Paradoxes in Probability How probability continues to amuse me! - - PowerPoint PPT Presentation

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Paradoxes in Probability How probability continues to amuse me! - - PowerPoint PPT Presentation

Apr 27, 2023 569 likes •1.94k views

Paradoxes in Probability How probability continues to amuse me! Let's play a game! Box A Box B Box A Box B Box A Box B x ice-creams Box A Box B Chosen but unobserved box x ice-creams Box A Box B Should you switch? 2x x ice-creams

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Paradoxes in Probability

How probability continues to amuse me!

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SLIDE 2

Let's play a game!

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SLIDE 3

Box A Box B

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Box A Box B

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Box A Box B

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Box A Box B x ice-creams

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Box A Box B x ice-creams

Chosen but unobserved box

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Should you switch?

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Box A Box B x ice-creams

2x

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SLIDE 10

Box A Box B x ice-creams

x/2

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Box A Box B x ice-creams

x/2

E[# ice-creams] = (0.5 * 2x) + (0.5 * x/2) = 1.25x

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SLIDE 12

Box A Box B x ice-creams

x/2

E[# ice-creams] = (0.5 * 2x) + (0.5 * x/2) = 1.25x

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SLIDE 13

Box A Box B x ice-creams

x/2

E[# ice-creams] = (0.5 * 2x) + (0.5 * x/2) = 1.25x

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SLIDE 14

Box A Box B x ice-creams

x/2

E[# ice-creams] = (0.5 * 2x) + (0.5 * x/2) = 1.25x

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SLIDE 15

Box A Box B x ice-creams

x/2

E[# ice-creams] = (0.5 * 2x) + (0.5 * x/2) = 1.25x

Hence you should switch!

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SLIDE 16

Here is the problem!

  • The first box was chosen randomly
  • The contents of the box were unobserved
  • Switching gives better expected value
  • Same argument applies to second box as well
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SLIDE 17

Here is the problem!

  • The first box was chosen randomly
  • The contents of the box were unobserved
  • Switching gives better expected value
  • Same argument applies to second box as well
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SLIDE 18

Here is the problem!

  • The first box was chosen randomly
  • The contents of the box were unobserved
  • Switching gives better expected value
  • Same argument applies to second box as well
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SLIDE 19

Here is the problem!

  • The first box was chosen randomly
  • The contents of the box were unobserved
  • Switching gives better expected value
  • Same argument applies to second box as well
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SLIDE 20

Here is the problem!

  • The first box was chosen randomly
  • The contents of the box were unobserved
  • Switching gives better expected value
  • Same argument applies to second box as well
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SLIDE 21

Keep switching forever!

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However...

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However...

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Let’s take a step back

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Let’s take a step back

Box A Box B

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Let’s take a step back

Box A Box B

Without extra information, the situation is symmetric

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SLIDE 27

Let’s take a step back

Box A Box B

Without extra information, the situation is symmetric Hence switching should not make any difference!

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SLIDE 28

What went wrong?

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Box A Box B

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A B

Box A Box B

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A B

A=y A=2y B=y B=2y

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SLIDE 32
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SLIDE 33

Recall: E[# ice-creams] = (0.5 * 2x) + (0.5 * x/2) = 1.25x

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Recall: E[# ice-creams in B] = (0.5 * 2x) + (0.5 * x/2) = 1.25x

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SLIDE 35

Recall: E[# ice-creams in B] = (0.5 * 2x) + (0.5 * x/2) = 1.25x

A general variable x = y or x = 2y

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SLIDE 36

Recall: E[# ice-creams in B] = (0.5 * 2x) + (0.5 * x/2) = 1.25x

A general variable x = y or x = 2y

If x = y, then B = x/2 is not possible If x = 2y, then B = 2x is not possible

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SLIDE 37

Recall: E[# ice-creams in B] = (0.5 * 2x) + (0.5 * x/2) = 1.25x

A general variable x = y or x = 2y

If x = y, then B = x/2 is not possible If x = 2y, then B = 2x is not possible

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SLIDE 38

We are considering possibilities that cannot exist!

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Solution?

