CS 573: Advanced AI Probabilities & Monte-Hall Problem.

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CS 573: Advanced AI Probabilities & Monte-Hall Problem

Transcript of CS 573: Advanced AI Probabilities & Monte-Hall Problem.

CS 573: Advanced AI

Probabilities & Monte-Hall Problem

CS 573 : Advanced AI Probabilities & Monte-Hall Problem 2

In a perfect simple world …

Actions have deterministic effects

States are completely observable

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But, this world is not perfect …

For example, a robot in the field

Actions -> uncertain effects Observations -> noises & errors Unpredictable events

• Rocks fall from sky• Robots gain intelligence and rebel

Uncertainties => Probabilities

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Notation

P(VAR = value)

The probability variable VAR takes on the given value

P(STUDENTS=20) = 0.93

P(Propositional_sentence)P(Propositional_sentence=true)

The probability the given propositional sentence is true

P(HAPPY) = 0.40 P(HAPPY AND HEALTHY) = 0.39

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Conditional probabilities

Rainy

0.36No Stars

0.44Rainy & No Stars

0.30

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Conditional probabilities

Rainy

0.36No Stars

0.44Rainy & No Stars

0.30

We know its rainy today, what’s the probability that last night there is no stars in sky?

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Conditional probabilities

Rainy

0.36No Stars

0.44Rainy & No Stars

0.30

P(No Stars | Rainy)

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Conditional probabilities

Rainy

0.36No Stars

0.44Rainy & No Stars

0.30

P(No Stars | Rainy) = P(Rainy & No Starts) / P(Rainy)

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Conditional probabilities

Rainy

0.36No Stars

0.44Rainy & No Stars

0.30

P(No Stars | Rainy) = P(Rainy & No Starts) / P(Rainy)

= 0.30 / 0.36

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Conditional probabilities

P(A | B) = P(A & B) / P(B)

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Axioms of probabilities

1. 0 <=P(A) <= 1

2. P(true) = 1; P(false) = 0

3. P(A or B) = P(A) + P(B) – P(A & B)

AB

A & B

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Example derivation

P(A) + P(NOT A) = 1

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Example derivation

P(A) + P(NOT A) = 1

P(A or B) = P(A) + P(B) – P(A & B)

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Example derivation

P(A) + P(NOT A) = 1

P(A or B) = P(A) + P(B) – P(A & B)

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Example derivation

P(A) + P(NOT A) = 1

P(A or B) = P(A) + P(B) – P(A & B)

P(A or NOT A) = P(A) + P(NOT A) – P(A & NOT A)

Let B be NOT A

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Example derivation

P(A) + P(NOT A) = 1

P(A or B) = P(A) + P(B) – P(A & B)

P(A or NOT A) = P(A) + P(NOT A) – P(A & NOT A)

Let B be NOT A

P(true) P(false)

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Example derivation

P(A) + P(NOT A) = 1

P(A or B) = P(A) + P(B) – P(A & B)

P(A or NOT A) = P(A) + P(NOT A) – P(A & NOT A)

Let B be NOT A

P(true) P(false)

1 = P(A) + P(NOT A) – 0

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Joint probability distribution

Happy Healthy P

true true 0.39

true false 0.01

false true 0.06

false false 0.54

Happy

Healthy

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Joint probability distribution

Happy Healthy P

true true 0.39

true false 0.01

false true 0.06

false false 0.54

Happy

Healthy

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Joint probability distribution

Happy Healthy P

true true 0.39

true false 0.01

false true 0.06

false false 0.54

Happy

Healthy

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Joint probability distribution

Happy Healthy P

true true 0.39

true false 0.01

false true 0.06

false false 0.54

Happy

Healthy

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Joint probability distribution

Happy Healthy P

true true 0.39

true false 0.01

false true 0.06

false false 0.54

Happy

Healthy

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Joint probability distribution

