Philosophical Foundations of Probability Theory

Question 1

What is a probability distribution

Answer 1

function mapping each proposition in the relevant representation language to a number between 0 and 1 conforming to the axioms K1 and K2

Question 2

What questions are we trying to answer here?

Answer 2

Who says degrees of belief should be probability distributions?
Who says that they should be revised by conditionalization?

Question 3

What is a Dutch Book?

Answer 3

Sequence of bets, each of which the agent is disposed to accept (given his degrees of belief) which taken together will cause the agent to lose money no matter what happens.

Question 4

What is the price of this ticket:
$1 if a
0 otherwise

Answer 4

$degree of belief in a

Question 5

How do we demonstrate this ticket is a dutch book?
p(a) = 0.6 bet 1: $1 if a– 60p
p(b) = 0.5 bet 2: $1 if b– 50p
p(! (a or b)) = 0.25 bet 3: $1 if !(a or b)– 25p
p(! (a and b)) = 0.7 bet 4: $1 if !(a and b)– 70p

Answer 5

create a table for ll possible combinations of a and b
Put the result of each bet for each case (+ or - based on what paid)
Calculate overall outcome for each row

Question 6

If a dutch book occurs, what do we know about the agents degrees of belief?

Answer 6

The degrees of belief violate K1 and K2

Question 7

What is the fair price of this ticket?
$1 if a and b
$0 if !a and b
money back if !b

Answer 7

$conditional belief in a given b

Question 8

$1 if a and b
$0 if !a and b
money back if !b
We have our price $x for this ticket.
We have the following tickets:
$1 if a and b             |          $x if !b
0 otherwise             |           0 otherwise
What is the price of these tickets?
Equate these to prove conditionalisation

Answer 8

$x . p(!b)

x = p(a and b) + x(a) p(!b)
= p(a and b) + x(a) (1 - p(b))
x . p(b) = p(a and b)
x = p(a and b) / p(b)

Question 9

What does a diachronic Dutch book argument show?

Answer 9

a justification for the policy of conditionalizing

Question 10

Explain the Diachronic dutch book argument

Answer 10

we want to say that if E has happened, p(A) = p(A | E).
If this is violated then p(A) < p(A | E).
We can show that this can create a dutch book through the following 3 bets:

Price $p(A | E)
$1 if A and E
0 if (!A) and E
Money back if (!E)

Price $a . p(E)
$a if E
0 otherwise

Price $p(A)
$1 if A
0 otherwise

From this we get payoffs for E and !E that show if p(A | E) != p(A) we get a dutch book

Question 11

How do we solve the monty hall problem?

Answer 11

Let A stand for the car being behind A
P(A and !B) = p(A) = 1/3
p(!B) = 1 - p(B) = 2/3
P(A | !B) = p(A and !B)/p(!B) = 1/2
But Monty opens one of the doors....so now we can say we know the car is not behind B:
p(K !B | A) = 1/2
p(K !B | !A) = 1/2
P(A | K !B) = p(K !B | A) P(A)   /    p(K !B | A)p(A) + p(K !B | !A)p(!A) = 1/3