Need clarification on Independent Events Probability question In the World Series of baseball, two teams A and B play a sequence of games against each other, and the first team that wins a total of four games becomes the winner
of the World Series. If the probability that team A will win any particular game against team B is 1/3, what is the probability that team A will win the World Series?
I can solve this easily my way, by summing the probabilities of winning on the 4th, 5th, 6th, and 7th game. My textbook offers a different solution and I'm trying to understand why it works. They say that 
$\sum \matrix{7 \\ i=4} \pmatrix{7 \\ i} \left(\displaystyle\frac{1}{3}\right)^i \left(\displaystyle\frac{2}{3}\right)^{7-i} $
Would anyone kindly explain why that is a valid approach?
 A: This problem can be formulated as the equivalent one:
Team A wins the majority of 7 games 
Thus, you compute the probability of winning 4, 5, 6 or 7 games out of 7. This yields the desired result. 
A: I am not a probability guru, but here is how I would read the solution:
Given two independent events, A and B, the probability of A happening is $P(A)$ and the probability of B happening is $P(B)$.


*

*The probability of A happening and B happening is $ P(A) \times P(B) $.

*The probability of B happening or B happening is $ P(A) + P(B) $.


$ \frac{1}{3} $ is the probability of winning a single game by itself.
$ \frac{2}{3} $ is the probability of loosing a single game by itself.
$ 7 \choose i $ (read as "7 choose i") is the number of ways $ i $ games could be won, out of 7 games played.  For example, $ 7 \choose 7 $ must be $ \{ win, win, win, win, win, win, win \} $.  $ 7 \choose 1 $ could be $ \{ win, loss, loss, loss, loss, loss, loss \} $ or $ \{ loss, win, loss, loss, loss, loss, loss \} $, etc.
The summation creates a string of or operations. Also, the $ 7 \choose i $ creates a similar string of or operations.
$ (\frac{1}{3})^i $ is the probability of winning $ i $ games.
$ (\frac{2}{3})^{7-i} $ is the probability of losing $ 7 - i $ games.
Therefore, $ (\frac{1}{3})^i (\frac{2}{3})^{7-i} $ is the probability of winning $ i $ games and losing $ 7 - i $ games.
Given the $ 7 \choose i $, the expression $ {7 \choose i} (\frac{1}{3})^i (\frac{2}{3})^{7-i} $ is the probability of winning some and losing some in one combination, or winning some and losing some in another combination, etc.
Going through the summation, we have:


*

*$ i = 4 $ means the probability of winning 4 games and losing 3 games.

*$ i = 5 $ means the probability of winning 5 games and losing 2 games.

*$ i = 6 $ means the probability of winning 6 games and losing 1 games.

*$ i = 7 $ means the probability of winning 7 games and losing 0 games.


Remember, the $ 7 \choose i $ creates an or condition across all win/loss combinations.  All combined, the expressions cover any possible win/loss conditions.
A: Your approach, as well as the textbook's approach, are equally valid. Here's how.
I'll just reframe the question to save some space and to avoid complexity but still maintaining the nature of the problem.
Suppose instead of 4, the winner is declared as soon as any team wins 3 games.
Let $p=\displaystyle\frac{1}{3}$ and $q=\displaystyle\frac{2}{3}$ such that $p+q=1$
Let W denote that team A wins a game and L denote that team A loses a game.
Following are the scenarios where team A wins (It's not very hard to figure out that playing 5 games are enough in any tournament to find a winning team).
Possible tournaments where A wins exactly 3 games ($\pmatrix{5 \\ 3}$ = 10 tournaments)


*

*W,W,W,L,L

*W,W,L,W,L

*W,W,L,L,W

*W,L,W,W,L

*W,L,W,L,W

*W,L,L,W,W

*L,W,W,W,L

*L,W,W,L,W

*L,W,L,W,W

*L,L,W,W,W
Possible tournaments where A wins exactly 4 games ($\pmatrix{5 \\ 4}$ = 5 tournaments)


*W,W,W,W,L

*W,W,W,L,W

*W,W,L,W,W

*W,L,W,W,W

*L,W,W,W,W
Possible tournaments where A wins exactly 5 games ($\pmatrix{5 \\ 5}$ = 1 tournament)


*W,W,W,W,W


The list mentioned above is the textbook's approach.
P(team A winning) = $\sum \matrix{5 \\ i=3} \pmatrix{5 \\ i} p^i * q^{5-i}$
Notice the bold W? Well, that is your approach. The bold W means that you are not considering the outputs after W (or, the teams don't play once team A wins 3 out of 5 games).
To see how it is equivalent, we'll rearrange the 3 lists into another set of lists (preserving the order number).

* Team A winning on the 3rd game 
1. W,W,W,L,L
 11. W,W,W,W,L
12. W,W,W,L,W
16. W,W,W,W,W

* Team A winning on the 4th game 
2. W,W,L,W,L
13. W,W,L,W,W
4. W,L,W,W,L
14. W,L,W,W,W
7. L,W,W,W,L
15. L,W,W,W,W

* Team A winning on the 5th game 
3. W,W,L,L,W 
5. W,L,W,L,W 
6. W,L,L,W,W 
8. L,W,W,L,W 
9. L,W,L,W,W 
10. L,L,W,W,W 
Let's calculate the probability when team A is winning on the 3rd game (This event occurs when either of 1, 11, 12 or 16 occurs).
P(team A winning on 3rd game) = $p^{3}*q*q+p^{3}*p*q+p^{3}*q*p+p^{3}*p*p$
P(team A winning on 3rd game) = $p^{3}*[q^{2}+2pq+p^{2}]$
P(team A winning on 3rd game) = $p^{3}$
The intuition here is that once team A has won on the 3rd game, the result of the remaining two games does not change the outcome of team A winning the tournament. Imagine telling the players to continue playing the games even after the match has been decided, just for fun.
Similarly, team A winning on 4th game happens when either of 2, 13, 4, 14, 7 or 15 occurs
P(team A winning on 4th game) = $p^{2}*q*p*[p+q] + p^{2}*q*p*[p+q] + p*q*p^{2}*[p+q]$
P(team A winning on 4th game) = $3*p^{3}*q$
Once team A has won on the 4th game, the result of the remaining game does not change the outcome of team A winning the tournament. Imagine telling the players to continue playing the game even after the match has been decided, just for fun.

On the same line, team A winning on 5th game happens when either of 3, 5, 6, 8, 9 or 10 occurs
P(team A winning on 5th game) = $6*p^{3}*q^{2}$

This gives: P(team A winning) = $p^{3} + 3 * p^{3}*q + 6*p^{3}*q^{2}$; which is equivalent to the formula derived using your approach, i.e.,
P(team A winning) = $\sum \matrix{4 \\ i=2} \pmatrix{i \\ 2} p^{3} * q^{i-2}$
Takeaway here is that there is no harm in teams playing all the 5 games (or even more number of games) once the winner is declared in the first 3 or 4 or 5 games because, for example, P(Team A will win the tournament by playing n more games | Team A has already won first 3 games) = 1
For more clarification refer to section 4 on page no. 9 here.