Happy Healthy P

true true 0.39

true false 0.01

false true 0.06

false false 0.54

P(NOT Healthy)

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Joint probability distribution

Happy Healthy P

true true 0.39

true false 0.01

false true 0.06

false false 0.54

P(NOT Healthy) = P(Healthy=false)

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Joint probability distribution

Happy Healthy P

true true 0.39

true false 0.01

false true 0.06

false false 0.54

P(NOT Healthy) = P(Healthy=false)

=P(Happy & NOT Healthy) + P(NOT Happy & NOT Healthy)

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Joint probability distribution

Happy Healthy P

true true 0.39

true false 0.01

false true 0.06

false false 0.54

P(NOT Healthy) = P(Healthy=false)

=P(Happy & NOT Healthy) + P(NOT Happy & NOT Healthy)

=0.01 + 0.54 = 0.55

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Bayes rule

P(A|B) = P(A) * P(B|A) / P(B)

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Bayes rule

P(A|B) = P(A) * P(B|A) / P(B)

P(A|B)

= P(A & B) / P(B) -> conditional probabilities

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Bayes rule

P(A|B) = P(A) * P(B|A) / P(B)

P(A|B)

= P(A & B) / P(B) -> conditional probabilities

P(A|B) * P(B) = P(A & B) -> *P(B)

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Bayes rule

P(A|B) = P(A) * P(B|A) / P(B)

P(B|A)

= P(A & B) / P(A) -> conditional probabilities

P(B|A) * P(A) = P(A & B) -> * P(A)

P(A|B)

= P(A & B) / P(B) -> conditional probabilities

P(A|B) * P(B) = P(A & B) -> *P(B)

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Bayes rule

P(A|B) = P(A) * P(B|A) / P(B)

P(B|A)

= P(A & B) / P(A) -> conditional probabilities

P(B|A) * P(A) = P(A & B) -> * P(A)

P(A|B)

= P(A & B) / P(B) -> conditional probabilities

P(A|B) * P(B) = P(A & B) -> *P(B)

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Bayes rule

P(A|B) = P(A) * P(B|A) / P(B)

P(B|A)

= P(A & B) / P(A) -> conditional probabilities

P(B|A) * P(A) = P(A & B) -> * P(A)

P(A|B) * P(B) = P(B|A) * P(A)

P(A|B)

= P(A & B) / P(B) -> conditional probabilities

P(A|B) * P(B) = P(A & B) -> *P(B)

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Bayes rule

P(A|B) = P(A) * P(B|A) / P(B)

P(disease|Symptom)

Disease

• Open your abdomen -> painful but accurate

• based on symptom -> less pain but less accurate

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Bayes rule

P(A|B) = P(A) * P(B|A) / P(B)

P(disease|Symptom)

Disease

• Open your abdomen -> painful but accurate

• based on symptom -> less pain but less accurate

= P(disease) * P(symptom|disease) / P(symptom)

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Bayes rule

P(A|B) = P(A) * P(B|A) / P(B)

P(disease|Symptom)

Disease

• Open your abdomen -> painful but accurate

• based on symptom -> less pain but less accurate

= P(disease) * P(symptom|disease) / P(symptom)

Population sampling

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Bayes rule

P(A|B) = P(A) * P(B|A) / P(B)

P(disease|Symptom)

Disease

• Open your abdomen -> painful but accurate

• based on symptom -> less pain but less accurate

= P(disease) * P(symptom|disease) / P(symptom)

Population sampling

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Bayes rule

P(A|B) = P(A) * P(B|A) / P(B)

P(disease|Symptom)

Disease

• Open your abdomen -> painful but accurate

• based on symptom -> less pain but less accurate

= P(disease) * P(symptom|disease) / P(symptom)

