Does observing life on Earth increase the probability of life elsewhere? Say I have an implausibly large sack of balls. All I know is that the balls are numbered randomly from $1$ to $n$. For all I know, any value of $n$ (a positive integer) is equally likely.
I reach into the sack and choose a ball randomly. The ball says $42$. Does this change at all the probabilities of the values of $n$ used to number the balls where $n \geq 42$? 
(Intuitively it might seem like $n$ is a low number in that if $n$ were very very large (say $2^{42}$) it seems implausible we would hit on a very low number from the first ball sampled. On the other hand, if $n$ is a very very large number, $42$ is equally as like as any ball to emerge.)

Another simplified version might be where the balls are either blue or red, but I don't know how many are blue or how many are red. The first ball I choose is blue. Does this increase the probability of observing further blue balls in later samples? 
(Again if there were only one blue ball, intuitively it seems unlikely we would choose it on the first sample. On the other hand, if there were only one blue ball, that ball is as equally likely to emerge as any on the first sample.)

It seems to be a question that crops up a lot. Like for example in the argument that well there's life here on Earth so it would be an improbable fluke if there were no life elsewhere. Of course this is a more complex question than just what colour the balls are, but the thrust of this argument seems to be a probabilistic one, like it boils down to the idea that we know there's one blue ball in the tiny sample we've seen, so there must be lots of blue balls in the implausibly large sack to explain that. 
I'm not convinced this latter argument makes sense, but on the other hand, I don't know how to reason about the problem or prove one way or the other whether seeing a blue ball early on affects the (relative) probability of the number of blue balls in the population. Hence I'm wondering, for example, if there's some sort of general theorem from probability that talks about this?
 A: This is a really interesting question. I suggest the following appoach using Bayes theorem.
Supose there exist n planets in total.
Define $E_r$ = event that there are exactly r planets with life(blue planets). You can check easily that the events are mutually exclusive and exhaustive.
A = event of observing one blue planet.
We shall calculate $P(E_r/ A)$= $\frac {P(E_r ).P(A/E_r)}{\sum P(E_i).P(A/E_i)}$
Assumig that the creator painted the planets randomly, what is the probabilty that r of them are blue?
Clearly its $P(E_r) = \frac{nCr}{2^n}$.
Also $P(A/E_r) = \frac{r}{n}$
SUbsituting,we will have
$P(E_r/A)= \frac{(n-1)!}{(r-1)!.(n-r)!.2^{n-1}}$
Suppose, that n is comparatively small, about a million. Note how negligibly small the probability of observing only one blue planet (r=1) becomes.
A: Let's look at your first problem, the one with the numbered balls.
Well, one problem with this problem is that there is no uniform distribution of all natural numbers. However we can consider the case where $n$ is uniformly distributed in the range $1$ to $N$, and see if we can make statements when $N$ goes to infinity.
So let's assume that we have a sack with $1\le n\le N$ numbered balls, and each value of $n$ in the range is initially equally likely, that is, we have an uniform prior for $n$. Now we draw at random (that is, again with uniform probability) a single ball from the sack, and get 42. The question is, what is the probability distribution for $n$ after drawing that ball.
According to Bayes' theorem, we have
$$P(n=n_0|\text{42 drawn}) =
\frac{P(n=n_0)P(\text{42 drawn}|n=n_0)}{\sum_k P(n=k)P(\text{42 drawn}|n=k)}$$
Now $P(n=k) = \frac{1}{N}$ and
$$P(\text{42 drawn}|n=k)=\begin{cases}
\frac{1}{k} & k\ge 42\\
0 & k<42
\end{cases}$$
Therefore for $n_0\ge 42$ we have
$$P(n=n_0|\text{42 drawn}) = \frac{1}{n_0\sum_{k=42}^N\frac{1}{k}}$$
Note that the sum in the denominator is independent of $n_0$ and basically just gives the normalization constant, so that the probabilities add up to $1$. Therefore the relevant information is:
$$P(n=n_0|\text{42 drawn}) \propto \frac{1}{n_0}$$
Therefore small values of $n_0$ (with the restriction $n_0\ge 42$, of course) are indeed favoured, but only very weakly; in particular, the probabilities still go to zero as $N\to\infty$.
Let's calculate the expectation value of $n$:
$$\langle n\rangle = \sum_{n_0=1}^N n_0\,P(n=n_0|\text{42 drawn}) = \frac{N-41}{\sum_{k=42}^N\frac{1}{k}}$$
Since the numerator grows linearly while the denominator grows logarithmically, this diverges for $N\to\infty$. The information we get from the single ball therefore is not sufficient to cut the expectation value down to a finite value, although it grows more slowly with $N$ than on the prior probability where it grows linearly with $N$.
