Shortcut trick for solving probabilities This question is a follow up from this answer + comments which I am really in love with. 
Following Steve's suggestion from the comments I decided to try it, even though at first I didn't believe that it can be the correct way.
Suppose that we have an urn with $5$ white balls and $7$ black balls. What is the probability that on the third extraction we get a black ball?
The first and the standard way that I've learn to this is by using Total Probability law.
I will denote $3$b as the third ball to be black, $2w=$ second ball to be white and so on.
$$P(3b)=P(3b,2b,1b)+P(3b,2b,1w)+P(3b,2w,1b)+P(3b,2w,1w)$$
$$=\frac{7}{12}\cdot \frac{6}{11}\cdot \frac{5}{10}+\frac{5}{12}\cdot \frac{7}{11}\cdot\frac{6}{10}+\frac{7}{12}\cdot\frac{5}{11}\cdot\frac{6}{10}+\frac{5}{12}\cdot\frac{4}{11}\cdot \frac{7}{10}$$
$$=\frac{210+210+210+140}{12\cdot 11\cdot 10}=\frac{3\cdot21+14}{132}=\frac{77}{132}$$
But the probability of drawing the black ball first is actually: $$P(1b)=\frac{7}{12}=\frac{7}{12}\cdot \frac{11}{11}=\frac{77}{132}$$
Which is equal to the result found the hard way.
Why does this phenomen happen? I can see that numerically it works, but I can't find an intuition behind it.
Does this shortcut trick have a name? In my school I never encountered this type of solving probabilities.
Also could you suggest me some more tricks like this? 
 A: Suppose you extract all of the balls and line them up. 
The probability that the third ball will be black will be the same in this situation, as it will for when we stop after drawing it.
Now, every individual ball will have an equal probability for being the third ball as for being the first ball.
Seven of the twelve balls are black. 
So the probability that the third ball will be black is $7/12$.
A: It is just due to the fact that the fact you already drew two balls doesn't matter because you didn't look at them.  You can imagine drawing three balls, then only looking at the third.  How is it different from drawing one and looking at it?  So the chance it will be black is the same for the third ball as the first.
A: Here’s an example that might help illuminate the idea in my answer to the other question. It’s somewhat convoluted, but maybe it’s fun to think about even if it doesn’t help.
Suppose an urn contains $10$ white balls and $11$ black balls. You draw a ball from the urn, but unbeknownst to you, someone has previously chosen one of the urn’s $21$ balls at random and cast a magic spell called $S(12,13)$ on it. The spell may cause the ball to change color one week after it’s removed.
Specifically, the spell $(12,13)$ causes a ball to end up white after one week with probability $12+r\over12+13+1$, where $r=0$ if the ball is initially black and $r=1$ if the ball is initially white. (It’s a weird spell, but if you read the question that began this discussion, you may see where I’m going with it.)
What is the probability $p$ that the ball you draw will be white a week after you draw it? 
Clearly, the probability that the ball is white when you draw it is $10\over21$, because the spell does nothing, and $10$ of the urn’s $21$ balls are white.
Denote by $M$ the event that you drew a magic ball, and denote by $W$ the event that you draw an initially-white ball. These are independent events, since the spell and your draw each target a random ball.
By the law of total probability, $p=P(M')P(W) + P(M)x$, where $x$ is the probability that a magic ball drawn from an urn with $10$ white and $11$ black balls will be white in a week.
What is $x$? If black magic balls are white in a week with probability $12+0\over26$ and white magic balls are white in a week with probability $12+1\over26$, then a ball drawn from an $10$-white-and-$11$-black distribution is white in a week with probability $12+{10\over21}\over26$.
Thus $p = P(M')P(W) + P(M)x={20\over21}\cdot{10\over21}+{1\over21}\cdot{12+{10\over21}\over26}={2731\over5733}\approx 0.4763649$.
