Ti = indicator random variable of the event
that i-th throw results in a tail
E[T] = E[T1] + … + E[T6] = 6*(1/2) = 3
P(T=3) = P(H=3) = binomial(6,3)/26 = 5/16 < 1/2
E[Xi,j] = ½
(consider only i<j)

X= Xi,j
1 i<j n
E[X] = n(n-1) /4

T = 1 + (1/2) * 0 + (1/2) * ( T + T )
T=1+T
There exists c such that
T(n)  T(n/2)+T(n/3)+c*n.
We need to show that there exists d such that
T(n)  d*n for all n.
Induction step:
T(n)  T(n/2) + T(n/3) + c*n 
d*n/2 + d*n/3 + c*n 
d*n + (c-d/6)*n  d*n,
taking d=6c.
l  m+1
if B  A[m] then
Reverse(a,b)
for i from a to a+b do
swap(A[i],A[a+b-i]);
Rotate(k)
Reverse(1,k)
Reverse(k+1,n)
Reverse(1,n)
1,….,k,k+1,….,n
k,….,1,k+1,….,n
k,….,1,n,….,k+1
k+1,….,n,1,….,k
1) find the median m of A
2)
m
m
m
sum S
3) if S\geq C then
recurse on A[n/2..n]
else
recurse on A[1..n/2] with C’=C-S
T(n) = T(n/2) + O(n)
3) if S\geq C then
recurse on A[n/2..n]
else
recurse on A[1..n/2] with C’=C-S
Coupon collector problem
n coupons to collect
What is the expected number of
cereal boxes that you need to buy?
Coupon collector problem
Assume that a dart
throw is uniform in
the circle. Let p be
The fraction occupied
by the bull’s eye.
Expected number of darts needed to hit the
bull’s eye ?
Coupon collector problem
Assume that a dart
throw is uniform in
the circle. Let p be
The fraction occupied
by the bull’s eye.
Expected number of darts needed to hit the
bull’s eye ?
1/p
What is the expected number of
boxes that I buy in k-th phase ?
k-th phase = when I have k different
Kinds of coupons.
E[X0] = 1
…
E[Xk] = ?
…
E[Xn-1] = n
What is the expected number of
boxes that I buy in k-th phase ?
k-th phase = when I have k different
Kinds of coupons.
E[X0] = 1
…
E[Xk] = n/(n-k)
…
E[Xn-1] = n
What is the expected number of
boxes that I buy in k-th phase ?
k-th phase = when I have k different
Kinds of coupons.
n-1
X=X0+X1+…+Xn-1
n
n
1
= 
=n
n-k
k
k=0
k=1
=  (n ln n)
What is the expected number of
boxes that I buy in k-th phase ?
k-th phase = when I have k different
Kinds of coupons.
n-1
X=X0+X1+…+Xn-1
n
n
1
= 
=n
n-k
k
k=0
k=1
E[X]=E[X0]+…+E[Xn-1]
=  (n ln n)
Harmonic numers
n
ln n 
1
 k
k=1
 1+ln n
Randomized algorithm for “median”
SELECT k-th element
for random x
1)
<x
L
=x
>x
R
2) recurse on the appropriate part
Randomized algorithm for “median”
Las Vegas algorithm
(never makes error, randomness
only influences running time)
The identity testing algorithm was
Monte Carlo algorithm with 1 sided
error.
Markov inequality
For non-negative random variable X:
P(X > a.E[X]) < 1/a
P(X  a.E[X])  1/a
Variance
For a random variable X:
V[ X ] = E[ (X-E[X])2 ]
What is the variance of
X=the number on a (6-sided) dice ?
Variance
For a random variable X:
V[ X ] = E[ (X-E[X])2 ]
Y = (X-E[X])2
P( Y > a.E[Y] ) < 1/a
P( (X-E[X])2 > a.V[X] ) < 1/a
P( (X-E[X])2 > b2.E[X]2 ) < V[X]/(b2 E[X]2)
P( |X-E[X]| > b.E[X] ) <
V[X]
E[X]2
*
1
b2
Chebychev’s inequality
P( |X-E[X]| > b.E[X] ) <
V[X]
E[X]2
*
P( (1-b)*E[X]  X  (1+b)*E[X] )
>1 -
V[X]
E[X]2
*
1
b2
1
b2