(b) Now consider a different random walk starting at the root of T , as described by thefollowing algorithm.
RandomTreeWalk(T ):v ← T.root
while v 6= Nullwith probability 1/3v ← v.left
else with probability 1/3v ← v.rightelsev ← v.parent

Find a tight bound on the worst-case expected running time of this algorithm, as afunction of n, and prove your answer is correct. What is the worst-case tree?

© Copyright 2018 Jeff Erickson.
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