Probability - Stanford University · Probability Will Monroe Summer 2017 with materials by Mehran...

Probability

Will MonroeSummer 2017

with materials byMehran Sahamiand Chris Piech

June 30, 2017

Today we will make history

Logistics: Office hours

Review: Permutations

The number of ways of orderingn distinguishable objects.

n ! =1⋅2⋅3⋅...⋅n=∏i=1

n

i

Review: Combinations

The number of unique subsets of size k from a larger set of size n. (objects are distinguishable, unordered)

(nk ) =n !

k !(n−k)!

choose k

n

Review: Bucketing

The number of ways of assigning n distinguishable objects to a fixedset of k buckets or labels.

kn

k buckets

n objects

Divider method

The number of ways of assigning n indistinguishable objects to a fixedset of k buckets or labels.

(n+(k−1)

n )

k buckets

n objects

(k - 1 dividers)

A grid of ways of counting

kn(nk )n !

n !k1! k2!…km !

(n+(k−1)

n )1 1

Creativity!– Split into cases– Use inclusion/exclusion– Reframe the problem

Ordering Subsets Bucketing

Alldistinct

Someindistinct

Allindistinct

Sample space

S = the set of all possible outcomes

Coin flip

Two coin flips

Roll of 6-sided die

# emails in day

Netflix hours in day

S = {Heads, Tails}

S = {(H, H), (H, T), (T, H), (T, T)}

S = {1, 2, 3, 4, 5, 6}

S = (non-neg. integers)ℕ

S = [0, 24] (interval, inclusive)

Events

E = some subset of S (E ⊆ S)

Coin flip is heads

≥ 1 head on 2 flips

Roll of die ≤ 3

# emails in day ≤ 20

“Wasted day”

E = {Heads}

E = {(H, H), (H, T), (T, H)}

E = {1, 2, 3}

E = {x | x ∈ , x ≤ 20}ℕ

E = [5, 24]

(Ross uses to mean )⊂ ⊆

Set operations

E ∪ F = {1, 2, 3}

E = {1, 2}

F ={2, 3}

S = {1, 2, 3, 4, 5, 6}

Union: outcomes that are in E or F

Set operations

EF = E ∩ F = {2}

E = {1, 2}

F ={2, 3}

S = {1, 2, 3, 4, 5, 6}

Intersection: outcomes that are in E and F

Set operations

Ec = {3, 4, 5, 6}

E = {1, 2}

F ={2, 3}

S = {1, 2, 3, 4, 5, 6}

Complement: outcomes that are not in E

A profound truth

https://medium.com/@timanglade/how-hbos-silicon-valley-built-not-hotdog-with-mobile-tensorflow-keras-react-native-ef03260747f3

A profound truth

Everything in the world is either

a hot dog

or

not a hot dog.

(E ∪ Ec = S)


Quiz question

https://b.socrative.com/login/student/

Room: CS109SUMMER17

Quiz questionWhich is the correct picture for EcFc (= Ec ∩ Fc)?

E F

S

E F

S

E F

S

E F

S

E F

S

A

B

C

D

E

Quiz question

E F

S

C

Which is the correct picture for EcFc (= Ec ∩ Fc)?

Quiz questionWhich is the correct picture for Ec ∪ Fc?

E F

S

E F

S

E F

S

E F

S

E F

S

A

B

C

D

E

Quiz questionWhich is the correct picture for Ec ∪ Fc?

E F

S

D


De Morgan's Laws“distributive laws” for set complement

E F

S

(E ∪ F)c = Ec ∩ Fc (E ∩ F)c = Ec ∪ Fc

E F

S

(⋃i Ei)c=⋂

i(Ei

c ) (⋂i Ei )c=⋃

i(Ei

c )

What is a probability?

A number between 0 and 1to which we ascribe meaning


Sources of probability

Experiment

Expert opinion Analytical solution

Datasets

https://b.socrative.com/login/student/

Meaning of probability

A quantification of ignorance

image: Frank Derks

P(E)=limn→∞

# (E)

n

What is a probability?

