Mathematical Paradoxes

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Mathematical Paradoxes

• That which is true is not always obvious.
• That which is obvious is not always true.

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Definition (from Wikipedia, the free encyclopedia)

• A paradox (Gk pa??d????, "aside belief") is an
  apparently true statement or group of statements
  that leads to a contradiction or a situation
  which defies intuition.
• Three types of paradoxes
• False Proofs
• Unexpected Facts
• Logical Dilemmas

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Unexpected facts or veridical paradoxes

• Results appears unrealistic.
• Results are correct.
• Results can be proven true.

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Examples

• The birthday paradox
• How many people must there be in a room to that
  states that the probability of having two or more
  people with the same birth date is 50?
• How about 90?
• How about 99.99996?
• How about 100?

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Probability results

• For 23 or more people, Probability gt 50.
• For 60 or more people, Probability gt 99.
• For 100 or more people Probability gt 99.9996.
• For 366 or more people, Probability 100.

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The Proof

• It is easier to first calculate the probability ?
  that all n birthdays are different. If n
  365, it is given by
• because the second person cannot have the same
  birthday as the first with probability (364/365),
  the third cannot have the same birthday as the
  first two (363/365), etc.
• The event of at least two of the n persons having
  the same birthday is complementary to all n
  birthdays being different. Therefore, its
  probability p(n) is

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The Graph
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Another Paradox

• The Monty Hall Paradox

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The Monty Hall problem

• Suppose you're on a game show, and you're given
  the choice of three doors Behind one door is a
  new automobile behind the others, nominal
  prizes. You pick a door, say No. 3, and the host,
  who knows what's behind each door, opens one of
  the other doors, say No. 2, which has a lesser
  prize. He then says to you, "Do you want to pick
  door No. 1 or do you want to keep your original
  choice?"
• Is it to your advantage to switch your choice?

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Answer

• YES! It is to your advantage to switch doors.
• Why? Because you double your chances of winning
  the automobile.
• The probability the automobile is behind door
  number 1 is 2/3.
• The probability it is behind door 3 is 1/3.

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Justification

• For i 1,2,3 let MiMonty picks door i,
  CiContestant picks door i, and AiAuto is
  behind door i.
• Assumptions Ai and Cj are independent events
  with P(Ai) P(Cj) 1/3 for all i, j.
• As in the original problem, the Contestant picked
  door 3 and Monty picked Door 2.
• Since Monty knows where the car is we have
  P(M2A1?C3) 1, P(M2A2?C3) 0, P(M2A3?C3)
  1/2.
• We want to findP(A1M2?C3) and compare it to
  P(A3M2?C3).

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A Little Probability Theory

• P(M2?C3) P(A1?M2?C3) P(A2?M2?C3)
  P(A3?M2?C3) (by the law of total probability)
  P(M2A1?C3)P(A1 ?C3)P(M2A2?C3)P(A2 ?C3)
  P(M2A3?C3)P(A3 ?C3) (conditional
  probabilities) 1?P(A1 ?C3) 0?P(A2 ?C3)
  (1/2)?P(A3 ?C3) 1?(1/9) 0 (1/2)?(1/9)
  (1/6).
• Probability that auto is behind door 1 given
  M2?C3 P(A1M2?C3) P(A1?M2?C3)/P(M2?C3)
  P(M2A1?C3)P(A1 ?C3 )/P(M2?C3) (Bayes Theorem)
  1?(1/9)/(1/6) 2/3.
• Probability that auto is behind door 3 given
  M2?C3 P(A3M2?C3) P(A3?M2?C3)/P(M2?C3)
  P(M2A3?C3)P(A3 ?C3 )/P(M2?C3)
  (1/2)?(1/9)/(1/6) 1/3.

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Sometimes questions seem too simple

• Another Paradox

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Boy or Girl

• The Boy or Girl problem is a well-known example
  in probability theory
• In a two-child family the older child is a boy.
  What is the probability that the younger child is
  a girl?
• In a two-child family one of the children is a
  boy. What is the probability that the other
  child is a girl?

