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?

Given that the first is a loss (T), we have one chance to win a 5-game series (THHHH), four chances to win a 6-game series (TTHHHH THTHHH THHTHH THHHTH), and ten chances to win a 7 game series.
That's $$(8/32)(1/4) + (20/64)(4/10) + (40/128)(10/20)\\ 
=  0.34375.$$
Simulation in R. Using a 'for' loop in hopes the steps will be clear.
It gives $0.344 \pm 0.001.$
set.seed(2021)
B = 10^6;  x.win = last = first.x = numeric(B)
for(i in 1:B) {
 x = cumsum( sample(0:1,7,rep=T) )
 y = 1:7 - x
 first.x[i] = x[1]
 Last =  min(which(pmax(x,y)==4))
 x.win[i] = (x[Last]==4)
 last[i] = Last
}
mean(last)
[1] 5.812322  # avg nr games played (evenly matched)
4*(2/16)+5*(8/32)+6*(20/64)+7*(40/128)
[1] 5.8125

mean(x.win[first.x==0])
[1] 0.3438682 
2*sd(x.win[first.x==0])/1000
[1] 0.0009499965