Axioms of probability

0≤P(E)≤1

P(S)=1

P(E∪F)=P(E)+P(F)If E∩F=∅ , then

(1)

(2)

(3)

(Sum rule, but with probabilities!)

Corollaries

P(Ec)=1−P(E)

P(E∪F)=P(E)+P(F)−P(EF)

If E⊆F , then P(E)≤P(F)

(Principle of inclusion/exclusion, but with probabilities!)

Principle of Inclusion/Exclusion

The total number of elements in two sets is the sum of the number of elements of each set,minus the number of elements in both sets.

|A∪B|=|A|+|B|−|A∩B|

3 4

3+4-1 = 6

Inclusion/exclusion withmore than two sets

|E∪F∪G|=

E

F G


|E∪F∪G|=|E|

E

F G

1

1

11


|E∪F∪G|=|E|+|F|

E

F G

1

2

12

11


|E∪F∪G|=|E|+|F|+|G|

E

F G

1

3

22

12 1


|E∪F∪G|=|E|+|F|+|G|

E

F G

1

2

21

12 1

−|EF|


|E∪F∪G|=|E|+|F|+|G|

E

F G

1

1

11

12 1

−|EF|−|EG|

https://www.flickr.com/photos/eurosporttuning/16442169022


|E∪F∪G|=|E|+|F|+|G|

E

F G

1

0

11

11 1

−|EF|−|EG|−|FG|


|E∪F∪G|=|E|+|F|+|G|

E

F G

1

1

11

11 1

−|EF|−|EG|−|FG|

+|EFG|


|⋃i=1

n

Ei|=∑r=1

n

(−1)(r+1) ∑

i1<⋯< ir

|⋂j=1

r

Ei j|

size ofthe union

sum oversubsetsizes

add orsubtract

(based on size)

size ofintersections

sum overall subsetsof that size


P(⋃i=1

n

Ei)=∑r=1

n

(−1)(r+1) ∑

i1<⋯< ir

P(⋂j=1

r

Ei j)

prob. ofOR

sum oversubsetsizes

add orsubtract

(based on size)

prob. ofAND

sum overall subsetsof that size


P(E∪F∪G)=P(E)+P(F)+P(G)

E

F G

1

1

11

11 1

−P(EF)−P(EG)−P(FG)

+P(EFG)

Break time!

Equally likely outcomes

Coin flip

Two coin flips

Roll of 6-sided die

S = {Heads, Tails}

S = {(H, H), (H, T), (T, H), (T, T)}

S = {1, 2, 3, 4, 5, 6}

P(Each outcome)=1|S|

P(E)=|E|

|S|(counting!)

Rolling two dice

P(D1+D2=7)=?D1 D2

How do I get started?

For word problems involving probability,start by defining events!

Rolling two dice

P(D1+D2=7)=?D1 D2

E: {all outcomes such that the sum of the two dice is 7}

Rolling two dice

P(D1+D2=7)=6

36=

16D1 D2

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

S = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

S = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),

S = {(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

S = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),

S = {(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

|S|=36

|E|=6

Rolling two (indistinguishable) dice?

P(D1+D2=7)=3

21=

17

?D1 D2

S = {(1, 1), (1, 2), (1, 3), (1, 4), (1, 5), (1, 6),

S = {(2, 1), (2, 2), (2, 3), (2, 4), (2, 5), (2, 6),

S = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6),

S = {(4, 1), (4, 2), (4, 3), (4, 4), (4, 5), (4, 6),

S = {(5, 1), (5, 2), (5, 3), (5, 4), (5, 5), (5, 6),

S = {(6, 1), (6, 2), (6, 3), (6, 4), (6, 5), (6, 6)}

|S|=21

|E|=3

[check your assumptions]

Getting rid of ORs

Finding the probability of an OR of events can be nasty. Try using De Morgan's laws to turn it into an AND!

P(A∪B∪⋯∪Z )=1−P(Ac Bc⋯Zc

)

E F

S

Running into a friend

image: torbakhopper / scott richard

21,000 people at StanfordYou're friends with 250

Go to an event with 70 other(equally-likely) Stanforders

What's P(at least one friend among the 70)?

