The limiting case of a discrete probability problem Say there are three jars, $j_1, j_2, j_3$ filled with different binary sequences of length two.  
The distribution of the binary sequences in each of the jars is given by the $p_i^k(1-p_i)^{n-k}$, where 
$p_i = \frac{i}{m + 1}$ where $m$ is the number of jars, $i$ is the jar index, $k $is number of 1$$'s and $n$ is the length of the string.  
So for three jars we have $p_1 = 0.25, p_2 = 0.5$, and $p_3 = 0.75$ for $j_1, j_2, j_3$ respectively.  
Here are the sequences and their probabilities for $j_1$ with $p_1 = 0.25$:
\begin{align*}
P(00) = 9 / 16 \\
P(10) = 3 / 16 \\  
P(01) = 3 / 16 \\  
P(11) = 1 / 16.
\end{align*}
If I tell you that I have selected a binary sequence and the first element is $1$ what is the E($p_i$)?
Well, this can be calculated by looking at each of the jars and adding up the probability of candidate sequences times the value of $p_i$.
Edit: I wasn't normalizing this conditionally space properly. I'm skipping a step which I'll explain, someone wants.
\begin{equation*}
E(p_i) = (4/24 * 1/4) + (8/24 * 1/2) + (12/24 * 3/4) = 14 / 24 = 0.58.
\end{equation*}
So the question is ... what is $E(p_i)$ when the numbers of jars goes to infinity (or alternatively, when $p$ can take on values between $0$ and $1$)? Also what happens when the size of the binary strings goes to infinity? Does it have an effect on the outcome? If it does, does the order we take the limits change the answer?
And most importantly what is the general case for when I have $s$ 1's and $r$ $0$'s?, with a continuous $p$ from $0$ to $1$ and infinite sequences?
 A: First, sticking to strings of length n=2:
$P(10)=p_i (1-p_i)=\frac{i}{m+1}\left (1-\frac{i}{m+1}  \right )=\frac{i(m+1-i)}{(m+1)^2}$ and $P(11)=p_i^2=\left (\frac{i}{m+1}  \right )^2=\frac{i^2}{(m+1)^2}$; 
$E(p_i)=\sum_{i=1}^{m}p_iP(p_i)=\sum_{i=1}^{m}\frac{i}{m+1}\left(\frac{P(10)+P(11)}{\sum_{i=1}^{m}(P(10)+P(11))} \right )$ $=\frac{\sum_{i=1}^{m}\frac{i}{m+1}\left(\frac{i(m+1-i)}{(m+1)^2}+\frac{i^2}{(m+1)^2} \right )}{\sum_{i=1}^{m}\left(\frac{i(m+1-i)}{(m+1)^2}+\frac{i^2}{(m+1)^2} \right )}$ $=\frac{\sum_{i=1}^{m}\frac{i}{m+1}\left(i(m+1-i)+i^2\right )}{\sum_{i=1}^{m}\left(i(m+1-i)+i^2\right )}$ $=\frac{\sum_{i=1}^{m}i^2}{(m+1)\sum_{i=1}^{m}i}$ $=\frac{\left(\frac{m(m+1)(2m+1)}{6} \right )}{(m+1)\frac{m(m+1)}{2}}$ $=\frac{2m+1}{3(m+1)}$; 
$\lim_{m\to\infty}E(p_i)=\lim_{m\to\infty}\frac{2m+1}{3(m+1)}=\frac{2}{3}$
For general n:
$P(1...)=p_i\sum_{k=0}^{n-1}\left({n-1 \choose k}p_i^k(1-p_i)^{n-1-k} \right )=p_i$ (yes, P(1...) does not depend on n at all!);
$E(p_i)=\sum_{i=1}^{m}p_iP(p_i)=\sum_{i=1}^{m}p_i\left(\frac{P(1...)}{\sum_{i=1}^{m}P(1...)} \right )=\sum_{i=1}^{m}p_i\left(\frac{p_i}{\sum_{i=1}^{m}p_i} \right )$ $=\frac{\sum_{i=1}^{m}p_i^2}{\sum_{i=1}^{m}p_i}=\frac{\sum_{i=1}^{m}\left(\frac{i}{m+1} \right )^2}{\sum_{i=1}^{m}\frac{i}{m+1}}$ $=\frac{\sum_{i=1}^{m}i^2}{(m+1)\sum_{i=1}^{m}i}=\frac{2m+1}{3(m+1)}$ (no surprise there, given that $P(1...)$ doesn't depend on n);
$\lim_{m\to\infty}E(p_i)=\lim_{m\to\infty}\frac{2m+1}{3(m+1)}=\frac{2}{3}$
