Probability...why are they different? Suppose that two teams are playing a series of games, each of which is independently won by team A with probability $p$ and by team B with probability $1-p$. The winner of the series is the first team to win $i$ games. 
If $i=5$, find the probability that a total of $8$ games are played. 
I got this solution for this one: $7 \choose 4$$p^5(1-p)^3$+ $7 \choose 4$ $p^3(1-p)^5$
But when $i=4$ and I have to find the probability that a total of $7$ games are played, I tried solving using the same method for the previous question, but it wasn't correct. The correct solution is supposed to be $6 \choose3$$p^3(1-p)^3$.
I don't understand why they are different,, can anyone clarify?
Thanks
 A: If a total of 7 games are played, before the seventh game, they must be tied at 3-3. What are the odds of that? This is the same question as "If 6 games are played, what is the chance team A wins 3 of them?", which, as you noted, is $\binom{6}{3} p^3(1-p)^3$.
A: If you try to solve the problem the same way, you get:
$$\binom{6}{3}p^4(1-p)^3+\binom{6}{3}p^3(1-p)^4 \,.$$
Now, $\binom{6}{3}p^3(1-p)^3$ is a common factor and you get
$$\binom{6}{3}p^4(1-p)^3+\binom{6}{3}p^3(1-p)^4 = \binom{6}{3}p^3(1-p)^3\,.$$
The results only look different ;)
You can actually reach the same conclusion faster: the probability that a team wins in 7 is the same as the probability that after $6$ games the score is 3-3.
A: From your description, I’m going to guess that you computed the answer to be
$$\binom63p^4(1-p)^3+\binom63p^3(1-p)^4\;.$$
If so, there’s nothing wrong with it except a failure to simplify it:
$$\begin{align*}
\binom63p^4(1-p)^3+\binom73p^3(1-p)^4&=\binom63p^3(1-p)^3\Big(p+(1-p)\Big)\\\\
&=\binom63p^3(1-p)^3
\end{align*}$$
What’s different about this problem is that no matter which team wins, after $6$ matches each team must have won $3$ times, so there’s no need to split the cases in which Team A wins from those in which Team B wins.