S: {all subsets of size 70 of 21,000 students}

E: {subsets of size 70 with at least one friend}

|S|=(21,00070 )

|E|=???


image: torbakhopper / scott richard

21,000 people at StanfordYou're friends with 250

Go to an event with 70 other(equally-likely) Stanforders

What's P(at least one friend among the 70)?

S: {all subsets of size 70 of 21,000 students}

E: {subsets of size 70 with at least one friend}Ec: {subsets of size 70 with no friends}|S|=(21,000

70 )

|EC|=(21,000−250

70 ) P(E)=1−P(EC)=1−

(20,75070 )

(21,00070 )

≈0.5512


https://cs109.stanford.edu/demos/serendipity.html

serendipity [ˌse.ɹənˈdɪ.pɨ.ɾi] n the faculty or phenomenon of finding valuable or agreeable things not sought for

Merriam-Webster

Getting rid of ORs

Finding the probability of an OR of events can be nasty. Try using De Morgan's laws to turn it into an AND!

P(A∪B∪⋯∪Z )=1−P(Ac Bc⋯Zc

)

E F

S

Defective chips

n

k

|S|=(nk)

|E|=1⋅(n−1k−1)

S: {all subsets of size k from n chips}

E: {all outcomes such that the defective chip is chosen}

Defective chips

n

k

|S|=(nk)

|E|=1⋅(n−1k−1)

P(E)=(n−1k−1)

(nk)=

(n−1)!(k−1)!(n−k )!

n !k !(n−k)!

=(n−1)! k !n!(k−1)!

=(n−1)!⋅k⋅(k−1)!n⋅(n−1)!(k−1)!

=kn

Poker straight

(Standard deck:)52 cards = 13 ranks × 4 suits

“Straight”: 5 consecutive ranks(suits can be different)

S: {all possible hands}

E: {all hands that are straights}

|S|=(525 )

|E|=10⋅45

startingrank

suit ofeach card

image: Guts Gaming (www.guts.com/en)

Poker straight

(Standard deck:)52 cards = 13 ranks × 4 suits

“Straight”: 5 consecutive ranks(suits can be different)

S: {all possible hands}

E: {all hands that are straights}

|S|=(525 )

|E|=10⋅45

P(E)=10⋅45

(525 )

≈0.00394

image: Guts Gaming (www.guts.com/en)

Birthdays

n people in a room.What is the probability none share the same birthday?

S: {all ways of giving all n people each a birthday}

E: {all ways of giving all n people different birthdays}

|S|=365n

|E|=365⋅364⋅…⋅(365−n+1)=365 !

(365−n)!=(365

n )⋅n !

P(E)=(365

n )⋅n !

365n

n = 23: P(E) < 1/2n = 75: P(E) < 1/3,000n = 100: P(E) < 1/3,000,000n = 150: P(E) < 1/3,000,000,000,000,000

Flipping cards● Shuffle deck.

● Reveal cards from the top until we get an Ace. Put Ace aside.

● What is P(next card is the Ace of Spades)?

● P(next card is the 2 of Clubs)?

https://bit.ly/1a2ki4G → https://b.socrative.com/login/student/Room: CS109SUMMER17

A) P(Ace of Spades) > P(2 of Clubs)

B) P(Ace of Spades) = P(2 of Clubs)

C) P(Ace of Spades) < P(2 of Clubs)





S: {all orderings of the deck}

E: {all orderings for which Ace of Spadescomes immediately after first Ace}

|S|=52!

|E|=51!⋅1P(E)=

51!52!

=152





S: {all orderings of the deck}

E: {all orderings for which 2 of Clubscomes immediately after first Ace}

|S|=52!

|E|=51!⋅1P(E)=

51!52!

=152





B) P(Ace of Spades) = P(2 of Clubs)

(We just made history)

Probability - Stanford University · Probability Will Monroe Summer 2017 with materials by Mehran...

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