The realistic probability of winning a best of 7 series (i.e., excluding all unreal and redundant combinations) where the first game is already lost? I want to know the probability of winning a best of 7 series (i.e., win 4 games out of 7) where the first game is already lost. There are some people who advocate taking all possible theoretical combinations, but that doesn't give us the true probability since some combinations are just not possible in real life (redundancies).
Here is a simple example using coin toss.
Example 1: Best two out of three Heads wins the series (no games have been played yet).
Option A1: Include only realistic combinations and remove redundancies:

Wins = {HH-, HTH, THH} ; Losses = {TT-, THT, HTT}
Here, the probability of winning the series is 3/6=1/2.

Option B1: Include all theoretical combinations:

Wins = {HHH, HHT, HTH, THH} ; Losses = {TTT, TTH, THT, HTT}
Here, the probability of winning the series is 4/8=1/2.

As you can see in the above example (when no games have been played yet) the probability of winning the series is the same for both options. So it doesn't really matter if we include the redundancies or not.
However, when the first game is already lost, the two options provide different probabilities of winning the series.
Example 2: Best two out of three Heads wins the series, but the first is already a Tail.
Option A2: Include only realistic combinations and remove redundancies:

Wins = {THH} ; Losses = {TT-, THT}
Here, the probability of winning the series is 1/3.

Option B2: Include all theoretical combinations:

Wins = {THH} ; Losses = {TTT, TTH, THT}
Here, the probability of winning the series is 1/4, not the same as the one before.

Doing this exercise by hand for a 7 game series is going to be a very lengthy process. I want to know how I can do this in R using simple functions or Monte Carlo simulations (no negative binomial).
Here is how I found the probability of winning the series given the first game is lost, but this includes all theoretical combinations/redundancies:
B <- 10000

set.seed(1)

results<-replicate(B,{x<-sample(0:1,6,replace=T)
sum(x)>=4})

mean(results)

Question: How do I find the realistic probability of winning a best of 7 series (i.e., excluding all unreal and redundant combinations) where the first game is already lost?

Summary
Your question is based on an incorrect premise: that impossible paths are somehow statistically invalid. Instead, the probabilities represented by those paths still add up to the probability of getting to the leg that stopped further play.
The simplest way to calculate the answer is to look at binomial distributions of all permutations, including the "impossible" routes you've ignored.
The answer is $\frac{11}{32}$, or $0.34375$, assuming the odds of winning are 50%.
Best 2 of 3 Coin Toss Corrected Examples
Here's the full range of possibilities for a best 2 of 3 coin toss:

The white nodes are undecided states. The blue node wins for heads before finishing all three rounds, while the yellow wins for tails prematurely. The purple and pink nodes are unreachable. The green and orange nodes represent wins decided by the third coin toss.
To find the odds that heads wins, we need to find the probability of each win condition, then add up those probabilities. One way is to look at the bottom row. There are two purple nodes and two green nodes, each of which represents a win. Each one has a 1/8, or 12.5%, probability. Adding the four of them gives us a total probability of 50%.
Another way is to short circuit at each node where the game is already decided. In this case, there are three nodes where heads wins. But note that the odds of each node are not the same. There's a 25% chance of winning at the blue node, then two chances of winning at the green nodes, with a 12.5% chance each. Although there are fewer nodes, then total is still a 50% chance of winning. Of importance is that the odds of getting to the blue node is exactly the sum of odds of each of the purple nodes.
So let's go to the circumstance where tails has already won the first round.

If we look at the bottom row again, we note that heads can only win at the green node with a 25% chance. Tails wins at the other three nodes with a 25% chance each.
If we short-circuit, there are only 3 possible games. But there's a 50% chance tails wins at the yellow node, a 25% chance tails wins at the orange node, and only a 25% chance heads wins at the green node.
While there are only 3 possible games, it's not a 1/3 chance of getting to the green node. This is because the green node requires winning two games in a row, the orange requires winning then losing, and the yellow node only requires losing the first game.
Winning a Best of N Game, Given You've Already Lost A Round
In any best of N game, the win condition is that you've won $\frac{N}{2}+1$ or more rounds. Calculating the odds given existing wins or losses really just boils down to counting how many more rounds there are, and how many you need to win from here.
In a best of 7 game, you need to win at least 4 rounds. If you've already lost 1 of 7, there are 6 more rounds, and you still need to win 4 of them. The math has already been done in Ben's answer, so I'm not going to replicate it here. We can also look at this math.se answer by Graham Kemp to a question nearly identical to yours. It's the same math as Ben's answer uses, but in a different format that might be helpful.

There are $6$ remaining games.  The desired criteria is that Heat wins
at least $4$, when given that Heat lost the first 1.  This is a binomial distribution; so named because of the use of the
binomial coefficient to count number of permutations of outcomes that
match the desired criteria.
The probability of exactly $k$ successes in $n$ trials with
probability $p$ of success in any trial, is:
$${n\choose k}p^k(1-p)^{n-k} \;=\; \,^n\mathrm{\large C}_k\;p^k(1-p)^{n-k} \;=\;\frac{n!}{k!(n-k)!}p^k(1-p)^{n-k}$$
So:
$\mathbb{\large P}(\text{win at least }4\text{ more of }6) = {6\choose 4}\left(\frac 12\right)^4\left(\frac 12\right)^2+{6\choose 5} \left(\frac 12\right)^5\left(\frac 12\right)^1+{6\choose 6} \left(\frac 12\right)^6\left(\frac 12\right)^0$.
$\therefore \mathbb{\large P}(\text{win at least }5\text{ more of }6) = \frac{1}{2^6}\left(\frac{6!}{4!2!}+\frac{6!}{5!1!}+\frac{6!}{6!0!}\right)$

We can plug that last formula into a calculator (such as WolframAlpha) to find that the answer is $\frac{11}{32}$, or $0.34375$.
