The paradox of calculating the probability when considering a set of trials differently?

Question

Probability theory

It is possible that I do not use the search skillfully enough to find an explanation of this issue available to me. I encountered it myself during the analysis of the foundations of the theory of probability with its seemingly simple model, but its simplicity turned out to be deceptive and the essence of the matter is this.
When tossing a symmetrical coin, we end up with an event consisting of two possible equally probable outcomes heads or tails zero or one to calculate the probable elementary outcomes: heads or tails (0 or 1). To calculate the probability of elementary outcomes of a series of N such trials, we need divide the probability of a true event, taken as a value equal to one, by the number of elementary outcomes of one trial, raised to a power with an indicator N. this is quite obvious, because the complete (reliable) group of events in this case is the Cartesian power of the set of outcomes of one toss
Ω = a^n where a={0,1} and P( Ω )=1
Accordingly: Ω ={(0,0),(0,1),(1,0),(1,1)} with the probability of elementary outcomes p(ω;)=1/||Ω||=1/4 for N=2; Ω ={(0,0,0),(0,0,1),(0,1,0),(0,1,1),(1,0,0),(1,0,1) ,(1,1,0),(1,1,1)} with the probability of elementary outcomes p(ω;)=1/||Ω||=1/8 for N=3; etc. (||.||-set size).
Consider a special case for N=4. Then Ω =a^4, ||Ω||=16 and p(ω;)=1/16
What is the probability of winning in such a game? Let's say we bet that the number of tails ("1") will exceed the number of heads ("0") and successively toss a coin. as a result of simple calculations, we can find out that there are 2 ^ 4 possible sequences, we have 5 outcomes, where the number of tails exceeds the number of heads (that is, their total sum in outcomes)
We also have 5 mutually inverse outcomes and 6 outcomes representing a "draw" in an equal sum of results. hence the probability of winning as well as the probability of defeat will be 5/16, the probability of draws is 6/16. all this seems to be quite obvious, but let's imagine that the course of the game goes beyond the vision of an observer or a player, and we only see its result. that is, we do not have knowledge about the sequence in which the coin fell out, we only know how many times it fell out tails and how many times it fell out heads. Does this change the essence of the game? all actions remain the same, but we see the whole game as one single test. In this case, our outcomes are:
1.0 heads: 4 tails
2.1 heads: 3 heads
3.2 heads: 2 tails
4.3 heads: 1 tails
5.4 heads: 0 tails
These 5 possible outcomes are our complete group of events (Ω). Of these, 2 outcomes are tails, 2 are heads, and 1 is a draw. therefore, the probability of winning in the case of a bet no more frequent tails, in this consideration is - 2/5.
Compared to the previous analysis with consideration of successive outcomes, we can see the following: 5\16 ≠2\5 (win, lose) and 6\16 ≠1\5 (draw).
If the essence of the game does not change depending on its observation and, accordingly, the probability of winning empirically remains the same, then, dear experts, why do two different, but seemingly correct methods of considering an event give different results? there must be a way to link these calculations in order to arrive at a more precise definition and understanding of the term "probability of an event". Or did I do something wrong?

Answer 1

Evaluate only by possible outcomes - this is a 50% chance that a brick will fall on your head in the next second. There are only two outcomes - either it will fall or it will not fall.
You see five possible outcomes. But do not take into account the probability of each of them.
Knowing how the results of these five outcomes are obtained, we add probabilities to them:
1. 0:4 - 1/16
2. 1:3 - 4/16
3. 2:2 - 6/16
4. 3:1 - 4/16
5. 4:0 - 1/16
We get the probability of winning 1/16 + 4/16 = 5/16

Answer 2

These 5 possible outcomes are our complete group of events (Ω). Of these, 2 outcomes are tails, 2 are heads, and 1 is a draw. hence the probability of winning in the case of a bet no more frequent tails, in this consideration is - 2/5.

These are not equally likely outcomes. For example, 0-4 has a probability of 1/16, etc.