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

sincsenekdq 2022-09-11 Answered
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 ( 4 0 ) ( 4 1 ) ( 4 2 ) ( 4 3 )
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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Answers (2)

Konner Parker
Answered 2022-09-12 Author has 12 answers
Step 1
If A wins after exactly k { 4 , 5 , 6 , 7 } 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 1 3 .
The probability that A wins 3 of the k 1 matches that precede match k is ( k 1 3 ) ( 1 3 ) 3 ( 2 3 ) k 4 .
Step 2
So if p k denotes the probability that A wins after exactly k { 4 , 5 , 6 , 7 } matches then:
p k = ( k 1 3 ) ( 1 3 ) 4 ( 2 3 ) k 4
The probability that A wins is:
k = 4 7 p k

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inhiba5f
Answered 2022-09-13 Author has 2 answers
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 ( 7 4 ) ( 1 / 3 ) 4 ( 2 / 3 ) 3 . The probability it wins 5 is ( 7 5 ) ( 1 / 3 ) 5 ( 2 / 3 ) 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 ( 1 / 3 ) 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 ( 4 3 ) ( 1 / 3 ) 3 ( 2 / 3 ) 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) ( 5 3 ) ( 1 / 3 ) 3 ( 2 / 3 ) 2 ( 1 / 3 ).
Now write down the probability of winning in 7 games, and add up.

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