my derivation of the Born rule

I just read (parts of) Sean Carroll's derivation of the Born rule, but I do not find it very convincing, because there is a much simpler, straightforward derivation available to resolve this problem of "self-locating uncertainty".

1) We shall use a "hardcore" many worlds interpretation, assuming that the world splits into a quasi-infinite number of branches at any time, which realizes all possible outcomes of quantum theory. We assume that those branches are all equally real and a simple counting argument shows that the Born rule does not hold for almost all of those branches. It follows that we do not live in one of those generic branches, which solves the first part of our self-location problem.

2) It is reasonable to assume that some of those infinitely many branches contain at least one quantum computer capable of simulating human life. Those computers will have to simulate quantum theory, but we can further assume that they will only keep one branch at a time in order to save resources. It is straightforward to assume that they are programmed to use the Born rule to select this branch randomly.

3) We observe the Born rule to great precision and it follows that we are the human beings simulated in one of those quantum computers. This finally resolves the self-location problem.

I would add that (some of) the simulated human beings will use the Copenhagen interpretation to explain what they experience; i.e. an interpretation which emphasizes the importance of the observer and her 'conscious experience'. Obviously, the simulated human beings are unaware that their 'conscious experience' is indeed a side effect of the procedure which selects the simulated branch randomly.

effective altruism

I mentioned Jess Riedel in the previous blog post. Here I want to highlight his list of organizations related to effective altruism.
While we contemplate how many worlds there are, we can try to improve the one we know - beyond posting hashtags on twitter.

preferred basis

I did write some comments on Scott's blog which might be interesting to those bothered by the 'preferred basis problem'. It begins here and references are made to this answer at physics.stackexchange by Jess Riedel and this paper by Dowker and Kent.

While I'm at it, I should also link to this paper about 'entanglement relativity' and a (claimed) inconsistency of the Everett interpretation. I am not sure if the argument is correct (decoherence might appear different for two different decompositions, but does this really prove anything?) and would appreciate any input.

added later: The back and forth in the comment thread ended (for now) with a homework exercise for mwi proponents.

added later: Btw another interesting paper from an Austrian team about decoherence due to classical, weak gravitation (i.e. on Earth).

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Btw this unrelated comment Scott made about the "arrow of time" was a bit shallow imho. My own view of the problem begins with this thought experiment.

a probability puzzle

No paradox and nothing profound here, just a little puzzle to pass the time until Monday.
I have two reasons for posting it: i) It is similar to some problems I have to deal with at work (*) and ii) it gives me an opportunity to link to the blog where I got it from (after the solution is revealed).

So Alice and Bob like to play a certain (card) game (if they are not busy with encryption problems and black hole entanglement). Everybody knows that Alice is slightly more skilled at this game and wins with probability 55%; However, she really likes to win and so Alice and Bob always play as many games as it takes for Alice to be ahead in winnings (x). So sometimes they play just one game (if Alice wins immediately) and sometimes many, but what is the expected number N of games the two will play (after a fresh start)?

(*) A similar problem I would be dealing with could be e.g. of the form "if I have an order sitting at the bid, how long will it take on average to get filled".

(x) Added later, just to clarify: Alice and Bob play N games until Alice wins one game more than Bob. E.g. Alice wins the 1st game; Or Bob wins the 1st and Alice wins the next 2 games; Or ...


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This puzzle is equivalent to a biased random walk of the difference D in winnings between Bob and Alice. It begins at D=0 and if D>0 it means that Bob is ahead; The random walk ends at D=-1 i.e. when Alice is ahead by one. So what is the expectation value E = E[N] of the length N of this random walk?

There are two ways to solve it. One can (try to) sum up all terms in the series of all possible events as described here. I assume this is how John von Neumann would have solved this puzzle.

Fortunately, there is a much easier solution for the rest of us and you can find it in the comments.
It gives us E = 1/(2p - 1) and with p=0.55 for Alice to win a single game we get E=10.

Notice that E diverges for p=1/2 and I find this somewhat counterintuitive, knowing that an unbiased random walk will visit every point D with probability 1.