I have following formula for calculation of expected value:
M(X)=np, where "n" is number of experiments and "p" is probability of success. I have fundamental problem with this formula: I always thought that expected value does NOT change with number of experiments, it's knowledge that we have in advance. For example, expected value for a six-sided dice is 3.5, and no number of experiments will change this. We can throw the dice once, five times, million times, etc, it won't change anything. We can even refuse to throw it at all, the expected value will be the same nonetheless.
I also have alternative formula that seems to share my sentiments: M(X)=Σxp
Thus: 1/6(1+2+3+4+5+6)=3.5
Now let's assume that we want to get 6, we'll count it as "success". Let's assume that I will throw the dice only once, it means n=1. Clearly, "p" (chance of success) is 1/6. M(X)=1*(1/6)=1/6.
This seems like blatant contradiction for me.
You're confusing two related things. The formula $np$ gives you the expected number of successes in $n$ trials. The formula $\sum xp(x)$ gives you the expected value of a single trial. These two things coincide for events with two outcomes and one trial, but in all other circumstances differ.
Consider a coin: The expected value of a single trial is obviously $1/2$. But if we flip it $1000$ times we expect to get $500$ heads.
You're also comparing apples to oranges. The expected number of successes in 1 roll of a die under your circumstances is in fact $1/6$. When you turn the die into a success/failure experiment in the way you describe, you're revaluing the numbers $1-5$ as "failure" rather than their numerical value. In your comparison, you failed to do this. If you assign a value of $0$ to $1-5$ and a value of $1$ to $6$, both formulae will yield $1/6$.