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    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.

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    Re: Basketball winning streak question

    If the probability of winning one game is \frac{1}{5} then assuming the games are independent, which is a big assumption, then the probability of winning two games in a row is \frac{1}{25} and the probability of winning three games in a row is \frac{1}{125}.

    The probability not winning 3 games in a row is \frac{124}{125} 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.

    .992^{80} = .526

    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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    Re: Basketball winning streak question

    Thank you for the reply Asterisk as it was extremely helpful.

    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

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    Re: Basketball winning streak question

    If we let E_i to be the event that winning at i, i+1, i+2 games, i = 1, 2, \ldots, 80
    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 80 \times \frac {1} {5^3} as the upper bound (the Boole's inequality)

    Next, we substract the sum of probabilities of \binom {80} {2} pairwise intersecting events to obtain a lower bound; i.e. substract

    79 \times \frac {1} {5^4} + 78 \times \frac {1} {5^5} + \left(\binom {80} {2}  - 79 - 78\right)\times \frac {1} {5^6}

    from the above answer.

    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.

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    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)

    P(a) = 1/125

    P(b) = 1/125

    P(b|a) = 1/5

    P(b) <> P(b|a) --> a and b are dependent

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