Predictions and Everett

Imagine this unfortunate sequence of events will certainly befall you in a classical universe:

1. You will be made to fall asleep.

2. Upon waking up, you will be shown a red square.

3. You will be made to fall asleep again.

4. While asleep, your memory will be reset to that which you had in step (1).

5. Upon waking up, you will be shown a green triangle.

6. You will be made to fall asleep for a third time.

7. While asleep, your memory will be reset again to that which you had in step (1).

8. Upon waking up, you will be shown a green circle.

9. You will then be permanently annihilated.

Questions:

10. How likely is it that you will be shown a green shape?

11. How likely is it that you will be shown a red shape?

The answers to these questions are obviously: one and one. You will be shown a green shape twice and a red shape one, and that’s certain.

Now consider a variant story where personal identity is not maintained in sleep. Perhaps each time in sleep the person who fell asleep will be annihilated and replaced by something that is in fact an exact duplicate, but that isn’t identical with the original according to the correct metaphysics of diachronic personal identity. (We can make this work on pretty much any metaphysics of diachronic personal identity. For example, we can make it work on a materialist memory theory as follows. We just suppose that before step (1), you happen to have three exact duplicates alive, who are not you. Then during the nth sleep cycle, the sleeper is annihilated, and a fresh brain is prepared and memories will be copied into it from your nth doppelganger. Since these memories don’t come from you, the resulting brain isn’t yours.)

And in the variant story, let’s ask the questions (10) and (11) again. What will the answers be? Again, it’s easy and obvious: zero and zero. You won’t be shown any shapes, because you will be annihilated in your sleep before any shapes are shown.

Now consider Everettian branching quantum mechanics. Suppose there is a quantum process that will result in your going to sleep in an equal superposition of states between having a red square, a green triangle and a green circle in front of your head, so that upon waking up an observation of the shape will be made. Now ask questions (10) and (11) again.

I contend that this is just as easy as in my classical universe story. Either the branching preserves personal identity or not. If it preserves personal identity, the answer to the questions is one and one. If it fails to preserve personal identity, the answer to the questions is zero and zero. The only relevant ontological difference between the quantum and classical stories is that in the quantum stories the wakeups might count as simultaneous while in the classical story the wakeups are sequential. And that really makes no difference.

In none of the four cases—the classical story with or without personal identity and the branching story with or without personal identity—are the answers to the questions 2/3 and 1/3. But those are in fact the right answers in the quantum case, contrary to the Everett model.

Now, one might object that we care more about decisions than predictions. Suppose that you have a choice between playing a game with one of two three-sided fair quantum dice:

• Die A is marked: red square, green triangle, green circle.

• Die B is marked: green square, red triangle, red circle.

And suppose pain will be induced if and only if the die comes up red. Which die should you prudentially choose for playing the game? Again, it depends on whether personal identity is preserved. If not, it makes no difference. If yes, clearly you should go for die A on the Everett model—and that is indeed the intuitively correct answer. But the reason for going for die A on the Everett model is different from the reason for going for it on a non-branching quantum mechanics. On the Everett model, the reason for going for die A is that it’s better to get pain once (die A) rather than twice (die B).

So far so good. But now suppose that you’ve additionally been told that if you go for die A, then before you roll A, an irrelevant twenty-sided die will be rolled. (This is a variant of an example Peter van Inwagen sent me years ago, which was due to a student of his.) Then, intuitively, if you go for die A, there will be twenty red branches and forty green branches on Everett. So on die A, you get pain twenty times if personal identity is preserved, and on die B you get pain only twice. And so you should surely go for die B, which is absurd.

One might reasonably object that there are in fact infinitely many branches no matter what. But then on the no-identity version, the choice is still irrelevant to you prudentially, while on the identity version, no matter what you do, you get pain infinitely many times no matter what you choose. And that doesn’t work, either. And if there is no fact about how many branches there will be, then the answer is just that there is no fact about which option is preferable on the identity version, and on the no-identity version, indifference still follows.

This is all basically well-known stuff. But I like the above way of making it vivid by thinking about classically sequentializing the story.