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Answer to question 1

• Assuming boy and girls are equally likely and the
  outcome of the second pregnancy is independent of
  the first, the probability that the second child
  is a girl is ½.

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Answer to question 2

• Assuming boy and girls are equally likely and the
  outcome of the second pregnancy is independent of
  the first, the probability that the other child
  is a girl is 2/3.

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Justification

• There are four possible combinations of children.
  Labeling boys B and girls G, and using the first
  letter to represent the older child, the sample
  space is
• (B,B), (B,G), (G,B), (G,G).
• Under the assumptions, the above four
  possibilities are equally likely.
• Since it is known that there is a boy in the
  family, only the first three events are under
  consideration.
• Of these 2/3 consist on a family with one boy and
  one girl, so the probability is 2/3.

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A similar paradox

• The three card paradox

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Three Card Problem

• Suppose you have a deck of 60 cards, consisting
  of
• 20 black cards that are black on both sides,
• 20 white cards that are white on both sides, and
• 20 mixed card that are black on one side and
  white on the other.
• The cards are shuffled. One is pulled out of the
  deck at random, and placed on a table observing
  only one of the sides. The side facing up is
  black. What is the probability that the other
  side is also black?

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Answer

• The probability that the other side is also black
  is 2/3.

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A Lie Not a Paradox

• If it appears too good to be true it is probably
  false.

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The Two Envelopes Problem

• The puzzle Let's say you are given two
  indistinguishable envelopes, each of which
  contains a positive sum of money. One envelope
  contains twice as much as the other. You may pick
  one envelope and keep whatever amount it
  contains. You pick one envelope at random but
  before you open it you're offered the possibility
  to take the other envelope instead.

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Now, suppose you reason as follows

• Denote by A the amount in my selected envelope.
• The probability that A is the smaller amount is
  1/2, and that it's the larger also ½.
• The other envelope may contain either 2A or A/2.
• If A is the smaller amount the other envelope
  contains 2A.
• If A is the larger amount the other envelope
  contains A/2.
• Thus, the other envelope contains 2A with
  probability 1/2 and A/2 with probability ½.
• So the expected value of the money in the other
  envelope is (1/2)(A/2) (1/2)(2A) (5/4)A.
• This is greater than A, so I gain on average by
  swapping.

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Follow Up

• After the switch I can denote that content B and
  reason in exactly the same manner as above.
• I will conclude that the most rational thing to
  do is to swap back again.
• To be rational I will thus end up swapping
  envelopes indefinitely.
• Since we cannot win arbitrarily large amounts
  given the amounts in the two envelopes something
  is wrong with the argument.

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Explanation

• The most common way to explain the problem (not a
  paradox) is to observe that A isn't a constant in
  the expected value calculation, step 7 above. In
  the first term A is the smaller amount while in
  the second term A is the larger amount. To mix
  different instances of a variable or parameter in
  the same formula like this shouldn't be
  legitimate, so step 7 is thus the proposed cause
  of the paradox.

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Paradoxes in Analysis

• Whats going on here?

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Tarski paradox

• Tarski paradox or Hausdorff-Banach-Tarski paradox
  is the famous "doubling the ball" paradox, which
  states that by using the axiom of choice it is
  possible to take a solid ball in 3-dimensional
  space, cut it up into finitely many
  (non-measurable) pieces and, moving them using
  only rotations and translations, reassemble the
  pieces into two balls of the same radius as the
  original.

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The Hausdorff paradox

• In mathematics, the Hausdorff paradox, named
  after Felix Hausdorff, states that if you remove
  a certain countable subset of the sphere S², the
  remainder can be divided into three subsets A, B
  and C such that A, B, C and B ?? C are all
  congruent. In particular, it follows that on S²
  there is no "finitely additive measure" defined
  on all subsets such that the measure of congruent
  sets is equal. In other words, one cannot define
  a probability space on the sphere S² which
  preserves congruence.

