If Teams A & B are contesting in a 7-game tournament where the winner must win 4 games overall and Team A has probability $p$ of winning against Team B,
(1) What is the probability that Team A will win in $6$ matches?
(2) What is the probability that Team A will win?
(3) If the same teams contested in a game of $n$ matches where the winner must win $k$ matches, where $k=\displaystyle\left\lfloor \frac{n}{2}\right\rfloor+1$, what is the probability that Team A will win?
For the (1) part, for Team A to win in $6$ matches, it must have won the other $3$ and lost $2$. This can be achieved in $\displaystyle{5 \choose3}=10$ ways. And because order is irrelevant, the overall probability of the first setup should be $\displaystyle{5 \choose 3}p^3(1-p)^2$, right? And then isn't the probability of Team A winning $\displaystyle{5 \choose 3}p^4(1-p)^2$? Something seems off here, but I can't quite put my finger on it.
For the (2) part, I shall look at the cases:
- Team A wins 4 times straight: Then the probability is $p_1$=$p^4$
- Team A loses once and wins the wins 4 times (order is irrelevant, but the loss cannot be the last match): Then the probability is $p_2=\displaystyle{4 \choose 3}p^4(1-p)=4p^4(1-p)$
- Team A loses twice and wins 4 times (again, order is irrelevant): $p_3=\displaystyle{5 \choose 3}p^4(1-p)^2=10p^4(1-p)^2$
- Team A loses thrice and wins 4 times (order is irrelevant): $p_4=\displaystyle{6 \choose 3}p^4(1-p)^3=20p^4(1-p)^3$
Then the overall probability is $p_f=\displaystyle\sum_{i=1}^{4}{p_i}=p^4(1+4(1-p)+10(1-p)^2+20(1-p)^3)$
For part (3), I am not really sure what to do. I believe that I must do something with the Binomial Distribution once again, but I am at a loss. Thank you in advance for your time and effort.
Best Answer
You may consider adopting a simpler formulation by assuming that the matches are fully played out regardless of the fact that team $A$ has already won, because subsequent wins/losses can't affect the result.
Taking $p,q$ for win, loss of team A, for part two, the only point you need to remember is that team $A$ can't be allowed to lose more than $3$ matches, thus
$$P(A\; wins) \large{= \sum_{i=0}^3\binom7{i}q^ip^{7-i}}$$
Similarly, for part three, Team A can afford to lose a maximum of $\lceil{(\frac{n}{2}\rceil} -1$ matches, thus
$$P(A\;wins)=\large{ \sum_{i=0}^{\lceil{\frac {n}{2}\rceil}-1}\binom{n}{i}q^ip^{n-i}}$$
Added note
If playing on after team A has already won seems paradoxical, look here where it has been explained in greater detail