How to calculate "at most" with special cases removed? I am trying to calculate the percentage of winning for a certain event but cannot find the right approach or an easier way to exclude special cases. 
Problem:: 
In many Trading Card Games (TCG) players are given the option to enter tournaments that reward them based on the number of wins they can achieve. Because said  tournaments charge an initial entry  fee and only give it  back if one reaches certain wins I am trying to determine the rate at which one comes out even or ahead. 
Tournament Rules::
A player is allowed to play up to incurring 3 losses or 7 wins whichever comes first. I am basing the following math on the probability that a deck has win rate of 50%.
At first I calculated it thru binomial probability, however due to the nature of the problem the number of trials changes. When you go 7-0 technically you don't play the 3 losses so that's 7 successes out of a number of trials of 7 right?
The probability of going 7-0 would then be.
.5 * .5 * .5 * .5 * .5 * .5 *.5 = 0.0078125 
I calculated the probabilities for going 7-1 at number of trials at 8, 7-2 at 9, 6-3 at 9, 5-3 at 8 and so forth until going 0-3 at 3. I know I did something wrong because when I added the probabilities it gave me ~1.57.  
When I asked a friend he suggested that I map out all the combinations for each scenario so I did that. By using Binomial Coefficient I then came up 238 possible combinations:
7-0     [1] 
7-1     [8]
7-2     [36]
6-3     [84]
5-3     [56]
4-3     [35]
2-3     [10]
1-3     [4]
0-3     [1]
Is this the correct approach?
Now even with this approach there are cases in the combinations that are invalid. For example in a 6-3 it gives me the combination of L-L-L-W-W-W-W, while this combination of Loss to Wins is correct it is not valid since the game would kick you out after the 3 losses. 
My head is stumped as to how to find the percentages and if someone could shed some light unto how to approach the problem I would gladly appreciate it. Thanks. 
 A: I'm not sure I understand the problem, but what I think you're saying is that you play a series of games, each of which you have a probability of 1/2 of winning, and the series ends whenever your cumulative score reaches -3 or +7, which ever comes first.
If this interpretation is correct, then another way to look at the problem is that you have 3 coins and "the system" has 7 coins.  You give the system a coin each time you lose, and the system gives you a coin each time you win.  The series is over when either of you runs out of coins.  You would like to know the probability that you win, i.e. you win all the system's coins.
This is a well-known problem in probability called the Gambler's Ruin Problem.  When Player 1 has $n_1$ coins and Player 2 has $n_2$ coins, it turns out that Player 1's probability of winning all of Player 2's coins is
$$P_1 = \frac{n_1}{n_1+n_2} \tag{*}$$
  In your case, we have $n_1=3$ and $n_2=7$, so your probability of winning is $3/10$.
One simple way to see (*) is to imagine that Player 1 has a team of $n_1$ little players on his side, and Player 2 has a team of $n_2$ little players.  Each game, two of the little players are chosen at random to compete, and each has a probability of $1/2$ of winning.  This goes on until the series ends.  Some one of the little players must be the winner of the final game, and by symmetry, all the little players are equally likely.  So the probability that the winner of the final game is one of Player 1's team is $n_1/(n_1+n_2)$.