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Origins of Probability

• Marquis de Laplace (1749-1827)
• Formalized the calculus of probability.
• Defined the probability p(A) as the ratio of the
  number of events that result in the outcome A to
  the total number of possible events.
• Coined the term equiprobable.
• Restrictions Required an experiment to consist
  of a finite number of possible outcomes.

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Extensions

• For experiments with an infinite number of
  possible outcomes, results were approximated by
  experimentation.
• Was not fully endorsed by all mathematicians.
• One problem that arose was coined Bertrand's
  paradox.

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The Problem

• Consider an equilateral triangle inscribed in a
  circle.
• Suppose a chord of the circle is chosen at
  random.
• What is the probability that the chord is longer
  than a side of the inscribed triangle?
• Different mathematicians were simulating this
  experiment with each obtaining different results.

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The analysis

• Joseph Louis François Bertrand (March 11, 1822
  April 5, 1900), a French mathematician began to
  analyze the various solution, to formalize the
  approaches and to summarize the results.
• He presented three techniques for selecting a
  random chord.
• The "random endpoints" method.
• The "random radius" method.
• The "random midpoint" method.
• The work was named Bertrands Paradox.

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The "random endpoints" method

• Randomly choose a point on the circumference of
  the circle.
• Randomly choose another point on the circle.
• Draw the chord joining it to the first point.
• The result is a random chord.

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The "random radius" method

• Choose, at random, a radius of the circle.
• Choose a random point on the radius.
• Construct the chord whose midpoint is the chosen
  point (such a chord is always perpendicular to
  the radius).
• The resulting chord is a random chord.

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The "random midpoint" method

• Choose a point at random anywhere within the
  circle.
• Construct a chord with the chosen point as its
  midpoint.
• The resulting chord is a random chord.

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The Analysis

• Bertrands Paradox

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The random endpoints" method

• Rotate the triangle so that the first point is at
  one vertex.
• If the second point is chosen between the
  endpoints of the side opposite the first point,
  the chord is longer than a side of the triangle.

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Random Endpoints

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Random Endpoints" method The Answer

• The length of this arc is one third of the
  circumference of the circle.
• Therefore the probability a random chord is
  longer than a side of the inscribed triangle is
  1/3.

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The "random radius" method

• Rotate the triangle so a side of the equilateral
  triangle is perpendicular to the radius.
• The cord with the midpoint on the radius will be
  perpendicular to the radius at the chosen point.
• The chord is longer than a side of the triangle
  if the chosen point on the radius is inside the
  triangle.

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Random Radius

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Random Radius" MethodThe Answer

• Notice that ½ of the radius is inside the
  triangle and ½ is outside.
• Therefore, the probability that the random ray is
  longer than a side of the inscribed triangle is
  ½.

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The "random midpoint" method

• Every chord has only one midpoint.
• Draw the chord having the randomly chosen point
  as its midpoint.
• The chord is longer than a side of the inscribed
  triangle if the chosen point falls within a
  concentric circle of radius ½.

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Random Point

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Random Midpoint" MethodThe Answer

• The area of the smaller circle is one fourth the
  area of the larger circle.
• Therefore the probability a random chord is
  longer than a side of the inscribed triangle is
  1/4.

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The selection methods can be visualized as
follows. Other than a diameter, a chord is
uniquely identified by its midpoint. Each of the
three selection methods presented above yields a
different distribution of midpoints. Methods 1
and 2 yield two different nonuniform
distributions, while method 3 yields a uniform
distribution.
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Explanation

• All answers are correct.
• What constitutes a random cord within a circle is
  not clear.
• Once a definition is accepted an appropriate
  method with a definitive answer can be developed.

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Infinity

• A problem in itself

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The Question

• Are there more Irrational Numbers than Rational
  Numbers?
• Be careful, if you answer the question you have
  to prove it.

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Consider the following

• True or False Between any two rational numbers
  there is always an irrational number.
• True or False Between any two irrational
  numbers there is always a rational number
• Both of these statements can be proven true.
• So, which set has more elements?

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The End