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?

If you assume that each game is independent, with an equal probability of each outcome, then the problem is quite simple (and as whuber points out in the comments, this is the classical "problem of points").  The solution is computed using the binomial distribution, which does consider all possible cases.
Under these assumptions, you can proceed easily for the general case where we have two players playing a tournament which is best-out-of-$n$ games (where $n$ is a positive odd number), where $0<\theta<1$ is the fixed probability that the first player will win a game.  Suppose that the current score is $p_1$ points to the first player and $p_2$ points to the second player (so we must have $p_1+p_2 \leqslant n$ with both values non-negative).  Let $X \sim \text{Bin}(n-p_1-p_2, \theta)$ denote the number of remaining games that will be won by the first player.  The total points for the first player will be $p_1+X$ and the total points for the second player will be $n-X-p_1$.  To facilitate our analysis, we define the bound:
$$B \equiv B(n,p_1) \equiv \Big\lfloor \frac{n}{2} - p_1 \Big\rfloor.$$
The probability that the first player wins is:
$$\begin{align}
\mathbb{P}(\text{Player 1 wins} | p_1,p_2) 
&= \mathbb{P}(X+p_1 > n-X-p_1) \\[16pt]
&= \mathbb{P}(2X > n-2p_1) \\[16pt]
&= 1-\mathbb{P}(2X \leqslant n-2p_1) \\[16pt]
&= 1-\mathbb{P} ( X \leqslant B(n,p_1) ) \\[12pt]
&= 1-\sum_{x=0}^{B(n,p_1)} \text{Bin}(x|n-p_1-p_2, \theta). \\[6pt]
&= 1-\sum_{x=0}^{B(n,p_1)} {n-p_1-p_2 \choose x} \theta^{x} (1-\theta)^{n-p_1-p_2-x}. \\[6pt]
\end{align}$$
It is simple to implement this as a function in R as follows.  (I have included checks on the inputs, but these can be removed if you prefer a more parsimonious version of the function.)
problem.of.points <- function(n, p1 = 0, p2 = 0, prob = 0.5, player = 1) {
  
  #Check inputs n, p1 and p2
  if (!is.numeric(n))        stop('Error: Input n must be a non-negative integer')
  if (!is.numeric(p1))       stop('Error: Input p1 must be a non-negative integer')
  if (!is.numeric(p2))       stop('Error: Input p2 must be a non-negative integer')
  if (length(n) != 1)        stop('Error: Input n must be a single value')
  if (length(p1) != 1)       stop('Error: Input p1 must be a single value')
  if (length(p2) != 1)       stop('Error: Input p2 must be a single value')
  nn  <- as.integer(n); pp1 <- as.integer(p1); pp2 <- as.integer(p2)
  if (n != nn)               stop('Error: Input n must be a non-negative integer')
  if (p1 != pp1)             stop('Error: Input p1 must be a non-negative integer')
  if (p2 != pp2)             stop('Error: Input p2 must be a non-negative integer')
  if (n < 0)                 stop('Error: Input n must be a non-negative integer')
  if (p1 < 0)                stop('Error: Input p1 must be a non-negative integer')
  if (p2 < 0)                stop('Error: Input p2 must be a non-negative integer')
  if (p1+p2 > n)             stop('Error: Cannot have p1 + p2 > n')
  if (!is.numeric(prob))     stop('Error: Input prob must be a probability')
  if (length(prob) != 1)     stop('Error: Input prob must be a single probability')
  if (prob < 0)              stop('Error: Input prob must be a probability')
  if (prob > 1)              stop('Error: Input prob must be a probability')
  if (!(player %in% c(1,2))) stop('Error: Input player must be 1 or 2')
  
  #Compute and return probability
  if (p1+p2 == n) {
    PROB <- as.numeric(p1 > p2) }
  if (p1+p2 < n)  {
    SIZE  <- n-p1-p2
    BOUND <- floor(n/2-p1)
    PROB  <- 1-pbinom(BOUND, size = SIZE, prob = prob) }
  if (player == 2) { PROB <- 1-PROB } 
  PROB }

For the case of a best-out-of-seven game with equal win probabilities for each player, and one lost game for the first player, the win-probability for the first player is:
problem.of.points(n = 7, p1 = 0, p2 = 1)
[1] 0.34375

The above gives a full solution to the classical problem of points, where the win probability is assumed to be known.  However, in practical cases, it is unrealistic to assume that the win probability of a game is known.  A more realistic approach is to treat the win probability as unknown, and use the observed outcomes of the initial games to make an inference about the unknown win probability in each game and the corresponding win probability for the set of games.  The simplest way to do this is with a Bayesian model where we use a beta prior for the win probability, leading to a binomial-beta model.
