Expected Value Reprise

CP Canoe Club Duck Derby



Prizes




Maximum of 2 000 tickets sold
$5/ticket
1) 12 VIP tickets to Cirque du Soleil ($2,000)
2) 32gb BlackBerry Playbook ($600)
3) $250
Should Mr. Lieff buy a ticket?
Expected Value Calculation

For 2 000 tickets



E(X) =1/2000 (2000) + 1/2000(600) + 1/2000(250)
= 1.425
Suppose Mr. Lieff receives a hot tip that only
1 500 tickets will be sold.


E(X) =1/1500 (2000) + 1/1500(600)+1/1500(250)
= 1.9
Probability Distributions and
Expected Value
Chapter 5.1 – Probability Distributions and
Predictions
MSIP / Home Learning: p. 277 #1-5, 9, 12, 13
Probability Distributions of a Discrete
Random Variable




a discrete random variable X is a variable
that can take on only a finite set of values
for example, rolling a die can only produce
numbers in the set {1,2,3,4,5,6}
rolling 2 dice can only produce numbers in
the set {2,3,4,5,6,7,8,9,10,11,12}
choosing a card from a standard deck
(ignoring suit) can produce only the cards in
the set {A,2,3,4,5,6,7,8,9,10,J,Q,K}
Probability Distribution Rolling A Die


the probability
distribution of a random
variable x, is a function
which provides the
probability of each
possible value of x
may be represented as a
table of values or a graph
ex, rolling a die:
outcome
1
=
2
3
4
5
6
probability <new>
1
2
3
4
5
6
1/6
1/6
1/6
1/6
1/6
1/6
Rolling A Die
Histogram
1
Count

0
1
2
3
4
outcome
5
6
7
Probability Distribution for 2 Dice
RollingDice
Two Dice
7
6
Count
5
4
3
2
1
2
sum
Histogram
4
6
8
sum
10
12
1
=
2
3
4
5
6
7
8
9
10
11
2
3
4
5
6
7
8
9
10
11
12
probability <new>
1/36
2/36
3/36
4/36
5/36
6/36
5/36
4/36
3/36
2/36
1/36
What would a probability distribution
graph for three dice look like?




We will try it! Using three dice, figure out how many
outcomes there are
Then find out how many possible ways there are to
create each of the possible outcomes
Fill in a table like the one below
Now you can make the graph
Outcome 3
# ways
1
4
5
6
7
8
9
…
Probability Distribution for 3 Dice
Outcome 3
4
5
6
7
8
9
# cases
3
6
10
15
21
28
1
10
So what does an experimental distribution
look like?


A simulated dice
throw was done a
million times using a
computer program
and generated the
following data
What is the most
common outcome?
Does this make
sense?
Line Scatter Plot
Rolling 3 Dice
140000
120000
100000
Freq

80000
60000
40000
20000
0
2
4
6
8
10
12
roll
14
16
18
20
Back to 2 Dice



What is the expected
value of throwing 2
dice?
How could this be
calculated?
So the expected
value of a discrete
variable X is the sum
of the values of X
multiplied by their
probabilities
E (sum of 2 dice)
1
2
3
1
 2   3   4   ...  12 
36
36
36
36
252

7
36
n
 E ( X )   xi P( X  xi )
i 1
Example 1a: tossing 3 coins
X
P(X)




0 heads 1 head 2 heads 3 heads
⅛
⅜
⅜
⅛
What is the likelihood of at least 2 heads?
It must be the total probability of tossing 2 heads
and tossing 3 heads
P(X = 2) + P(X = 3) = ⅜ + ⅛ = ½
so the probability is 0.5
Example 1b: tossing 3 coins
X
P(X)





0 heads 1 head 2 heads 3 heads
⅛
⅜
⅜
⅛
What is the expected number of heads?
It must be the sums of the values of x multiplied
by the probabilities of x
0P(X = 0) + 1P(X = 1) + 2P(X = 2) + 3P(X = 3)
= 0(⅛) + 1(⅜) + 2(⅜) + 3(⅛) = 1½
So the expected number of heads is 1.5
Combinations

Recall that C(n, r) is the number of ways r
objects can be chosen from n when order
doesn’t matter.
Example 2a: Selecting a Committee of
three people from a group of 4 men and 3
women






What is the probability of having at least one
woman on the team?
There are C(7,3) or 35 possible teams
C(4,3) = 4 have no women
C(4,2) x C(3,1) = 6 x 3 = 18 have one woman
C(4,1) x C(3,2) = 4 x 3 = 12 have 2 women
C(3,3) = 1 has 3 women
Example 2a cont’d: selecting a committee




X
0 women
1 woman
2 women
3 women
P(X)
4/35
18/35
12/35
1/35
What is the likelihood of at least one woman?
It must be the total probability of all the cases
with at least one woman
P(X = 1) + P(X = 2) + P(X = 3)
= 18/35 + 12/35 + 1/35 = 31/35
Example 2b: selecting a committee




X
0 women
1 woman
2 women
3 women
P(X)
4/35
18/35
12/35
1/35
What is the expected number of women?
0P(X = 0) + 1P(X = 1) + 2P(X = 2) + 3P(X = 3)
= 0(4/35) + 1(18/35) + 2(12/35) + 3(1/35)
= 1.3 (approximately)
MSIP / Home Learning

p. 277 #1-5, 9, 12, 13
Pascal’s Triangle
Chapter 5.2 – Probability Distributions and Predictions
Due now: p. 277 #1-5, 9, 12, 13
MSIP/Home Learning: p. 289 #1, 2aceg, 6-8, 11-12
How many routes are there to the top
right-hand corner?




you need to move up 4
spaces and over 5
spaces
This is the same as
rearranging the letters
NNNNEEEEE
This can be calculated
by C(9,4) or C(9,5)
= 126 ways
Permutation AND Combination?

