# Many-worlds defeats Doomsday argument (improved version)?

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. Commented Dec 3, 2017 at 2:56
• 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. Commented Dec 3, 2017 at 3:04