1. Extending AnsProlog to reason with probabilities Chitta Baral Arizona State university (joint work with Michael Gelfond, and Nelson Rushton)

2. Motivation: The Monty Hall problem A player is given the opportunity to select one of three closed doors, behind one of which there is a prize, and the other 2 rooms are empty. Once the player has made a selection, Monty is obligated to open one of the remaining closed doors, revealing that it does not contain the prize. Monty gives a choice to the player to switch to the other unopened door if he wants. Question: Does it matter if the player switches, or – Which unopened door has the higher probability of containing the prize?

3. Go to White board http://www.remote.org/fr ederik/projects/ziege/be weis.html

4. Illustration-1 First, let us assume that the car is behind door no. 1. – We can do this without reducing the validity of our proof, because if the car were behind door no. 2, we only had to exchange all occurrences of "door 1" with "door 2" and vice versa, and the proof would still hold. – The candidate has three choices of doors. – Because he has no additional information, he randomly selects one. – The possibility to choose each of the doors 1, 2, or 3 is 1/3 each: Candidate chooses p – Door 1 1/3 – Door 2 1/3 – Door 3 1/3 – Sum1

5. Illustration-2 Going on from this table, we have to split the case depending on the door opened by the host. – Since we assume that the car is behind door no. 1, the host has a choice if and only if the candidate selects the first door - because otherwise there is only one "goat door" left! – We assume that if the host has a choice, he will randomly select the door to open. – Candidate chooses Host opens p – Door 1Door 2 1/6 – Door 1Door 3 1/6 – Door 2Door 3 1/3 – Door 3 Door 2 1/3 – Sum1

6. Illustration-3 candidate who always sticks to his original choice no matter what happens: – Candidate chooses Host opens final choice win p – Door 1 Door 2 Door 1 yes 1/6 – Door 1 Door 3 Door 1 yes 1/6 – Door 2Door 3 Door 2 no 1/3 – Door 3 Door 2 Door 3 no 1/3 – Sum1Sum of cases where candidate wins1/3 candidate who always switches to the other door whenever he gets the chance: – Candidate choosesHost opensfinal choicewin p – Door 1Door 2 Door 3no 1/6 – Door 1Door 3 Door 2no 1/6 – Door 2Door 3 Door 1yes 1/3 – Door 3Door 2 Door 1yes 1/3 – Sum1Sum of cases where candidate wins 2/3

7. Key Issues The existing languages of probability do not really give us the syntax to express certain knowledge about the problem Lot of reasoning is done by the human being Our goal: Develop a knowledge representation language and a reasoning system such that once we express our knowledge in that language the system can do the desired reasoning P-log is such an attempt

8. Representing the Monty Hall problem in P-log. doors = {1, 2, 3}. open, selected, prize, D: doors ~can_open(D)  selected = D. ~can_open(D)  prize = D. can_open(D)  not ~can_open(D). pr(open=D | c can_open(D), can_open(D1), D =/= D1 ) = ½ – By default pr(open = D | c can_open(D) ) = 1 when there is no D1, such that can_open(D1) and D =/= D1. random(prize), random(selected). random(open: {X : can_open(X)}). pr(prize = D) = 1/3. pr(selected = D) = 1/3. obs(selected = 1). obs(open = 2). obs(~prize = 2).