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B Stat Problem that I solved using simulations; verify answer

  1. Nov 5, 2016 #1

    rollingstein

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    Someone posed this Stat / Probablity problem to me:

    Two teams A vs B play a tournament of seven games against each other. First team to won 4 games wins the tournament. The games are split over each team's home stadium as A A B B B A A.

    Historically the home team wins a game 55% of the time. What's the probability that team A wins the tournament?


    Now, I simulated this in R as a series of events drawn from three binomial distributions. I get a 0.52 win probability for team A. i.e. A small advantage remains for the team that starts the tourney since it gets 4 home games vs three for the other team.

    Just wanted to double check this answer if it is amenable to an analytic solution.

    I tried increasing the number of simulations and the answer of 0.52 is fairly stable but not sure whether I've converged onto the asymptotically right answer or not.
     
    Last edited: Nov 5, 2016
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  3. Nov 5, 2016 #2

    mfb

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    The tournament ends the earliest after game 4 (AABB) and the latest after game 7 (AABBBBA). In every case, B has at least as many home games as A. B must have an advantage.

    Do you mean A A B B B A A? Then I see that A can have an advantage, although B has a smaller advantage from AABBB.

    You can calculate the analytic solution by looking at a few cases.
     
  4. Nov 5, 2016 #3

    rollingstein

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    Yes. Thanks @mfb. My bad. I do mean A A B B B A A

    I've corrected my post.
     
  5. Nov 5, 2016 #4

    mfb

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    Rounded to whole percent, I can confirm 52%, but I think you should give at least one more digit, otherwise the advantage of A is poorly estimated.
     
  6. Nov 5, 2016 #5

    rollingstein

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    Thanks again @mfb!

    I get 51.56612% actually. But I wasn't sure how many digits to use to avoid overstating the accuracy of sims.

    Out of curiosity, what's the exact analytical answer just to see how close I got. :)

    uCLQNaR.png
     
  7. Nov 5, 2016 #6

    mfb

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    51.57197%, by going through the cases of (A won x, B won y).
     
  8. Nov 5, 2016 #7

    rollingstein

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    How did you solve it! I'm so stuck.

    I brute force listed all 128 ways the game outcomes may turn out. i.e. a 128 x 7 grid of A / B combinations.

    Of course not all tournaments will run 7 games.

    But I'm stuck. What next?
     
  9. Nov 6, 2016 #8

    rollingstein

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    Ok, I did get your answer but by an ugly brute-force method:

    My approach:

    Out of 128 total permutations I discarded the ones in which A lost by counting only for those rows with 4 or more "A"-wins.

    Next, I manually replaced each "A"-win element by either a 0.55 or a 0.45 depending on which of the 7 game columns it was in.

    Now multiplied the 7 probablities and summed over all 64 games.

    I get 0.5157197.

    Is there a smarter, more elegant approach?
     
  10. Nov 6, 2016 #9

    mfb

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    P(B won 3 games, A won 2 games | 5 games played) = 0.55 * P(B won 2 games, A won 2 games | 4 games played) + 0.45 * P(B won 3 games, A won 1 game | 4 games played)

    A 5x5 matrix is sufficient. It does not matter how "both won two games out of the first 4" occurred, for example, as long as you know the probability of this intermediate score.
     
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