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SLIDE 40
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E[# ice-creams in B] = (0.5 * 2y) + (0.5 * y) = 1.5y

If x = y, # ice-creams in B = 2y and if x = 2y, # ice-creams in B = y

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SLIDE 42

E[# ice-creams in B] = (0.5 * 2y) + (0.5 * y) = 1.5y

If x = y, # ice-creams in B = 2y and if x = 2y, # ice-creams in B = y

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SLIDE 43

E[# ice-creams in B] = (0.5 * 2y) + (0.5 * y) = 1.5y

If x = y, # ice-creams in B = 2y and if x = 2y, # ice-creams in B = y with probability 0.5

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SLIDE 44

E[# ice-creams in B] = (0.5 * 2y) + (0.5 * y) = 1.5y

If x = y, # ice-creams in B = 2y and if x = 2y, # ice-creams in B = y with probability 0.5

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SLIDE 45

E[# ice-creams in B] = (0.5 * 2y) + (0.5 * y) = 1.5y

If x = y, # ice-creams in B = 2y and if x = 2y, # ice-creams in B = y with probability 0.5 with probability 0.5

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SLIDE 46

E[# ice-creams in B] = (0.5 * 2y) + (0.5 * y) = 1.5y

If x = y, # ice-creams in B = 2y and if x = 2y, # ice-creams in B = y with probability 0.5 with probability 0.5

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SLIDE 47

E[# ice-creams in B] = (0.5 * 2y) + (0.5 * y) = 1.5y

If x = y, # ice-creams in B = 2y and if x = 2y, # ice-creams in B = y with probability 0.5 with probability 0.5

E[# ice-creams in A] = (0.5 * 2y) + (0.5 * y) = 1.5y

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SLIDE 48

E[# ice-creams in B] = (0.5 * 2y) + (0.5 * y) = 1.5y

If x = y, # ice-creams in B = 2y and if x = 2y, # ice-creams in B = y with probability 0.5 with probability 0.5

E[# ice-creams in A] = (0.5 * 2y) + (0.5 * y) = 1.5y

Nothing can be gained by switching!

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SLIDE 49

Paradox Resolved!

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SLIDE 50
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Friendship Paradox

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N people

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N people

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N people

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Each person shouts the name of someone else Pairs get “linked”

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Each person shouts the name of someone else Pairs get “linked”

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Each person shouts the name of someone else Pairs get “linked”

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Very Important Fact

Links are Symmetric

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Very Important Fact

Links are Symmetric

A and B get linked as long as one of them shouts other person’s name

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Problem Setup

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Problem Setup

  • The linking phase is over
  • A person is chosen at random (say person X)
  • Objective:

Find the probability that the friend chosen by X has more friends than X does?

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Problem Setup

  • The linking phase is over
  • A person is chosen at random (say person X)
  • Objective:

Find the probability that the friend chosen by X has more friends than X does?

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Problem Setup

  • The linking phase is over
  • A person is chosen at random (say person X)
  • A Basic Question:

Find the probability that the friend chosen by X (say person Y) has more friends than X does?

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A Simple Solution

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A Simple Solution

P(Y has more friends than X) = 0.5

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A Simple Solution

P(Y has more friends than X) = 0.5

Since links are symmetric!

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A Simple Solution

P(Y has more friends than X) = 0.5

Since links are symmetric! Otherwise, most of the people will have more/less friends than their friends have on average which violates the symmetric nature of links

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SLIDE 68

Okay! Let’s move on to more complex questions!

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But wait a minute...

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Simulation

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Simulation (Cont.)

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Simulation (Cont.)

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Simulation (Cont.)

Can compare these two numbers

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Simulation (Result)

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Simulation (Result)

P(num_links_friend > num_links_person) = 0.65 (n = 50)

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Simulation (Result)

P(num_links_friend > num_links_person) = 0.65 (n = 50) But we agreed that it should be 0.5!

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What went wrong (again)?

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Sampling Bias

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Sampling Bias

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Sampling Bias

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Find the probability of a bus being crowded!

