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?
 A: I think the crux of the matter is that you are misapplying Laplace's classic definition of probability of an event as the ratio of the favorable cases to the total cases, assuming that all cases are equally likely.
In your example of best two out of three, the first toss being tails, you say there are only three possible cases, THH, THT, TT and that heads wins only in the first case, so the probability of success, applying Laplace's definition is $\frac13$.  The problem is that you have not checked the condition in italics above.
The probability of any $3$-toss sequence is $\left(\frac12\right)^3=\frac18$ whereas the probability of any two-toss sequence is $\left(\frac12\right)^2=\frac14$
In this case, we must take the ratio of the sum of the probability of all favorable cases to the sum of the probabilities of all cases:
$$\frac{1/8}{1/8+1/8+1/4}=\frac14$$
You may recognize this as the formula for the conditional probability that heads wins, given that the first toss is a tail.
As I mentioned in a comment, this is obviously correct; heads wins if and only if the next two tosses are heads, which has probability $\frac14$.
The method you don't like, is much more convenient than the method you suggest, even when carried out correctly.  Suppose that a game is being played where the first player to $21$ points wins and the present score is $11$ to $7$.  What is the probability that the player who is ahead now will win, if each player has a 50% probability of winning each game?  Listing all the possible outcomes would be tedious, to say the least.  But we know the match can't last more than $41$ games, so there are at most $23$ left to play.  We ask for the probability that the leader wins at least $10$ of $23$ games.  There's a lot of arithmetic needed to answer this question, true, but the approach is straightforward, and we can do in on Wolfram Alpha, for example.
A: Your problem seems to be that you're assuming all possibilities are equally likely, but that's not the case.
In option A2, TT- happens more often than THT or THH.  When the second coin is tossed, it comes up as T half the time, and H the other half. So in half of the games played, TT- is the outcome.  On the other hand, you can only get to THT or THH when the second toss is H.  That's already only half of the games, and once you get H on the second toss, you have another coinflip left, so each of he options THT and THH comes up in a quarter of the games.
So you can use either option to write down the winning combinations, but you have to be more careful in thinking about how likely each of them is in order to come up with probabilities.