As a permutation:


However, 4 are identical N moves and 5 are
identical E moves




There are 9 moves  9!
Divide by 4! due to identical arrangements of Ns
Divide by 5! due to identical arrangements of Es
 9! ÷ 4!5! = 126
As a combination:



There are 9 moves required _ _ _ _ _ _ _ _ _
Choose 4 to be Ns, the rest are Es  C(9, 4) = 126
Choose 5 to be Es, the rest are Ns  C(9, 4) = 126
Pascal’s Triangle
1
1
1
1
the outer values
are always 1
 the inner values
are determined by
adding two values
diagonally above

2 1
1 3 3 1
1 4 6 4 1
 1 5 10 10 5 1
Pascal’s Triangle
sum of each row is a power of 2
 1
 1 = 20
 1 1
 2 = 21
 1 2 1
 4 = 22
 1 3 3 1
 8 = 23
 1 4 6 4 1
 16 = 24
 1 5 10 10 5 1
 32 = 25
 1 6 15 20 15 6 1
 64 = 26

Pascal’s Triangle

1


1 1

 1 2 1
 1 3 3 1
 1 4 6 4 1
 1 5 10 10 5 1 
 1 6 15 20 15 6 1

Uses?
binomial theorem
combinations!
 e.g. choose 2 items from 5
 go to the 5th row, the 2nd
number = 10 (always start
counting at 0)
modeling the electrons in
each shell of an atom (google
‘Pascal’s Triangle electron’)
Pascal’s Triangle – Cool Stuff

1
1 1
 1 2 1
 1 3 3 1
 1 4 6 4 1
 1 5 10 10 5 1
 1 6 15 20 15 6 1





each diagonal is summed
up in the next value below
and to the left
called the “hockey stick”
property
there may even be music
hidden in it
http://www.geocities.com/Vi
enna/9349/pascal.mid
Music: Three Parts in D Alternating Octatonic





The piece of music has three parts:
1) A treble part where the center numbers 1, 2 and 6 are the tonic D
of the D alternating octatonic scale. The other vertical lines of
numbers represent one scale step; a descending scale step to the
left and an ascending scale step to the right. All ways of reaching all
the numbers in the triangle are mapped to consecutive sixteenths,
starting from the top and going down south west first to every new
row, inwards. This part is repeated.
2) The main treble part. It has the structure A1 B A2 B. It starts with
the left side of the symmetric triangle, including the center (A1), the
numbers are the number of scale steps to ascend (south east
arrows) or descend (south west arrows). After an intermission (B)
where this part doubles the first part, the right side is initiated (A2)
and it is the mirror image of A1.
3) A walking bass part (quarter notes), same as the first treble part
but consecutive equal tones are tied. This part is repeated.
Playing time: 4' 47".
Pascal’s Triangle – Cool Stuff



numbers
divisible by 5
similar
patterns exist
for other
numbers
http://www.shodor.
org/interactivate/ac
tivities/pascal1/
Pascal’s Triangle can also be seen in terms
of combinations
n=0
 n = 1
 n = 2
 n = 3
 n = 4
 n = 5
 n = 6
0
 
0
 1   1
   
 0   1

 2
 
0
 2
 
1
 2
 
 2
 3   3  3   3
       
 0  1  2   3
 4  4  4  4  4
         
 0 1  2  3  4
5
 
0
6
 
0
5  5 5  5 5
         
1  2  3  4   5
6  6 6  6 6
6
         
 
5
1  2  3  4  
6
Pascal’s Triangle - Summary




symmetrical down the middle
outside number is always 1
second diagonal values match the row
numbers
sum of each row is a power of 2



sum of nth row is 2n
Begin count at 0
number inside a row is the sum of the two
numbers above it
And one more thing…



remember that for the inner numbers in the
triangle, any number is the sum of the two
numbers above it
for example 4 + 6 = 10
this suggests the following:
 4  4  5
       
1  2  2

which is an example of Pascal’s Identity
 n   n   n  1
   
  

 r   r  1  r  1 
For Example…
 6  6 7 
       
 3 4  4 
8   8   9 
       
5   6   6 
How can this help us solve our original
problem?

so by overlaying
Pascal’s Triangle
over the grid we
can see that there
are 126 ways to
move from one
corner to another
1
5
15
35
70 126
1
4
10 20
35
56
1
3
6
10
15
21
1
2
3
4
5
6
1
1
1
1
1
How many routes pass through the green
square?



to get to the green
square, there are
C(4,2) ways (6 ways)
to get to the end from
the green square
there are C(5,3) ways
(10 ways)
in total there are 60
ways
How many routes do not pass through the
green square?



there are 60 ways
that pass through the
green square
there are C(9,5) or
126 ways in total
then there must be
126 – 60 = 66 paths
that do not pass
through the green
square
MSIP / Home Learning



Read the examples on pp. 281-287
The example starting on the bottom of
page 287 is important
page 289 #1, 2aceg, 6-8, 11-12