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This is a more carefully argued version of my previous post Many-worlds Interpretation defeats the Doomsday argument?

**Standard Doomsday Argument**

Let N be the total number of humans who will ever live.

Let n be our present birth rank along the chronological list of all the humans who will ever live.

According to Bayes' Theorem, the posterior probability of N given n, P(N|n), is given by

P(N|n) = P(n|N) P(N) / P(n).

If we assume prior complete ignorance of n and N then we should use the improper prior probability distributions P(N)=1/N and P(n)=1/n.

This leaves us with the conditional probability of our present position n given some final total human population N, P(n|N).

The Doomsday argument assumes that only one future will exist so that, a priori, we are equally likely to find ourselves at any position n from 1 to some particular value N. Therefore we assume P(n|N) = 1/N.

Bayes' theorem then gives

P(N|n) = P(n|N) P(N) / P(n) = (1/N) * (1/N) / (1/n) = n / N^2

which is a properly normalized probability distribution that allows us to estimate the unique value N given our present position n.

**Many-worlds Doomsday Argument**

But, as mentioned above, the uniform conditional probability distribution P(n|N)=1/N implicitly assumes one future with some particular final total population size N.

If the many worlds interpretation is true then our present position n is consistent with many actually existing futures with different values of N.

We can express Bayes' theorem in terms of non-exclusive "weights" W rather than exclusive probabilities P:

W(N|n) = W(n|N) W(N) / P(n).

We assume that multiple values of the final population size N will exist in the future so that we have weights W(N|n), W(n|N) and W(N) whereas only one value of our present position n exists with prior probability P(n).

The weight of our present position n given all future values of final total population size N, W(n|N), is given by

W(n|N) = Sum[N=n to infinity] P(n|N) W(N)

where P(n|N)=1/N is the standard Doomsday argument conditional probability of our position n given a particular total population size N.

If we assume the initial prior weight for N given by W(N)=1/N then

W(n|N) = Sum[N=n to infinity] (1/N) * (1/N) = 1 / n

where we have approximated the sum with an integral.

Therefore if we assume the prior probability for n, P(n)=1/n, then Bayes' theorem for the many worlds scenario is given by

W(N|n) = W(n|N) W(N) / P(n) = (1/n) W(N) / (1/n) = W(N).

Therefore the posterior weights for the final total population sizes N, W(N|n), given our present position n, is exactly the same as our a priori weights for N, W(N).

Therefore our state of knowledge about the weights of N has not changed, after we have learned of our present position n, and thus the Doomsday argument does not work in the many worlds scenario.

Does this make sense?

In your formula, N is alternatively used as an index in the sum and as a condition for W(n/N) which is a bit confusing. This condition never appears in the right hand part. It seems that you're rather calculating W(n). You should explain how W(n/N) is to be interpreted in MWI. – Quentin Ruyant – 2017-12-03T02:56:41.363

Take a subjective interpretation of the kind: W(N/n) is how much I should bet on the future branches with total population N, given actual population n. Then the reverse W(n/N) could be: how much I would bet that the actual population is n if all future branches had population N. I'm not sure that the right hand part expresses this. – Quentin Ruyant – 2017-12-03T03:04:35.913