Transcript PARADOXES IN MATHEMATICS

PARADOXES IN MATHEMATICS
WHAT IS A PARADOX?
A
paradox is an argument that produces an
inconsistency within logic. Most logical paradoxes
are known to be invalid arguments however there
are exceptions.
A few TYPES OF PARADOXES
 SELF-REFERENCE
 CONTRADICTION
 VICIOUS
CIRCULARITY ALSO KNOWN AS
INFINITE REGRESS
SOME FAMILIAR
EXAMPLES
SELF-REFERENCE PARADOX
 THE
 IS
BARBER PARADOX
THE ANSWER TO THIS QUESTION NO?
CONTRADICTION PARADOX
 THIS
STATEMENT IS FALSE
 NOTHING
IS IMPOSSIBLE
VICIOUS CIRCULARITY PARADOX
 “THE
FOLLOWING SENTENCE IS TRUE”
“THE PREVIOUS SENTENCE IS FALSE”
 WHAT
HAPPENS WHEN PINOCCHIO SAYS
“MY NOSE WILL GROW NOW?”
Why use paradoxes in math?

They are good in promoting critical
thinking .
 Some
paradoxes have revealed errors in
definitions assumed to be rigorous, and
have caused axioms of mathematics and
logic to be re-examined. One example is
Russell’s paradox.
TRHEE FAMOUS PARADOXES IN
PROBABILITY
 THE
THREE DOGS PARADOX
 BERTRAND’S
 THE
BOX PARADOX
TWO DOGS PARADOX
THE THREE DOGS PARADOX
There are three dogs, equally likely to be male or female.
If one of them is male, what is the probability that all of
them are male?
are male?
PROBABILITY OF EACH DOG to
be male
 D1=
1/2 (male or female)
 D2= 1/2 (male or female)
 D3= 1/2 (male or female)
But we are told that one of the dogs is male, say D1,
so we are left with D2 and D3 to be males. so now we
only have D1 and D2 unknown.
D2= 1/2
D3=1/2
From here, can we say that since the probability D2
and D3 to be both males is (1/2)(1/2)= 1/4 therefore
the answer is 1/4?
ANOTHER APPROACH
Let S be a sample space for the three dogs, then.
S={MMM, MMF, MFM,FMM, FFM, FMF, MFF}
Note: {FFF} possibility is ruled out since we know that
at least one of the dogs is male.
Looking at the sample space we created. What is the
probability of having all three dogs be male?
Conclusion for the three
dogs paradox
1.
The probability for 3 dogs to be male is (1/8)
2.
The probability for 2 dogs to be male is (1/4)
3.
The probability for 3 dogs to be male given
that one is male is (1/7).
Hence the paradox.
BERTRAND’S BOX PARADOX
There are 3 jewelry boxes. Each box has 2
drawers with one gemstone in each drawer as
follows.
1. B1- Diamond, Diamond
2. B2- Diamond, Emerald
3. B3- Emerald, Emerald
One box is chosen at random and one of its
drawers opened. If a diamond is found what is
the probability that the other drawer of this box
has the other diamond?
ATTEMPT TO THE SOLUTION
Let B=Box, D=Diamond, E= Emerald
B1- D,D
B2- D,E
B3- E,E
Since we are looking for the P(D,D) and we know
we found one diamond, the B-(E,E) option is
ruled out. We have left B1 and B2 therefore it
seems like we have a 50% chance on each of the
boxes because we could easily have the (D,E) or
the (D,D), right?
A DIFFERENT ATTEMPT
Let’s assign individual probabilities to each jewelry box.
Using probability notation to get a diamond on each box.
 If we have the (D,D) box the P(D,D)= 1
 If we have the (D,E) box the P(D,E)=1/2
 if we have the (E,E) box the P(D,E) =0
If we want the probability of the 2 diamonds on the same
box, we must use the diamond path by Bayes theorem.
P(D,D)= [P(D,D)*P(B1)] /
[P(D,D)*P(B1)+P(D,E)*P(B2)+P(E,E)*P(B3)]
Picture approach
Box
Diamond
Found
Box
Box
EMERALD
EMERALD
Probability finding a
second diamond
Found P(0)
P(1/3)
P(0)
P(1/3)
LAST BUT NOT LEAST
The two dogs paradox
There are two dogs. One of them is
male and was born on Sunday. What is
the probability that the other dog is
male? Assume that male and female
dogs are equally likely to be born.
We have two dogs, and we know that one is a
male.
We know they are equally likely to be born male
and female. We have one more dog, so this dog
can be male or female which give us the answer
that the probability that this dog is male is .5 =
1/2.
The problem also mentions that the dog was born
on Sunday. Should that change anything?
Let’s see.
A MORE DETAILED LOOK AT THE
PROBLEM
We have two genders, Male and Female.
We have Seven days of the week, Monday, Tuesday,
Wednesday, Thursday, Friday, Saturday and Sunday.
If we create a sample space for the two dogs we would
have something like:
S={SMSM, SMSF, SFSM,SMMF,SFMM,….,SFSF}
In total our sample space would be 14^(2)=196 paired
by gender/day. But this configuration includes all
possibilities, so we need to take some out.
Lets take out the samples that do not feature a male
born on Sunday.
 Female/Female=49
 Non Sunday male/Non Sunday male=36
 Female/Non Sunday male =42
 Non Sunday male/Female =42
49+36+42+42= 169
This gives us 169 outcomes that we can rule out since
these don’t have a male born on Sunday.
We have
196-169= 27
samples with a male born on Sunday. These could
be our possibilities.
Now we can count the samples with one male dog
born on Sunday.
 Sunday male/Non Sunday male =6
 Non Sunday male/Sunday male=6
 Sunday male/Female =7
 Female/Sunday male=7
 Sunday male/Sunday male=1
The probability that the second dog is male is
produced by counting the days in which the two dogs
where male.
The tree samples in which the two dogs were male
given that at least one was born on Sunday are:
 Sunday male/Non Sunday male =6
 Non Sunday male/Sunday male=6
 Sunday male/Sunday male=1
We have
6+6+1=13
Since our possibilities were 27 we get the final result
(13/27)< (1/2)
Hence it matters that a dog is born on Sunday,
Hence the paradox.