This is the third part of my argument for the importance of perspective disagreement in the sleeping beauty problem.
Different part of the argument can be found here: I, II, III, IV.
In this part I would give a simple example to show how two agents can reach to an agreement in a typical probability problem. It would also highlight why such an agreement cannot be reached in the sleeping beauty problem.
Balls In Urns(BIU):
Suppose there is a urn filled with either 2 blue balls and 1 red ball(BBR) or 2 red balls and a blue ball(BRR) with equal chances. Andy randomly picked 2 balls from the urn and finds one ball of each colour. He correctly conclude the probability of BBR is 1⁄2, same as the probability of BRR. Afterwards Bob asked for a red ball and was given one, he then randomly picked 1 ball from the 2 remaining balls in the urn and saw a blue one. He correctly concluded the probability of BBR is 2⁄3. It turns out however Andy and Bob actually saw the exact same 2 balls. The two of them are free to communicate and argue. Suppose both of them are rational can they reach to an agreement? Who should change his answer?
We can use a frequentist approach to solve this problem. Suppose the experiment is repeated many times with equal number of BBR and BRR. We can count the total number of occurrences when they both see the same 1 red and 1 blue balls. The relative frequency of BBR and BRR among these occurrences would indicate the correct probability. Both Andy and Bob should have no problem agreeing with this method. Here it becomes apparent that the exact procedure of the experiment determines whose initial probability is correct and who need to adjust his answer. More specifically how is the red ball given to Bob determined.
Scenario 1: The red ball picked by Andy is always given to Bob. In this case Andy is correct, Bob should adjust his answer of BBR from 2⁄3 to 1⁄2. This is because for both BBR and BRR Andy has the same chance to pick a red and a blue ball. Given the same red ball Andy has Bob would have equal chance to pick the same blue ball again regardless the colour of the last ball left. The relative frequency of BBR and BRR given the occurrences (that they both have the same 1 red and 1 blue balls), would be about the same.
Scenario 2: Any red ball in the urn can be given to Bob. In this case Bob’s initial judgement is correct and Andy would have to change his probability for BBR to 2⁄3. Because all else equal, Bob is twice more likely to have the same red ball as Andy if there is only 1 red ball in the urn. The relative frequency of BBR to BRR with the occurrences would be 2:1.
Since only one out of the two Scenarios can be true Andy and Bob must agree with each other as long as there is no ambiguity about the experiment procedure.
However, if we duplicate Bob (either by cloning or memory wiping) in the case of BRR and give each Bob a different red ball suddenly both Scenarios become true. To Andy the red ball he picked will always be given to Bob. To Bob the red ball given to him can be any one from the urn. Neither person would have reason to adjust his own probability once knowing they have the same balls while fully knowing the experiment procedure. So Andy would stick his probability of BBR to 1⁄2 and Bob to 2⁄3. In this case the the two person having exact same information will remain in disagreement even if they are free to communicate. I believe the parallel between this duplication and sleeping beauty problem is obvious enough to show why the selector and beauty, just as Andy and Bob, can be in disagreement.
Next part of my argument discuss why Elga’s counter argument to traditional halfers is valid. Which concludes SSA as incorrect. And why by considering the importance of perspective disagreement I must concluded double-halving as the correct answer.
Sleeping Beauty Problem Can Be Explained by Perspective Disagreement (III)
This is the third part of my argument for the importance of perspective disagreement in the sleeping beauty problem.
Different part of the argument can be found here: I, II, III, IV.
In this part I would give a simple example to show how two agents can reach to an agreement in a typical probability problem. It would also highlight why such an agreement cannot be reached in the sleeping beauty problem.
Balls In Urns(BIU):
Suppose there is a urn filled with either 2 blue balls and 1 red ball(BBR) or 2 red balls and a blue ball(BRR) with equal chances. Andy randomly picked 2 balls from the urn and finds one ball of each colour. He correctly conclude the probability of BBR is 1⁄2, same as the probability of BRR. Afterwards Bob asked for a red ball and was given one, he then randomly picked 1 ball from the 2 remaining balls in the urn and saw a blue one. He correctly concluded the probability of BBR is 2⁄3. It turns out however Andy and Bob actually saw the exact same 2 balls. The two of them are free to communicate and argue. Suppose both of them are rational can they reach to an agreement? Who should change his answer?
We can use a frequentist approach to solve this problem. Suppose the experiment is repeated many times with equal number of BBR and BRR. We can count the total number of occurrences when they both see the same 1 red and 1 blue balls. The relative frequency of BBR and BRR among these occurrences would indicate the correct probability. Both Andy and Bob should have no problem agreeing with this method. Here it becomes apparent that the exact procedure of the experiment determines whose initial probability is correct and who need to adjust his answer. More specifically how is the red ball given to Bob determined.
Scenario 1: The red ball picked by Andy is always given to Bob. In this case Andy is correct, Bob should adjust his answer of BBR from 2⁄3 to 1⁄2. This is because for both BBR and BRR Andy has the same chance to pick a red and a blue ball. Given the same red ball Andy has Bob would have equal chance to pick the same blue ball again regardless the colour of the last ball left. The relative frequency of BBR and BRR given the occurrences (that they both have the same 1 red and 1 blue balls), would be about the same.
Scenario 2: Any red ball in the urn can be given to Bob. In this case Bob’s initial judgement is correct and Andy would have to change his probability for BBR to 2⁄3. Because all else equal, Bob is twice more likely to have the same red ball as Andy if there is only 1 red ball in the urn. The relative frequency of BBR to BRR with the occurrences would be 2:1.
Since only one out of the two Scenarios can be true Andy and Bob must agree with each other as long as there is no ambiguity about the experiment procedure.
However, if we duplicate Bob (either by cloning or memory wiping) in the case of BRR and give each Bob a different red ball suddenly both Scenarios become true. To Andy the red ball he picked will always be given to Bob. To Bob the red ball given to him can be any one from the urn. Neither person would have reason to adjust his own probability once knowing they have the same balls while fully knowing the experiment procedure. So Andy would stick his probability of BBR to 1⁄2 and Bob to 2⁄3. In this case the the two person having exact same information will remain in disagreement even if they are free to communicate. I believe the parallel between this duplication and sleeping beauty problem is obvious enough to show why the selector and beauty, just as Andy and Bob, can be in disagreement.
Next part of my argument discuss why Elga’s counter argument to traditional halfers is valid. Which concludes SSA as incorrect. And why by considering the importance of perspective disagreement I must concluded double-halving as the correct answer.