A non standard way to a probability problem We have teams $A$ and $B$.Every team has equal chance to win. A team wins after they have won $4$ matches. Team $A$ won the first match, what is the probability, that team $B$ won the whole game.
My try:
So team $A$ needs $3$ more wins to win the game. Team $B$ needs $4$ more wins to win the game. So the winner will be determined in maximum of $7$ games. So now I tried to do it my own way (I dont know if it is usual to do it this way)
Let's say that we are looking at the wins of team $A$. (They need $3$ more wins).
$$...A_i,...A_j,...A_n$$,
where $3$ dots symbolize the wins of team $B$. So $$(i-1) + (j-i-1) + (n-j-i-1) \leq 3 $$
(Because there must not be more than 3 losses, because at 4 team $B$ wins the game )
we get:
$$n-i \leq 6 \quad \text{and} \quad 0<i<n $$
If we look what integer tuples can satisfy the upper inequalities we get: $$N = \{(1,6),(2,6), (3,6),(4,6),(5,6),\\(1,5),(2,5),(3,5),(4,5),(1,4),(2,4),(3,4),(1,3),(2,3),(1,2)\}$$
Where $N$...possible winnings of team $A$
So we can count the number of elements ($15$) an that would be our number in the numerator, where in the denominator is the number of elements in our sample space.
To calculate the sample space I would do:
$$|\omega| = 1 + (\binom{4}{3} - 1)\\ + (\binom{5}{3} - \binom{4}{3} - 1)\\ + (\binom{6}{3} - (1 + (\binom{4}{3} - 1) + (\binom{5}{3} - \binom{4}{3} - 1)))\\ + (\binom{7}{3} -(1 + (\binom{4}{3} - 1)+ (\binom{5}{3} - \binom{4}{3} - 1) + (\binom{6}{3} - (1 + (\binom{4}{3} - 1)+ (\binom{5}{3} - \binom{4}{3} - 1)))) )$$
I guess we could also write this with a recursive formula. However we get $|\omega|=35$.
(What I did here is that $1$ is for $3$ consecutive wins, $\binom{4}{3} - 1$ are other possible matches in 4 games and so on.)
So if my procedure is correct the probability is $$P(\text{team B wins the whole game}) = 1 - \frac{|N|}{|\omega|} = 1 - \frac{15}{35} = \frac{20}{35}$$
However is my process atleas partially correct (I do not know the solution).

To more simply calculate the odds of this problem, we can relate it to flipping a coin. A heads gives team $A$ the victory, while a tails gives team $B$ the victory. To solve the problem, we are looking for the probability that at least $4$ of $6$ flips are tails. This is not difficult to calculate.
Total possible outcomes: $2^6=64$
6 tails: $_6C_6=1$ way; $\dfrac1{64}$
5 tails: $_6C_5=6$ ways; $\dfrac{6}{64}$
4 tails: $_6C_4=15$ ways; $\dfrac{15}{64}$
Total: $\dfrac{1+6+15}{64}=\dfrac{22}{64}=\dfrac{11}{32}$. This gives a $34.375\%$ chance that team B manages to win the tournament.