Sampling Bias

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Sampling Bias

Strategy

1. Choose a person uniformly at random from the city 2. Ask the person if he/she used a bus today a. If no, discard the response b. If yes, ask the person if the bus was crowded 3. Calculate probability as:

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Sampling Bias

Strategy

1. Choose a person uniformly at random from the city 2. Ask the person if he/she used a bus today a. If no, discard the response b. If yes, ask the person if the bus was crowded 3. Calculate probability as:

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Sampling Bias

Strategy

1. Choose a person uniformly at random from the city 2. Ask the person if he/she used a bus today a. If no, discard the response b. If yes, ask the person if the bus was crowded 3. Calculate probability as:

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SLIDE 85

Sampling Bias

Strategy

1. Choose a person uniformly at random from the city 2. Ask the person if he/she used a bus today a. If no, discard the response b. If yes, ask the person if the bus was crowded 3. Calculate probability as:

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SLIDE 86

Sampling Bias

Strategy

1. Choose a person uniformly at random from the city 2. Ask the person if he/she used a bus today a. If no, discard the response b. If yes, ask the person if the bus was crowded 3. Calculate probability as:

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SLIDE 87

Sampling Bias

Strategy

1. Choose a person uniformly at random from the city 2. Ask the person if he/she used a bus today a. If no, discard the response b. If yes, ask the person if the bus was crowded 3. Calculate probability as:

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SLIDE 88

Sampling Bias

A person from a crowded bus is more likely to contribute to the ratio in step 3

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Sampling Bias

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Sampling Bias

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Sampling Bias

Incorrect conclusion: Buses are crowded with probability 1

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Where is the sampling bias in our

  • riginal problem?
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If a person has a lot of friends then it is more likely that a randomly chosen person is their friend!

Where is the sampling bias in our

  • riginal problem?
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Paradox Resolved!

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Monty Hall Problem

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Monty Hall Problem

“You made a mistake, but look at the positive side. If all those PhDs were wrong, the country would be in some very serious trouble”

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Monty Hall Problem

“You made a mistake, but look at the positive side. If all those PhDs were wrong, the country would be in some very serious trouble” “I must admit I doubted you until my fifth grade math class proved you right”

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Monty Hall Problem

A B C

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Monty Hall Problem

A B C

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Monty Hall Problem

A B C

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Monty Hall Problem

A B C

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Monty Hall Problem

A B C

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Should you switch?

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Monty Hall Problem

A B C

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Monty Hall Problem

A B C

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Monty Hall Problem

A B C

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Monty Hall Problem

A B C

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Switching does not make a difference!

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Simulation

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Simulation

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Simulation

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Simulation

Can use this to calculate probability of winning if we switch

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You win 66% of the times Switching does make a difference!

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You win 66% of the times Switching does make a difference!

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You win 66% of the times Switching does make a difference!

Can explain this by drawing the correct probability tree

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Other Famous Problems

  • St. Petersburg Lottery
  • Infinite Monkey Theorem
  • Necktie Paradox
  • Will Rogers Phenomenon
  • And many more...
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Other Famous Problems

  • St. Petersburg Lottery
  • Infinite Monkey Theorem
  • Necktie Paradox
  • Will Rogers Phenomenon
  • And many more...
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SLIDE 120

Other Famous Problems

  • St. Petersburg Lottery
  • Infinite Monkey Theorem
  • Necktie Paradox
  • Will Rogers Phenomenon
  • And many more...
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Other Famous Problems

  • St. Petersburg Lottery
  • Infinite Monkey Theorem
  • Necktie Paradox
  • Will Rogers Phenomenon
  • And many more...
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SLIDE 122

Other Famous Problems

  • St. Petersburg Lottery
  • Infinite Monkey Theorem
  • Necktie Paradox
  • Will Rogers Phenomenon
  • And many more...
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SLIDE 123

Other Famous Problems

  • St. Petersburg Lottery
  • Infinite Monkey Theorem
  • Necktie Paradox
  • Will Rogers Phenomenon
  • And many more...
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SLIDE 124
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SLIDE 125

Why Paradoxes?

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SLIDE 126

Why Paradoxes?

  • Intellectually Stimulating
  • Highlight loopholes in our understanding
  • I like messing around with people!
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Why Paradoxes?

  • Intellectually Stimulating
  • Highlight loopholes in our understanding
  • I like messing around with people!
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Why Paradoxes?

  • Intellectually Stimulating
  • Highlight loopholes in our understanding
  • I like messing around with people!
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One Last Paradox

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One Last Paradox

This statement is false.

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One Last Paradox

This statement is false.

If it is, then it is not and if it is not, then it is!

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One Last Paradox

This statement is false.

If it is, then it is not and if it is not, then it is!

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One Last Paradox

This statement is false.

If it is, then it is not and if it is not, then it is!

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Thank You!

Shubham Gupta

Statistics and Machine Learning Group, Department of Computer Science and Automation, Indian Institute of Science, Bangalore (560012) shubham.gupta@csa.iisc.ernet.in