Patients sampling

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The “Monte-Hall” Problem

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The “Monte-Hall” Problem

Your selection: DOOR 2

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The “Monte-Hall” Problem

Your selection: DOOR 2

P(You-get-prize) = 1/3

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The “Monte-Hall” Problem

Your selection: DOOR 2

P(You-get-prize) = 1/3

Host open DOOR 1

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The “Monte-Hall” Problem

Your selection: DOOR 2

P(You-get-prize) = 1/3

Host open DOOR 1

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The “Monte-Hall” Problem

Before DOOR 1 open:

P(prize=1) = P(prize=2) = P(prize=3) = 1/3

After DOOR 1 open:

P(prize=1)=0, P(prize=2) = P(prize=3) = 1/2

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The “Monte-Hall” Problem

There are three objects

1. Empty-A

2. Empty-B

3. Prize

Three Doors

1. DOOR-1

2. DOOR-2

3. DOOR-3

1. You choose Empty-A

2. You choose Empty-B

3. You choose Prize

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The “Monte-Hall” Problem

There are three objects

1. Empty-A

2. Empty-B

3. Prize

Three Doors

1. DOOR-1

2. DOOR-2

3. DOOR-3

1. You choose Empty-A host reveal Empty-B switch->win

2. You choose Empty-B

3. You choose Prize

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The “Monte-Hall” Problem

There are three objects

1. Empty-A

2. Empty-B

3. Prize

Three Doors

1. DOOR-1

2. DOOR-2

3. DOOR-3

1. You choose Empty-A host reveal Empty-B switch->win

2. You choose Empty-B host reveal Empty-A switch->win

3. You choose Prize

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The “Monte-Hall” Problem

There are three objects

1. Empty-A

2. Empty-B

3. Prize

Three Doors

1. DOOR-1

2. DOOR-2

3. DOOR-3

1. You choose Empty-A host reveal Empty-B switch->win

2. You choose Empty-B host reveal Empty-A switch->win

3. You choose Prize host reveal Empty-A or B switch->lose

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The “Monte-Hall” Problem

Bayes rule explanation

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The “Monte-Hall” Problem

Notation:

Oi : Open DOOR i

Xi : Prize is behind DOOR i

Choose Door 3

P(X2) = 1/3

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The “Monte-Hall” Problem

Notation:

Oi : Open DOOR i

Xi : Prize is behind DOOR i

Choose Door 3

P(X2) = 1/3

P(O1)=P(O2)=1/2

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The “Monte-Hall” Problem

Notation:

Oi : Open DOOR i

Xi : Prize is behind DOOR i

Choose Door 3

P(X2) = 1/3

P(O1)=P(O2)=1/2

• X1 -> O2

• X2 -> O1

• X3 -> O1 or O2

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The “Monte-Hall” Problem

Notation:

Oi : Open DOOR i

Xi : Prize is behind DOOR i

Choose Door 3

P(X2) = 1/3

P(O1)=P(O2)=1/2

Host open DOOR 1, P(X2 | O1) = ?

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The “Monte-Hall” Problem

Notation:

Oi : Open DOOR i

Xi : Prize is behind DOOR i

Choose Door 3

P(X2) = 1/3

P(O1)=P(O2)=1/2

Host open DOOR 1, P(X2 | O1) = ?

P(X2|O1)

= P(O1|X2)*P(X2)/P(O1)

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The “Monte-Hall” Problem

Notation:

Oi : Open DOOR i

Xi : Prize is behind DOOR i

Choose Door 3

P(X2) = 1/3

P(O1)=P(O2)=1/2

Host open DOOR 1, P(X2 | O1) = ?

P(X2|O1)

= P(O1|X2)*P(X2)/P(O1)

= 1 *

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The “Monte-Hall” Problem

Notation:

Oi : Open DOOR i

Xi : Prize is behind DOOR i

Choose Door 3

P(X2) = 1/3

P(O1)=P(O2)=1/2

Host open DOOR 1, P(X2 | O1) = ?

P(X2|O1)

= P(O1|X2)*P(X2)/P(O1)

= 1 * (1/3) / (1/2)

= 2/3