Mathematics for Computer Science
MIT 6.042J/18.062J
Great
Expectations
Copyright © Albert R. Meyer, 2008. All rights reserved.
May 12, 2008
lec 14M.1
Carnival Dice
choose a number from 1 to 6,
then roll 3 fair dice:
win $1 for each match
lose $1 if no match
Copyright © Albert R. Meyer, 2008. All rights reserved.
May 12, 2008
lec 14M.6
Carnival Dice
Example:
choose 5, then
roll 5,4,6: win $1
roll 5,4,5: win $2
roll 5,5,5: win $3
Copyright © Albert R. Meyer, 2008. All rights reserved.
May 12, 2008
lec 14M.7
Carnival Dice
Is this a
fair game?
Copyright © Albert R. Meyer, 2008. All rights reserved.
May 12, 2008
lec 14M.8
Carnival Dice
# matches probability
$ won
0
125/216
-1
1
75/216
1
2
15/216
2
3
1/216
3
Copyright © Albert R. Meyer, 2008. All rights reserved.
May 12, 2008
lec 14M.10
Carnival Dice
so every 216 games, expect
0 matches about 125 times
1 match about 75 times
2 matches about 15 times
3 matches about once
Copyright © Albert R. Meyer, 2008. All rights reserved.
May 12, 2008
lec 14M.11
Carnival Dice
So on average expect to win:
NOT fair!
125  ( 1)  75  1  15  2  1  3
216
17

 8 cents
216
Copyright © Albert R. Meyer, 2008. All rights reserved.
May 12, 2008
lec 14M.13
Carnival Dice
You can “expect” to lose 8 cents
per play. But you never actually
lose 8 cents on any single play,
this is just your average loss.
Copyright © Albert R. Meyer, 2008. All rights reserved.
May 12, 2008
lec 14M.14
Expected Value
The expected value of
random variable R is
the average value of R
--with values weighted
by their probabilities
Copyright © Albert R. Meyer, 2008. All rights reserved.
May 12, 2008
lec 14M.15
Expected Value
The expected value of
random variable R is
E[R]::= ∑vPr{R = v}
so for D ::= $win in Carnival
17
E[D] = 
216
Copyright © Albert R. Meyer, 2008. All rights reserved.
May 12, 2008
lec 14M.16
Sums vs Integrals
We get away with sums
instead of integrals because
the sample space is assumed
countable:
={
,
0
Copyright © Albert R. Meyer, 2008. All rights reserved.
,
,
,
}
1
n
May 12, 2008
lec 14M.17
Expected Value
also called
mean value, mean, or
expectation
Copyright © Albert R. Meyer, 2008. All rights reserved.
May 12, 2008
lec 14M.18
Expectation
for indicator I
E[I] = 1 pr{I=1} +
0 pr{I=0}
E[I] = pr{I=1}
Copyright © Albert R. Meyer, 2008. All rights reserved.
May 12, 2008
lec 14M.19
Mean Time to Failure
•biased coin with
pr{Head} = p
•flip until first Head
•expected #flips?
Copyright © Albert R. Meyer, 2008. All rights reserved.
May 12, 2008
lec 14M.20
Mean Time to Failure
pr{1st H on flip 1} = p
pr{1st H on flip 2} = (1-p)p
2
pr{1st H on flip 3} = (1-p) p
Copyright © Albert R. Meyer, 2008. All rights reserved.
May 12, 2008
lec 14M.21
Mean Time to Failure
E[# flips until first head]
n-1
(1-p) p
= ∑n>0 n
n
= p ∑n 0 (n+1)(1-p)
1
2
= p (1/(1-(1-p)) 
p
Copyright © Albert R. Meyer, 2008. All rights reserved.
May 12, 2008
lec 14M.22
Mean Time to Failure
application: Space station Mir
say has 1/150,000 chance of
exploding in any given hour
After how may hours do
we expect it to explode?
150,000 hours 17 years
Copyright © Albert R. Meyer, 2008. All rights reserved.
May 12, 2008
lec 14M.23
Law of Total Expectation
conditional expectation:
E[R | A] ::  v  pr{R  v | A}
E[R]  E[R | A]  pr{A}
 E[R | A]  pr{A}
good for reasoning by cases
Copyright © Albert R. Meyer, 2008. All rights reserved.
May 12, 2008
lec 14M.24
Expected time to Gamble
e(n) ::= E[time with stake n]
e(0) = 0
e(T) = 0
E[time with stake n | win 1st bet]
= 1 + e(n+1)
so by Total Expectation
e(n) = p (1 + e(n+1))
+ (1-p) (1 + e(n-1))
Copyright © Albert R. Meyer, 2008. All rights reserved.
May 12, 2008
lec 14M.25
Expected time to Gamble
e(n) ::= E[time with stake n]
e(0) = 0
e(T) = 0
e(n) 1  p
e(n  1) 

 e(n  1)  1
p
p
we know how to solve this!
Copyright © Albert R. Meyer, 2008. All rights reserved.
May 12, 2008
lec 14M.26
Linearity of Expectation
A,B random variables, a,b
constants
aE[A] + bE[B]
= E[aA + bB]
Copyright © Albert R. Meyer, 2008. All rights reserved.
May 12, 2008
lec 14M.30
Expected returned hats
Say n people with hats leave
their hats at a hat-check
station. The hats get totally
scrambled randomly. How
many hats do we expect will
be returned to their owners?
Copyright © Albert R. Meyer, 2008. All rights reserved.
May 12, 2008
lec 14M.32
Linearity of Expectation
Let Ri be indicator for ith hat
being returned to its owner
Then E[# hats returned] =
E[∑i Ri] = ∑i E[Ri] =
∑i Pr{Ri=1} = ∑i 1/n =
n(1/n) = 1
Copyright © Albert R. Meyer, 2008. All rights reserved.
May 12, 2008
lec 14M.33
Expectation & Independence
for independent R,S
E[R S] = E[R] E[S]
Copyright © Albert R. Meyer, 2008. All rights reserved.
May 12, 2008
lec 14M.34
Team Problems
Problems
1 5
Copyright © Albert R. Meyer, 2008. All rights reserved.
May 12, 2008
lec 14M.36