Here is a calculation regarding observers placing bets in the many-worlds scenario.
Let us assume that an observer has some current rank n in the human race (there have been n-1 humans born before him/her).
Let us assume a many futures scenario with the total population sizes N>=n weighted by the function W(N>=n) given by
W(N>=n) = n^d d / N^(1+d)
where we can let d approach zero. As I understand it this scale-free distribution is the broadest normalized function that we can use (as opposed to a uniform distribution over all N>=n, for example, which cannot be normalized).
The observer bets that the total population size N that he shall subjectively experience will be less than or equal to m * n (given that versions of himself will live long enough to see the end of the human race).
The fraction of the future versions of the observer that will be correct is given by
P(N <= m * n) = Integral (N=n to m * n) n^d d / N^(1+d) dN = 1 – 1/m^d
For any value of m, if we let d->0 then P(N <= m * n) -> 0.
Thus the fraction of future versions of the observer who observe N <= n * m is vanishingly small. Almost all future versions of the observer lose their money.
Thus the doomsday argument doesn’t work if there will be many actually occurring futures.
This is in contrast to the Doomsday argument in the standard scenario of one actually occurring future. In that case the Doomsday Argument says that the fraction of observers who correctly predict that N <= n * m, P(N <= n * m) = 1 - 1/m. (To derive this result consider an observer's fractional position f along the human race. The probability that f is larger or equal to some ratio r, P(f>=r), is given by P(f>=r) = 1 - r. Substitute f=n/N and r = 1/m so that we obtain P(N <= m * n) = 1 - 1/m).