1. ## Basketball winning streak question

Can anyone please help me figure out how it was concluded that a basketball team with a projected .200% winning percentage has a 47% chance of winning three straight games at some point during an 82-game schedule?

Thank you.

2. ## Re: Basketball winning streak question

If the probability of winning one game is then assuming the games are independent, which is a big assumption, then the probability of winning two games in a row is and the probability of winning three games in a row is .

The probability not winning 3 games in a row is or .992

If we break up the 82 game into 27 groups of three, the probability of not winning three games in a row 27 times in a row is .992^27 = .805. So the probability of winning 1 or more of those groups is 1 - .805 = .195 or 19.5%

But the team could also win game 2,3,4 or games 3,4,5. SO we can group the 82 games into (1,2,3) (2,3,4) ... (80,81,82). So there are 80 groups of 3 games where the team has an opportunity to win or lose all 3 games.

1 - .526 = .474 or 47.4%

but... (1,2,3) and (2,3,4) are not independent. So I think the answer is between 19.5% and 47.4%

That's about as far as I got.

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4. ## Re: Basketball winning streak question

It is interesting that you are from Philly because the question was actually based upon how the 76ers winning their first three games (I am a Sixer's fan). The reason I said a .200% winning team is because Vegas had them lined at 16.5 season O/U wins.

I am pretty sure you nailed the answer here with:

1 - .526 = .474 or 47.4%.

As I read on a stats blog, under a slightly different context that the answer is 47%.

http://broomonthewarpath.sportsblog...._a_3_game.html

However, I am not sure why each game wouldn't be considered independent though; is that because the 3-game samples bleed into each other?

I love learning the methodology behind these answer and I wish there was some online calculator where I could plug in the numbers and have it spit out the answer

5. ## Re: Basketball winning streak question

If we let to be the event that winning at games,
then we are calculating the union of these 80 non-disjoint event.

Theoretically you can always apply the Inclusion-Exclusion prinicple; but the counting process is tedious.

Anyway, applying the Inclusion-Exclusion principle partially can also gives you upper and lower bounds alternatively, approaching the true answer.
Asterisk has already gives you a start which is the 1st, crude estimate of the required probability. These are the celebrated Bonferroni Inequalities:

http://en.wikipedia.org/wiki/Boole%2...i_inequalities

E.g. for the 1st time we have as the upper bound (the Boole's inequality)

Next, we substract the sum of probabilities of pairwise intersecting events to obtain a lower bound; i.e. substract

As you can see in general the magnitude of the terms decrease as the number of intersection increase and thus the bound is becoming more accurate.

6. ## Re: Basketball winning streak question

Winning (1,2,3) and (2,3,4) are not independent

Let a = probability winning (1,2,3)
Let b = probability winning (2,3,4)

a and b are dependent

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