Note that if we draw a second ball, then the probabilities should be $\sim \frac{1}{k^2}$, which gives a convergent series. Therefore drawing two balls should be sufficient to force a finite probability even in the limit $N\to\infty$, and therefore probably also a finite expectation value (but at the moment I'm too lazy to calculate that, especially given that it is already far past midnight and I should go to bed).
A: This article is about problems of this sort, it generalizes the traditional ad hoc method where you assume the "Self Sampling Assumption" (SSA: one should reason as if one were a random sample from the set of all observers in one’s reference class) and the "Self Indication Assumption" (SIA: we should take our own existence as evidence that the number of observers in our reference class is more likely to be large than small). In case of the Doomsday argument, the SSA and SIA cancel each other out exactly, but as the article points out, invoking SSA and SIA is a rather ad hoc thing to do, it's better to simply take into account all the available information.
A: Earlier I was trying to tackle this from an intuitive perspective so perhaps it's worth posting some ideas (as thinking out loud).
Coming at this from someone who knows little about probability/probability theory, the only way I can see to reason about this problem is to think in an intuitive way about simulations and just counting cases.
Let's take the case of blue and red balls for example. Let us assume we have $X$ balls and that $B$ of those are blue. To keep this as general as possible, we don't know the value for $B$ (other than $B\geq 1$; put another way, we have no idea of what "prior" probability a ball has of being blue) nor for $X$ (other than $X\geq B$). 
However, for argument's sake, let's fix some arbitrary large value for $X$. So given $X$, now we can try run a great many simulations and count how many times, for various values of $b \leq X$, we sample a blue ball on the first go. Let's say we run $s$ simulations for each value of $b$ ($s \gg X$), drawing a first ball and keeping track of how many times it's blue.
As $s$ approaches infinity for a given value of $b$, the number of times we will sample blue in the first ball will equal $\frac{b}{X}$. Across all values of $b$, we get will have $sX$ simulations in total. The number of times a blue ball will be sampled first will be $\frac{sX}{2}$. 
Okay now we assume that we know that a blue ball was sampled first and we look at our $sX$ simulations to see in what ratio of simulations that happened for various values of $b$. 
Thus if we know that a blue ball is sampled first, then $s$ of those cases will occur when $b=X$, $s-1$ cases when $b=X-1$ and more generally, $s-n$ cases when $b=X-n$.
Since we're interested in probabilities, rather than counting cases, we can look at ratios. In total, we can see that in half the cases, a blue ball is sampled first. For a given value of $b$, $\frac{1}{X}$ cases will have been run for that value (either blue or not). The ratio of cases where the first ball sampled was blue will be $\frac{b}{X^2}$. So for example, given that the first ball sampled is blue, that leaves us with $\frac{1}{2}$ of the original cases, of which $\frac{1}{X}$ cases are explained by all balls being blue ($b=X$).
This is not so satisfying though since it always goes back to the value of $X$. But we can try to find a way to find a general conclusion or an "invariant": let's try to see in terms of "cumulative probability" at what point 50% of the cases where blue is drawn first are covered. In other words, we're looking for $\beta$ such that $\Sigma_{b \leq \beta} \frac{b}{X^2} = \frac{1}{4}$. We can consider $\beta$ as something of a "tipping point", meaning that knowing that blue was sampled first, the value of $b$ being below $\beta$ or above $\beta$ are equally likely (where we would expect it, for sure, to be somewhere above $\frac{X}{2}$). In fact, as $X$ approaches $\infty$, the value for $\beta$ converges to $\frac{X}{\sqrt{2}}$.
So for example, if $X=100000$, and we know that a blue ball is drawn first, this tells us that $P(B>70711\mid \text{blue ball drawn first}) \approxeq 0.5$. Also that ratio is fixed as $\frac{X}{\sqrt{2}}$, so given $X=10000$, $P(B>7071\mid \text{blue ball drawn first}) \approxeq 0.5$.

However, this all presumes I guess what one would call a "uniform distribution" over the possible values that $B$ might take. That seems like the most natural assumption to take when no other information is given, but I rather tend to think that it's just as unknown as the value for $B$, and hence really, no information can be gained from knowing that a blue ball is drawn first unless one assumes something quite strong: that any value of $B$ is somehow equally likely. Under that assumption, there's a 50/50 chance that $\sim$70.7\% or more of balls are blue according to the above line of reasoning.
What I have difficulty grappling with now – on a more philosophical level – is just how reasonable it is to assume – in the absence of further information – that any value of $B$ is equally likely. It's tempting to assume this to even start to make progress, but equally it seems to be something we do not know and thus cannot use.
(Comments are very welcome.)
