Sufficient Statistics - Relating the Intuition with the Mathematical Definition I believe the heuristic definition of a Sufficient Statistic makes sense to me - when you take a sample in order to make an inference about the parameter related to the probability distribution, and you reduce that realised sample to a (sufficient) statistic - you do not lose any information or "inference power" regarding the parameter. (However, we still do not know what the actual true value of the parameter is, but I could make an estimate of the range of values that the parameter may be in just as well with the statistic as I could with the original sample).
(Please correct me if you feel there is anything inaccurate about what I said above)
However where I am struggling to connect the dots is how the mathematical definition of Sufficient statistics is saying same thing.

Definition: A statistic $t = T(X)$ is sufficient for underlying parameter $θ$ precisely if the conditional probability distribution of the data $X$, given the statistic $t = T(X)$, does not depend on the parameter $θ$. (Wikipedia)

Why does $\theta$ disappearing in the conditional probability distribution mean the same thing as $T(X)=t$ and $X=x$ having the same "inference power" about $\theta$.
The intuition I have interpreted about this is "Knowing $T(X)$ means you know everything about $\theta$ from the information supplied by the sample $X$" - but my confusion is why does this mean that the conditional distribution should not have $\theta$ once we have $T(X)$ -> it's not like we know the actual value of $\theta$ when we obtain $T(X)$, right? I just imagined $T(X)$ is all that is required when we try to make an inference - a "guess" about the range of values $\theta$ may be in. But what does this have to do with the conditional probability being independent of $\theta$?
 A: I think a common way to motivate the mathematical definition is the following. 
Say you have the sufficient statistic $T(X)$, and I just have the data/random sample $X$.
By the mathematical definition of sufficiency, 
$$ p(X|T(X), \theta) = P(X|T(X)).$$
The r.h.s. is a probability distribution you, again per definition and in theory, know, and can use to do inference. In particular, if you draw a sample $X' \sim  P(X'|T(X))$, then this sample is also a draw from the distribution of interest, in particular, the distribution that depends directly on $\theta$. But you have now recovered a sample using just the sufficient statistic that contains as much information as the original sample $X$. Thus just using your sufficient statistic you can infer just as much (confidence intervals, for example) as me who enjoyed access to the full sample $X$, and it is therefore sufficient to know $T(X)$. 
A: 
"A statistic t=T(X) is sufficient for underlying parameter θ precisely if the conditional probability distribution of the data X, given the statistic t = T(X), does not depend on the parameter θ."

If the sampling distribution for some data $X$ does not depend on $\theta$ then how can that data say anything about $\theta$? 
It would be like estimating some value by observing something irrelevant (that does not depend on the value to be estimated). 
This is a general statement. In this case we have more specifically as data 'the rest of the data conditional on the sufficient statistic'. And that is confusing because the sample distribution for the rest of the data does depend on the parameter to be estimated. It is only that the conditional distribution of that data does not depend on the parameter to estimated.

example 3 (different outcomes of data, but with the same probability for a given $\theta$) 
(edit: based on the comments I came up with a much more simple/intuitive explanation)
Say you do a urn problem trying to estimate the fraction of blue balls in an urn. You perform an experiment by drawing balls with replacement. 
Say you got "$x_1 = red$, $x_2 = blue$, $x_3 = red$, $x_4 = blue$, $x_5 = blue$"
That is 3 blue balls in total (the total is the sufficient statistic). You could base on this a point estimate of 0.6 for the fraction of blue balls in the urn. (in reality you should take a bigger sample if you want to make a confidence interval with a narrow bandwith, but that makes this example difficult to write down) 
Now, does it matter (for the fraction) which particular balls $x_i$ where blue (beyond the fact that we already know the total number 3)? Is the estimate gonna be different for "$x_1 = blue$, $x_2 = red$, $x_3 = red$, $x_4 = blue$, $x_5 = blue$" or any other different observation that also has 3 blue balls in total? Each of these outcomes, with a total of 3 blue balls, are equally possible. So they will not give more information about the fraction of balls in the urn.
We could tabulate all the different outcomes and how the probability of observing them depends on $\theta$ (the fraction of blue in the vase)
observation        probability of observing given theta

bbbbb              (1-theta)^0(theta)^5

rbbbb              (1-theta)^1(theta)^4
brbbb              (1-theta)^1(theta)^4
bbrbb              (1-theta)^1(theta)^4
bbbrb              (1-theta)^1(theta)^4
bbbbr              (1-theta)^1(theta)^4

rrbbb              (1-theta)^2(theta)^3
rbrbb              (1-theta)^2(theta)^3
rbbrb              (1-theta)^2(theta)^3
rbbbr              (1-theta)^2(theta)^3
brrbb              (1-theta)^2(theta)^3
brbrb              (1-theta)^2(theta)^3
brbbr              (1-theta)^2(theta)^3
bbrrb              (1-theta)^2(theta)^3
bbrbr              (1-theta)^2(theta)^3
bbbrr              (1-theta)^2(theta)^3

rrrbb              (1-theta)^3(theta)^2
rrbrb              (1-theta)^3(theta)^2
rbrrb              (1-theta)^3(theta)^2
brrrb              (1-theta)^3(theta)^2
rrbbr              (1-theta)^3(theta)^2
rbrbr              (1-theta)^3(theta)^2
brrbr              (1-theta)^3(theta)^2
rbbrr              (1-theta)^3(theta)^2
brbrr              (1-theta)^3(theta)^2
bbrrr              (1-theta)^3(theta)^2

brrrr              (1-theta)^4(theta)^1
rbrrr              (1-theta)^4(theta)^1
rrbrr              (1-theta)^4(theta)^1
rrrbr              (1-theta)^4(theta)^1
rrrrb              (1-theta)^4(theta)^1

rrrrr              (1-theta)^5(theta)^0

Notice that in this table above there are groups of potential observations/outcomes for which the probability to be observed have exactly the same dependency on $\theta$. This means that it doesn't matter whether you observe rbrbb or brrbb, they relate to $\theta$ in the same way. All the observations with three blue balls can be considered to provide the same information about $\theta$. 
This is sort of what the sufficient statistic does. It groups together the observations whose Likelihood dependency on $\theta$ is the same.

I have deleted examples 1 and 2, because it makes the post very large, but you can still see them in the history of this post
A: Here is a very simple example that might make things clear.  For a normal model $X\sim N(\theta,1)$ with a sample size of $n=1$, $X$ is sufficient for $\theta$.  Without conditioning on $T(X)=X$, then $P(X\le x)=\Phi(x-\theta)$ depends on $\theta$.  If I condition on $T(X)=X=c$ then
$$
P(X\le x|X=c)=\left\{\begin{array}{ll}
0 & \text{if } x < c\\
1 & \text{if } x \ge c
\end{array}
\right.$$
does not depend on $\theta$.  This shows that once $T(X)$ is known there is no other information available for estimating $\theta$.  If there was additional information available then $P(X\le x|T(X)=c)$ would depend on $\theta$ just like $P(X\le x)$ did (and certainly would not be degenerate).  Let me known if I have made any mistakes.
