# Binomial probability for sporting event where the order of the first 3 outcomes does not matter

Binomial probability for sporting event where the order of the first 3 outcomes does not matter
There is a sporting event where team A has 1/3 chance of winning and team B has a 2/3 chance of winning. In order to win a team needs to be the first to win 4 matches. There are no ties.
The question: What is the probability that team A will win?
My thoughts:
It will take no more than 7 matches for team A or B to win.
Neither team will win within 3 matches, so the order for the first 3 matches does not matter.
The possible scores where A wins are $\left(4-0\right)$ $\left(4-1\right)$ $\left(4-2\right)$ $\left(4-3\right)$
So I need to find the binomial probability for each possibility and sum them. However, the order of the first 3 matches does not affect the outcome, so they should not be included in the nCr part.
That's where I'm stuck as to how I need to write out nCr for each possibility whilst taking into account that the order of the first 3 matches does not matter.
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Konner Parker
Step 1
If A wins after exactly $k\in \left\{4,5,6,7\right\}$ matches then A must win match k and must win 3 of the $k-1$ matches that precede match k.
The probability that A wins the k-th match is $\frac{1}{3}$.
The probability that A wins 3 of the $k-1$ matches that precede match k is $\left(\genfrac{}{}{0}{}{k-1}{3}\right){\left(\frac{1}{3}\right)}^{3}{\left(\frac{2}{3}\right)}^{k-4}$.
Step 2
So if ${p}_{k}$ denotes the probability that A wins after exactly $k\in \left\{4,5,6,7\right\}$ matches then:
${p}_{k}=\left(\genfrac{}{}{0}{}{k-1}{3}\right){\left(\frac{1}{3}\right)}^{4}{\left(\frac{2}{3}\right)}^{k-4}$
The probability that A wins is:
$\sum _{k=4}^{7}{p}_{k}$

inhiba5f
Step 1
We describe two different approaches. Assume independence (this is not reasonable, in many games there is a home rink/field advantage).
Way 1: Imagine that whatever happens, the teams keep on playing until 7 games have been played played. Then Team A wins the real Stanley Cup finals if and only if in the modified series A wins 4 or more games of the 7 games. The probability A wins 4 games in the modified series is $\left(\genfrac{}{}{0}{}{7}{4}\right)\left(1/3{\right)}^{4}\left(2/3{\right)}^{3}$. The probability it wins 5 is $\left(\genfrac{}{}{0}{}{7}{5}\right)\left(1/3{\right)}^{5}\left(2/3{\right)}^{2}$. And so on. Add up.
Step 2
Way 2: We look at the real series, and calculate the probability the series lasts 4 games and A wins it, the probability the series lasts 5 games and A wins it, and so on up to 7.
The probability A wins in 4 is $\left(1/3{\right)}^{4}$.
For winning in 5 games, A must win 3 of the first 4, and then win the fifth. The probability of winning 3 of the first 4 games is $\left(\genfrac{}{}{0}{}{4}{3}\right)\left(1/3{\right)}^{3}\left(2/3{\right)}^{1}$. Multiply by 1/3 to find the probability of winning in 5 games.
To win in 6 games, A must win 3 of the first 5, and then win. The probability is (53)$\left(\genfrac{}{}{0}{}{5}{3}\right)\left(1/3{\right)}^{3}\left(2/3{\right)}^{2}\left(1/3\right)$.
Now write down the probability of winning in 7 games, and add up.