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?
