Intuitional Nash mixed equilibrium explanation I would like to get some help in understanding the underlying logic of a mixed Nash equilibrium. I'll present how I have perceived it - not in math language, but still (I hope) in a well-defined manner.
When I started reading about the nonmixed Nash equilibrium, I had no problems at all understanding both the logic and its implementation. Long story short, try to have the best result for each possible case.
Things changed with the mixed strategies though. Obviously they were created for the cases that an answer could not be provided by the nonmixed theory. But I cannot understand the following core concept:

When no clear strategy exists, spread out probability in such a way
  that [expected result]*[probability used] is the same for each case.

I have little doubt that I have understood this point well. It looks super arbitrary to me, without any logic. Now to be fair, it is well defined, and thus less arbitrary than...nothing. Still, when talking about optimization, we usually have very clear goals. Isn't the Nash equilibrium supposed to make us..."win"?
Let me strengthen my statement with the following example. I think it's called something along the lines of "clean synchronization". A journalist selects two random people from the road. He gives them a piece of paper, and without any form of communication, they are to write an "A" or "B" and give it to the journalist. 


*

*If both give "A", each of them gets 100 dollars

*If both give "B", each of them gets 900 dollars

*If they give different, they both get nothing


We need a mixed strategy here, as there is no selection that is better in both cases - when opponent gives A, you should give A for the maximum result; and vice Versa for B.
So what is the mixed strategy? Implementing the rule I've quoted, we come to the following result:
90% of the time select "A". 0.9 * 100 = 90
10% of the time select "B". 0.1 * 900 = 90
This is very counterintuitive (read: wrong) for me. I find it almost certain that both players would choose B in reality. And note here, that by "reality" I am not implying the part that people are emotional or unpredictable and so on - just some mathematically-inclined thinking and wanting the best result, which seems fully compatible with the assumptions taken in game theory.
This theory is too established to be so easily disproven, so I am obviously missing something. Maybe a domain of application? Can you please clear this out?
 A: What the Wikipedia article you mention in comments says is not exactly what's presented as a quote in this question. The article says (my emphasis):

In order for a player to be willing to randomize, their expected payoff for each strategy should be the same.

If we consider your game, there are two pure Nash equilibria: both choose A or both choose B. There is also a mixed Nash equilibrium, where (as you calculate) both choose A with probability 0.9 and B with probability 0.1. (Despite the asymmetry, this is rather similar to the driving game of the Wikipedia article).
However, the mixed Nash equilibrium is not inherently "better" in some way than the pure Nash equilibria which also exist.
The uniformity of expected payoff allows you to find the mixed equilibrium. That's all it does. In this case, since one of the Nash equilibria is best for each of the players individually, if they're both rational then they will both play the strategy which corresponds to that equilibrium, which is B.
(This is assuming that they both reckon their payoff simply as the money they will receive from the journalist).
