Symmetry in Probability Around a Particular Phenomenon in Time? This has been hurting my brain substantially, recently. I'm not sure if I'm failing to make connections or if I see connections but am weary of their relevance.
In my text the author claims that events occurring in the future are just as relevant as events before the said event. He writes that
"the results of later draws have precisely the same relevance as do the results of earlier ones! Even though performing the later draw does not physically affect the number Mk of red balls, information about the result of a later draw has the same effect on our state of knowledge about what could have been taken on the kth draw, as does information about an earlier one."
_Probability Theory: The Logic of Science, E.T. Jaynes
Surely this doesn't refer to future events? I can see how knowing that I'll get a red ball in 4 draws would affect my knowledge of the next 3 draws, but where does something like this become prevalent in everyday use?
Is it just a mathematical proof or does it actually hold some value in actual application? Also, correct me if I'm completely missing the point as I've allowed myself to get really confused and it may not even be that complicated.
EDIT: The proof I followed resulted in:
P(Rj|Rk) = P(Rk|Rj) so it makes mathematical sense but I'm still confused on the intuitive aspect.
 A: My interpretation on my frequencist basis: Past and future lose its causal relationship in probability theory.
Consider the following example. 

We have a friend who goes to the pub some evenings. Also, he is in bad mood or in good in some mornings. 

We observe our friend at $N$ evenings and at the following mornings. Our data is collected in a  table:
$$
\begin{matrix}
\text{day}&\text{Evening}&\text{Next Morning}\\
1^{st}&\text{InP}&\text{BM}\\
2^d&\text{InP}&\text{GM}\\
3^d&\text{NInP}&\text{BM}\\
4^{th}&\text{InP}&\text{BM}\\
\vdots\\
N-1^{st}&\text{NInP}&\text{GM}\\
N^{th}&\text{InP}&\text{BM}\\
\end{matrix}
$$
(I hope that the notations are clear.)
Assume that the summary of our experimental results is as follows:


*

*Number of experiments: $N=2000$

*Number of nites spent In the Pub followed by Good Mood in the next morning: $200$.

*Number of nites spent In the Pub followed by Bad Mood in the next morning: $600.$

*Number of mornings in Good Mood $300$

*Number of mornings in Bad Mood $1700$

*Number of nites spent In the Pub $800$

*Number of nites spent Not In the Pub $1200$


Based on our observations we can calculate the following conditional probabilities:
$$P(\text{BM}|\text{InP})=\frac{P(\text{BM}\cap \text{InP})}{P(\text{InP})}=\frac{\frac{600}{2000}}{\frac{800}{2000}}=\frac{3}{4}\tag 1.$$
and
$$P(\text{InP}|\text{BM})=\frac{P(\text{BM}\cap \text{InP})}{P(\text{BM})}=\frac{\frac{600}{2000}}{\frac{1700}{2000}}=\frac{6}{17}\tag 2.$$
These conditional probabilities have a very good frequencist basis: In the future we will measure approximately the same conditional relative frequencies.
So we can make forward predictions and back ward "post"-dictions if we observe our friend in the evening or in the morning. Causality or the direction of the time arrow have nothing to do with our pre- or post-dictions.
A: In the context of the question (drawing coloured balls from an urn without replacement, so with the hypergeometric distribution), the argument is that knowledge of the colour of the $k$th ball drawn affects the conditional probability of the colour of the $j$th ball drawn, whether $j \gt k$ or $j \lt k$.
In a sense this is obvious, in that knowing a ball was drawn later means that it could not have been drawn at the point of interest.  The point being made is not that later draws affect the earlier draw, but that information about later draws affect knowledge about earlier draws.  This is applicable in general, especially in Bayesian analysis.
As a side comment, Jaynes generally made sensible comments about Bayesian probability, though his style appeals to some people more than others.  When he combined this with maximum entropy arguments, his methods were more controversial and less obviously accepted. 
A: Suppose you have say $5$ red balls and $8$ green balls, and some one draws them all at random, without replacement, giving some ordering of all the reds and greens.
If the person tells you that the first ball was red, then from your point of view, the probability that the second ball is red is $\frac{4}{12}$.  
On the other hand, if the person tells you that the second ball is red (with no information about any of the other draws), you would know that the first ball would be one of the other $4$ red balls, or one of the $8$ green balls.  So again from your point of view, the probability of the first ball being red is $\frac{4}{12}$. 
A: Say the urn contains three red balls and three green.
The conditional probability that the second is red, given that the first is red, is $\dfrac 2 5.$
The conditional probability that the first is red, given that the second is red, is also $\dfrac 2 5.$
There is complete symmetry here.
And similarly if one has different numbers of marbles and more trials than the first and second are involved.
Let $X_k$ be the number of red marbles drawn on the $k$th trial, so that $X_k = \text{either $0$ or $1$}$ for $k= 1,2,3,4,5,6.$ Then the sequence $X_1,X_2,X_3,X_4,X_5,X_6$ is what is called an exchangeable sequence of random variables, i.e. no matter what order you put them in, the distribution of the whole sequence remains the same.
